Exercise 1: Powers of sine and cosine: the three cases

Every product $\sin^{m}x\cos^{n}x$ is handled according to the PARITY of the two exponents, and there are only three situations. You split off one factor from the ODD power, because that factor will supply the $du$ ; if both exponents are even, no factor can be split off and the powers must be reduced.

a) Compute $\int\sin^{5}x\cos^{2}x\,dx$ , stating which power you attack and why.
b) Compute $\int\cos^{5}x\,dx$ .
c) Compute $\int\sin^{4}x\,dx$ by power reduction. How many times must the power be lowered?
d) Compute $\int\sin^{2}x\cos^{2}x\,dx$ in the shortest possible way.
e) Compute $\int_{0}^{\pi/2}\sin^{6}x\,dx$ and state the decision rule of the whole exercise in two sentences.

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Answers

a) $-\frac{\cos^{3}x}{3}+\frac{2\cos^{5}x}{5}-\frac{\cos^{7}x}{7}+C$
b) $\sin x-\frac{2\sin^{3}x}{3}+\frac{\sin^{5}x}{5}+C$
c) $\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C$ , two reductions
d) $\frac{x}{8}-\frac{\sin 4x}{32}+C$
e) $\frac{5\pi}{32}\approx 0.4909$ ; odd: split off, even: reduce

a)

The sine carries the odd power: split off one factor from it. $\int\sin^{4}x\cos^{2}x\cdot\sin x\,dx=\int(1-\cos^{2}x)^{2}\cos^{2}x\sin x\,dx$ . With $u=\cos x$ and $du=-\sin x\,dx$ : $-\int(1-u^{2})^{2}u^{2}du=-\int(u^{2}-2u^{4}+u^{6})du$ , that is $-\frac{\cos^{3}x}{3}+\frac{2\cos^{5}x}{5}-\frac{\cos^{7}x}{7}+C$ . You attack the odd power because, once one factor has been taken from it, what remains is an EVEN power, the only kind the identity $\sin^{2}=1-\cos^{2}$ can convert.

b)

$\int\cos^{4}x\cdot\cos x\,dx=\int(1-\sin^{2}x)^{2}\cos x\,dx$ . With $u=\sin x$ : $\int(1-u^{2})^{2}du=u-\frac{2u^{3}}{3}+\frac{u^{5}}{5}$ , that is $\sin x-\frac{2\sin^{3}x}{3}+\frac{\sin^{5}x}{5}+C$ . A cosine alone, to an odd power, is handled exactly like the mixed case: it is the case $n=0$ .

c)

No odd power: the power must be reduced with $\sin^{2}x=\frac{1-\cos 2x}{2}$ . $\int\sin^{4}x\,dx=\frac{1}{4}\int(1-2\cos 2x+\cos^{2}2x)\,dx$ . The term $\cos^{2}2x$ forces a SECOND reduction: $\cos^{2}2x=\frac{1+\cos 4x}{2}$ . This gives $\frac{1}{4}\left[x-\sin 2x+\frac{x}{2}+\frac{\sin 4x}{8}\right]=\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C$ . Rule: a power $2k$ calls for $k$ successive reductions, and the term in $x$ that comes out of them explains why the integral of an even power is never periodic.

d)

The short route goes through the double angle BEFORE reducing: $\sin^{2}x\cos^{2}x=\left(\sin x\cos x\right)^{2}=\frac{\sin^{2}2x}{4}=\frac{1-\cos 4x}{8}$ . Hence $\int\sin^{2}x\cos^{2}x\,dx=\frac{x}{8}-\frac{\sin 4x}{32}+C$ , in one line instead of three. Recognizing $\sin x\cos x=\frac{\sin 2x}{2}$ always saves one reduction.

e)

An even power on a standard interval: Wallis's formula gives $\int_{0}^{\pi/2}\sin^{n}x\,dx=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots\frac{1}{2}\cdot\frac{\pi}{2}$ for even $n$ , here $\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}=\frac{5\pi}{32}\approx 0.4909$ . Decision rule: if one of the two exponents is ODD, split off one factor from it and substitute the other function; if BOTH are even, reduce the powers with the double-angle formulas, as many times as needed.

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Exercise 2: Secant and cosecant: the two antiderivatives you cannot guess

$\int\sec x\,dx$ does not follow from any formula in the table: it takes an artificial multiplication, found in the 17th century for the needs of Mercator's map projection. The same trick, transposed, gives the cosecant.

a) Prove that $\int\sec x\,dx=\ln\left|\sec x+\tan x\right|+C$ by multiplying the integrand by $\dfrac{\sec x+\tan x}{\sec x+\tan x}$ . What structure then appears?
b) Obtain $\int\csc x\,dx$ the same way and write the result in two equivalent forms.
c) Show that $\ln\left|\sec x+\tan x\right|$ and $-\ln\left|\sec x-\tan x\right|$ are the same function, without differentiating.
d) Compute $\int_{0}^{\pi/4}\sec x\,dx$ and give a decimal value.
e) A student writes $\int\sec x\,dx=\frac{\sec^{2}x}{2}+C$ by analogy with $\int x\,dx$ . Refute this, then say which of the four integrals $\sec x$ , $\sec^{2}x$ , $\sec x\tan x$ , $\sec^{3}x$ are immediate and why.

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Answers

a) $\ln|\sec x+\tan x|+C$ , form $\frac{f'}{f}$
b) $-\ln|\csc x+\cot x|+C=\ln|\csc x-\cot x|+C$
c) Product equal to $1$ : opposite logarithms
d) $\ln(\sqrt{2}+1)\approx 0.8814$
e) False; $\sec^{2}x$ and $\sec x\tan x$ are immediate

a)

$\int\sec x\cdot\frac{\sec x+\tan x}{\sec x+\tan x}dx=\int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx$ . Now the numerator is exactly the DERIVATIVE of the denominator, since $(\sec x+\tan x)'=\sec x\tan x+\sec^{2}x$ . This is the form $\int\frac{f'}{f}$ , hence $\ln\left|\sec x+\tan x\right|+C$ . The whole trick is to manufacture a denominator whose derivative is the numerator.

b)

Multiply by $\frac{\csc x+\cot x}{\csc x+\cot x}$ : the numerator becomes $\csc^{2}x+\csc x\cot x$ , that is, the NEGATIVE of the derivative of $\csc x+\cot x$ , since $(\csc x+\cot x)'=-\csc x\cot x-\csc^{2}x$ . Hence $\int\csc x\,dx=-\ln\left|\csc x+\cot x\right|+C$ , also written $\ln\left|\csc x-\cot x\right|+C$ , or $\ln\left|\tan\frac{x}{2}\right|+C$ , this last form being the most convenient for evaluating at bounds.

c)

Because $(\sec x+\tan x)(\sec x-\tan x)=\sec^{2}x-\tan^{2}x=1$ : the two quantities are RECIPROCALS of each other. Since the logarithm of a reciprocal is the negative of the logarithm, $\ln\left|\sec x-\tan x\right|=-\ln\left|\sec x+\tan x\right|$ , which gives the stated equality. Two formulas that look different in two different textbooks can therefore denote exactly the same antiderivative.

d)

$\left[\ln\left|\sec x+\tan x\right|\right]_{0}^{\pi/4}=\ln\left(\sqrt{2}+1\right)-\ln\left(1+0\right)=\ln\left(\sqrt{2}+1\right)\approx 0.8814$ .

e)

The refutation is immediate by differentiation: $\left(\frac{\sec^{2}x}{2}\right)'=\sec x\cdot\sec x\tan x=\sec^{2}x\tan x$ , which is not $\sec x$ . The analogy with $\int x\,dx$ fails because $\sec x$ is not the variable of integration: the power rule applies only to $x$ , never to a function. Of the four, TWO are immediate and can be read straight off the table of derivatives, backwards: $\int\sec^{2}x\,dx=\tan x+C$ and $\int\sec x\tan x\,dx=\sec x+C$ . The other two are not: $\sec x$ requires the trick of a), and $\sec^{3}x$ a cyclic integration by parts.

Exercise 3: Tangent and secant: the two cases and the stubborn case

The family $\tan^{m}x\sec^{n}x$ obeys two rules, based on the two derivatives $(\tan x)'=\sec^{2}x$ and $(\sec x)'=\sec x\tan x$ , and on the identity $\sec^{2}x=1+\tan^{2}x$ . Only one case escapes both rules.

a) Compute $\int\sec^{4}x\,dx$ . Which factor do you split off, and which substitution follows?
b) Compute $\int\tan^{3}x\sec^{3}x\,dx$ . Which factor do you split off this time?
c) Compute $\int\tan^{4}x\,dx$ . Why does neither rule apply directly, and what move replaces them?
d) Compute $\int\sec^{3}x\,dx$ by parts, setting $u=\sec x$ and $dv=\sec^{2}x\,dx$ .
e) Derive the reduction formula $\int\sec^{n}x\,dx=\frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}x\,dx$ and check that it gives back the result of d).

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Answers

a) $\tan x+\frac{\tan^{3}x}{3}+C$
b) $\frac{\sec^{5}x}{5}-\frac{\sec^{3}x}{3}+C$
c) $\frac{\tan^{3}x}{3}-\tan x+x+C$
d) $\frac{1}{2}(\sec x\tan x+\ln|\sec x+\tan x|)+C$
e) Reduction formula for $\sec^{n}$ ; $n=3$ gives back d)

a)

The power of the secant is EVEN: split off $\sec^{2}x$ , which will be the $du$ . $\int\sec^{2}x\cdot\sec^{2}x\,dx=\int(1+\tan^{2}x)\sec^{2}x\,dx$ . With $u=\tan x$ and $du=\sec^{2}x\,dx$ : $\int(1+u^{2})du=u+\frac{u^{3}}{3}$ , that is $\tan x+\frac{\tan^{3}x}{3}+C$ .

b)

The power of the tangent is ODD: this time split off the block $\sec x\tan x$ , the derivative of the secant. $\int\tan^{2}x\sec^{2}x\cdot\sec x\tan x\,dx=\int(\sec^{2}x-1)\sec^{2}x\cdot\sec x\tan x\,dx$ . With $u=\sec x$ : $\int(u^{2}-1)u^{2}du=\frac{u^{5}}{5}-\frac{u^{3}}{3}$ , that is $\frac{\sec^{5}x}{5}-\frac{\sec^{3}x}{3}+C$ .

c)

Here $n=0$ : there is no secant from which to split off a $\sec^{2}x$ , and the power of the tangent is even, so $\sec x\tan x$ cannot be split off either. The move is to MANUFACTURE the missing secant with $\tan^{2}x=\sec^{2}x-1$ : $\int\tan^{2}x(\sec^{2}x-1)dx=\int\tan^{2}x\sec^{2}x\,dx-\int\tan^{2}x\,dx=\frac{\tan^{3}x}{3}-(\tan x-x)+C$ . You go down two powers at a time this way, until you reach $\int\tan^{2}x=\tan x-x$ or $\int\tan x=-\ln|\cos x|$ .

d)

With $u=\sec x$ , $du=\sec x\tan x\,dx$ , $dv=\sec^{2}x\,dx$ and $v=\tan x$ : $I=\sec x\tan x-\int\sec x\tan^{2}x\,dx=\sec x\tan x-\int\sec x(\sec^{2}x-1)dx=\sec x\tan x-I+\int\sec x\,dx$ . The integral you are looking for reappears on both sides: $2I=\sec x\tan x+\ln\left|\sec x+\tan x\right|$ , hence $I=\frac{1}{2}\left(\sec x\tan x+\ln\left|\sec x+\tan x\right|\right)+C$ . This is the CYCLIC case, the only one in the family that does not reduce to a substitution.

e)

Generalize d) by setting $u=\sec^{n-2}x$ and $dv=\sec^{2}x\,dx$ : $\int\sec^{n}x\,dx=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x\tan^{2}x\,dx$ . Replacing $\tan^{2}x$ with $\sec^{2}x-1$ , the original integral comes back with the coefficient $-(n-2)$ : $(n-1)\int\sec^{n}x\,dx=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}x\,dx$ , which gives the stated formula. For $n=3$ : $\frac{\sec x\tan x}{2}+\frac{1}{2}\int\sec x\,dx$ , exactly the result of d). For $n=4$ , it also gives back a), which provides an independent check.

Exercise 4: Cotangent and cosecant: the mirror image, and its signs

The family $\cot^{m}x\csc^{n}x$ is the mirror image of the previous one, with $(\cot x)'=-\csc^{2}x$ and $(\csc x)'=-\csc x\cot x$ and the identity $\csc^{2}x=1+\cot^{2}x$ . The two minus signs are the only difficulty, and they are enough to cost a lot of marks.

a) Compute $\int\cot x\,dx$ and compare with $\int\tan x\,dx$ .
b) Compute $\int\cot^{3}x\,dx$ .
c) Compute $\int\cot^{4}x\csc^{4}x\,dx$ .
d) Compute $\int\csc^{3}x\,dx$ by parts, then check that the result is indeed the mirror image of $\int\sec^{3}x\,dx$ .
e) Draw up the table of the six basic trigonometric derivatives and of the matching antiderivatives, showing exactly where the minus signs are.

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Answers

a) $\ln|\sin x|+C$ versus $-\ln|\cos x|+C$
b) $-\frac{\cot^{2}x}{2}-\ln|\sin x|+C$
c) $-\frac{\cot^{5}x}{5}-\frac{\cot^{7}x}{7}+C$
d) $\frac{1}{2}(-\csc x\cot x-\ln|\csc x+\cot x|)+C$
e) The co functions carry the minus sign

a)

$\int\cot x\,dx=\int\frac{\cos x}{\sin x}dx=\ln\left|\sin x\right|+C$ , a direct $\frac{f'}{f}$ form. For the tangent, $\int\tan x\,dx=\int\frac{\sin x}{\cos x}dx=-\ln\left|\cos x\right|+C$ , also written $\ln\left|\sec x\right|+C$ . The minus sign comes from the derivative of the cosine being negative: it is the prototype of every sign error in the cotangent-cosecant family.

b)

$\int\cot x(\csc^{2}x-1)dx=\int\cot x\csc^{2}x\,dx-\int\cot x\,dx$ . For the first one, $u=\cot x$ and $du=-\csc^{2}x\,dx$ give $-\frac{\cot^{2}x}{2}$ . Hence $-\frac{\cot^{2}x}{2}-\ln\left|\sin x\right|+C$ . Note the two minus signs, one from the substitution, the other from a).

c)

The power of the cosecant is even: split off $\csc^{2}x$ . $\int\cot^{4}x(1+\cot^{2}x)\csc^{2}x\,dx$ ; with $u=\cot x$ and $du=-\csc^{2}x\,dx$ : $-\int(u^{4}+u^{6})du=-\frac{u^{5}}{5}-\frac{u^{7}}{7}$ , that is $-\frac{\cot^{5}x}{5}-\frac{\cot^{7}x}{7}+C$ . The rule is the one for the secant, up to an overall sign.

d)

With $u=\csc x$ and $dv=\csc^{2}x\,dx$ , so $v=-\cot x$ : $I=-\csc x\cot x-\int\csc x\cot^{2}x\,dx=-\csc x\cot x-\int\csc x(\csc^{2}x-1)dx=-\csc x\cot x-I+\int\csc x\,dx$ . Hence $I=\frac{1}{2}\left(-\csc x\cot x-\ln\left|\csc x+\cot x\right|\right)+C$ . This is indeed the exact mirror image of $\int\sec^{3}x\,dx=\frac{1}{2}\left(\sec x\tan x+\ln\left|\sec x+\tan x\right|\right)$ , with the opposite overall sign: every time you go from a function to its cofunction, a minus sign appears.

e)

Derivatives: $(\sin)'=\cos$ ; $(\cos)'=-\sin$ ; $(\tan)'=\sec^{2}$ ; $(\cot)'=-\csc^{2}$ ; $(\sec)'=\sec\tan$ ; $(\csc)'=-\csc\cot$ . Matching antiderivatives: $\int\cos=\sin$ ; $\int\sin=-\cos$ ; $\int\sec^{2}=\tan$ ; $\int\csc^{2}=-\cot$ ; $\int\sec\tan=\sec$ ; $\int\csc\cot=-\csc$ . Then the ones that are not immediate: $\int\tan=-\ln|\cos|$ and $\int\cot=+\ln|\sin|$ , as well as $\int\sec=\ln|\sec+\tan|$ and $\int\csc=-\ln|\csc+\cot|$ . The memory rule fits in one sentence: EVERY function whose name starts with “co” (cosine, cotangent, cosecant) carries a minus sign in its derivative, and therefore in the antiderivative of its cofunction.

Exercise 5: The identities that replace a technique

Some trigonometric integrals seem to call for a heavy change of variable when an identity makes them immediate. We will use $1+\cos x=2\cos^{2}\frac{x}{2}$ , $1-\cos x=2\sin^{2}\frac{x}{2}$ , and multiplication by the conjugate.

a) Compute $\int\frac{dx}{1+\cos x}$ with the half-angle identity.
b) Compute $\int\frac{dx}{1-\cos x}$ .
c) Compute $\int\frac{dx}{1+\sin x}$ by multiplying by the conjugate $1-\sin x$ .
d) Compute $\int_{0}^{\pi}\sqrt{1-\cos x}\,dx$ .
e) A student computes $\int_{0}^{2\pi}\sqrt{1+\cos x}\,dx$ and finds zero. Where is the error, and what is the true value?

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Answers

a) $\tan\frac{x}{2}+C$
b) $-\cot\frac{x}{2}+C$
c) $\tan x-\sec x+C$
d) $2\sqrt{2}\approx 2.828$
e) Absolute value forgotten; $4\sqrt{2}\approx 5.657$

a)

$1+\cos x=2\cos^{2}\frac{x}{2}$ , so $\int\frac{dx}{2\cos^{2}\frac{x}{2}}=\frac{1}{2}\int\sec^{2}\frac{x}{2}dx=\tan\frac{x}{2}+C$ , the factor 2 from the inner derivative cancelling the $\frac{1}{2}$ . Three lines, where the Weierstrass substitution would have taken ten.

b)

Symmetrically, $1-\cos x=2\sin^{2}\frac{x}{2}$ gives $\frac{1}{2}\int\csc^{2}\frac{x}{2}dx=-\cot\frac{x}{2}+C$ .

c)

$\int\frac{1-\sin x}{(1+\sin x)(1-\sin x)}dx=\int\frac{1-\sin x}{\cos^{2}x}dx=\int\left(\sec^{2}x-\sec x\tan x\right)dx=\tan x-\sec x+C$ . The conjugate turns a stubborn denominator into $\cos^{2}x$ , and both pieces are then antiderivatives from the table. It is the same idea as the denominator $\sec x+\tan x$ in the integral of the secant.

d)

$\sqrt{1-\cos x}=\sqrt{2}\left|\sin\frac{x}{2}\right|$ , and on $[0, \pi]$ the argument $\frac{x}{2}$ stays in $[0, \frac{\pi}{2}]$ , where the sine is positive: the absolute value drops. $\int_{0}^{\pi}\sqrt{2}\sin\frac{x}{2}dx=\sqrt{2}\left[-2\cos\frac{x}{2}\right]_{0}^{\pi}=\sqrt{2}(0+2)=2\sqrt{2}\approx 2.828$ .

e)

The error is writing $\sqrt{1+\cos x}=\sqrt{2}\cos\frac{x}{2}$ and forgetting the ABSOLUTE VALUE. On $[0, 2\pi]$ , the argument $\frac{x}{2}$ runs over $[0, \pi]$ and the cosine changes sign there at $\frac{\pi}{2}$ , that is, at $x=\pi$ . Without the absolute value, the second half of the integral exactly cancels the first and you get zero, which is absurd for a positive function. You must split: $\int_{0}^{2\pi}\sqrt{2}\left|\cos\frac{x}{2}\right|dx=2\int_{0}^{\pi}\sqrt{2}\cos\frac{x}{2}dx=2\sqrt{2}\left[2\sin\frac{x}{2}\right]_{0}^{\pi}=4\sqrt{2}\approx 5.657$ . General rule: $\sqrt{u^{2}}=\left|u\right|$ , never $u$ , and you must ALWAYS check the sign on the interval before simplifying a root.

Exercise 6: The root of a squared minus x squared: the sine

In front of $\sqrt{a^{2}-x^{2}}$ , set $x=a\sin\theta$ with $\theta\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ : the root becomes $a\cos\theta$ , positive on that whole interval, and the radical disappears. The return to the variable $x$ is read off a right triangle with hypotenuse $a$ .

a) Compute $\int\frac{dx}{\sqrt{9-x^{2}}}$ and say why the result contains no root.
b) Compute $\int\frac{x^{2}}{\sqrt{4-x^{2}}}dx$ , showing the return to $x$ through the triangle in detail.
c) Compute $\int\frac{\sqrt{9-x^{2}}}{x^{2}}dx$ .
d) Compute $\int_{0}^{2}\sqrt{4-x^{2}}\,dx$ by changing the bounds, then recover the result by geometry.
e) Why do we require $\theta\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and not just any interval? What would happen with $\theta$ in the second quadrant?

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Answers

a) $\arcsin\frac{x}{3}+C$
b) $2\arcsin\frac{x}{2}-\frac{x\sqrt{4-x^{2}}}{2}+C$
c) $-\frac{\sqrt{9-x^{2}}}{x}-\arcsin\frac{x}{3}+C$
d) $\pi$ , a quarter of the disk of radius 2
e) Inverse defined, $\cos\theta\geq 0$ , $x$ runs over $[-a, a]$

a)

$x=3\sin\theta$ , $dx=3\cos\theta\,d\theta$ and $\sqrt{9-x^{2}}=3\cos\theta$ . The integral becomes $\int\frac{3\cos\theta}{3\cos\theta}d\theta=\int d\theta=\theta+C=\arcsin\frac{x}{3}+C$ . The result contains no root because the substitution was used precisely to eliminate it: the inverse sine is what keeps track of it, and that is the deep reason why inverse trigonometric functions show up everywhere in this chapter.

b)

$x=2\sin\theta$ : $\int\frac{4\sin^{2}\theta}{2\cos\theta}\cdot 2\cos\theta\,d\theta=4\int\sin^{2}\theta\,d\theta=2\theta-2\sin\theta\cos\theta+C$ . The right triangle with angle $\theta$ , opposite side $x$ and hypotenuse 2 gives $\sin\theta=\frac{x}{2}$ , $\cos\theta=\frac{\sqrt{4-x^{2}}}{2}$ and $\theta=\arcsin\frac{x}{2}$ . Hence $2\arcsin\frac{x}{2}-\frac{x\sqrt{4-x^{2}}}{2}+C$ .

c)

$x=3\sin\theta$ : $\int\frac{3\cos\theta}{9\sin^{2}\theta}\cdot 3\cos\theta\,d\theta=\int\frac{\cos^{2}\theta}{\sin^{2}\theta}d\theta=\int\cot^{2}\theta\,d\theta=\int(\csc^{2}\theta-1)d\theta=-\cot\theta-\theta+C$ . The triangle gives $\cot\theta=\frac{\sqrt{9-x^{2}}}{x}$ , hence $-\frac{\sqrt{9-x^{2}}}{x}-\arcsin\frac{x}{3}+C$ . Notice that Part A of the set has just been put to use: a trigonometric substitution almost always ends with an integral of trigonometric powers.

d)

With $x=2\sin\theta$ , the bound $x=0$ becomes $\theta=0$ and the bound $x=2$ becomes $\theta=\frac{\pi}{2}$ . $\int_{0}^{\pi/2}2\cos\theta\cdot 2\cos\theta\,d\theta=4\int_{0}^{\pi/2}\cos^{2}\theta\,d\theta=4\cdot\frac{\pi}{4}=\pi$ . Geometrically, $y=\sqrt{4-x^{2}}$ is the upper half of the circle of radius 2, and the interval $[0, 2]$ cuts out a QUARTER of the disk: its area is $\frac{\pi(2)^{2}}{4}=\pi$ , which confirms the computation without writing a single antiderivative.

e)

For three reasons that go together. First, the inverse sine is defined as an inverse only on that interval: without this restriction, $\theta$ could not be recovered uniquely from $x$ . Next, the cosine is POSITIVE or zero there, which lets you write $\sqrt{a^{2}-x^{2}}=a\cos\theta$ without an absolute value; in the second quadrant the cosine is negative and you would have to write $-a\cos\theta$ , with a sign that would spread through the whole computation. Finally, $x=a\sin\theta$ runs there over exactly $[-a, a]$ , that is, the whole domain of the integrand, once and only once.

Exercise 7: The root of a squared plus x squared: the tangent

In front of $\sqrt{a^{2}+x^{2}}$ , set $x=a\tan\theta$ with $\theta\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ : the identity $1+\tan^{2}\theta=\sec^{2}\theta$ turns the root into $a\sec\theta$ , and $dx=a\sec^{2}\theta\,d\theta$ .

a) Compute $\int\frac{dx}{x^{2}\sqrt{x^{2}+4}}$ .
b) Compute $\int\frac{dx}{\left(x^{2}+9\right)^{3/2}}$ and comment on how unexpectedly simple the result is.
c) Compute $\int\sqrt{x^{2}+9}\,dx$ . Which integral from Part A shows up along the way?
d) Evaluate $\int_{0}^{3}\sqrt{x^{2}+9}\,dx$ and give a decimal value.
e) The same integral can be done with $x=3\sinh u$ . Write the first two lines of that computation and say in which case this route is preferable.

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Answers

a) $-\frac{\sqrt{x^{2}+4}}{4x}+C$
b) $\frac{x}{9\sqrt{x^{2}+9}}+C$
c) $\frac{x\sqrt{x^{2}+9}}{2}+\frac{9}{2}\ln|x+\sqrt{x^{2}+9}|+C$ , via $\int\sec^{3}$
d) $\approx 10.33$
e) $9\int\cosh^{2}u\,du$ : a power reduction

a)

$x=2\tan\theta$ , $dx=2\sec^{2}\theta\,d\theta$ , $\sqrt{x^{2}+4}=2\sec\theta$ : $\int\frac{2\sec^{2}\theta}{4\tan^{2}\theta\cdot 2\sec\theta}d\theta=\frac{1}{4}\int\frac{\sec\theta}{\tan^{2}\theta}d\theta=\frac{1}{4}\int\frac{\cos\theta}{\sin^{2}\theta}d\theta=-\frac{1}{4\sin\theta}+C$ . The triangle with legs $x$ and 2 and hypotenuse $\sqrt{x^{2}+4}$ gives $\sin\theta=\frac{x}{\sqrt{x^{2}+4}}$ , hence $-\frac{\sqrt{x^{2}+4}}{4x}+C$ .

b)

$x=3\tan\theta$ : $\left(x^{2}+9\right)^{3/2}=27\sec^{3}\theta$ , so $\int\frac{3\sec^{2}\theta}{27\sec^{3}\theta}d\theta=\frac{1}{9}\int\cos\theta\,d\theta=\frac{\sin\theta}{9}+C=\frac{x}{9\sqrt{x^{2}+9}}+C$ . The result is ALGEBRAIC, with no logarithm and no inverse tangent, although the integrand looked worse than that of $\int\frac{dx}{x^{2}+9}$ , which does give an inverse tangent. The lesson: a higher power in the denominator often simplifies the computation rather than complicating it.

c)

$\int 3\sec\theta\cdot 3\sec^{2}\theta\,d\theta=9\int\sec^{3}\theta\,d\theta$ , that is, exactly the cyclic integral of Part A. Using its result: $\frac{9}{2}\left(\sec\theta\tan\theta+\ln\left|\sec\theta+\tan\theta\right|\right)$ . With $\tan\theta=\frac{x}{3}$ and $\sec\theta=\frac{\sqrt{x^{2}+9}}{3}$ : $\frac{x\sqrt{x^{2}+9}}{2}+\frac{9}{2}\ln\left|x+\sqrt{x^{2}+9}\right|+C$ , the constant $-\frac{9}{2}\ln 3$ having been absorbed into $C$ .

d)

At $x=3$ : $\frac{3\sqrt{18}}{2}+\frac{9}{2}\ln\left(3+\sqrt{18}\right)\approx 6.364+8.910=15.274$ . At $x=0$ : $0+\frac{9}{2}\ln 3\approx 4.944$ . Difference: about $10.33$ .

e)

$x=3\sinh u$ gives $dx=3\cosh u\,du$ and $\sqrt{x^{2}+9}=3\cosh u$ , since $\cosh^{2}u-\sinh^{2}u=1$ and the hyperbolic cosine is always positive. The integral becomes $9\int\cosh^{2}u\,du=\frac{9}{2}\int(1+\cosh 2u)du$ , that is, an elementary POWER REDUCTION, where the trigonometric route forced the cyclic integral of $\sec^{3}$ . The hyperbolic route is therefore clearly preferable as soon as the root sits in the numerator, and it gives the logarithm directly since $\mathrm{arcsinh}x=\ln\left(x+\sqrt{x^{2}+1}\right)$ . The trigonometric route remains preferable when the question imposes a triangle or when the other factors of the integrand are themselves trigonometric.

Exercise 8: The root of x squared minus a squared: the secant and its two branches

In front of $\sqrt{x^{2}-a^{2}}$ , set $x=a\sec\theta$ : the root becomes $a\left|\tan\theta\right|$ . It is the only one of the three substitutions whose domain splits into TWO pieces, $x\geq a$ and $x\leq -a$ , and that is where nearly all the errors hide.

a) Compute $\int\frac{dx}{\sqrt{x^{2}-4}}$ for $x>2$ , and express the result without an inverse trigonometric function.
b) Compute $\int\frac{\sqrt{x^{2}-9}}{x}dx$ for $x>3$ .
c) Why do we write $\sqrt{x^{2}-a^{2}}=a\left|\tan\theta\right|$ and not $a\tan\theta$ ? Give the interval of $\theta$ chosen for each of the two branches.
d) Check that the antiderivative of a) remains valid on the negative branch by computing $\int_{-4}^{-3}\frac{dx}{\sqrt{x^{2}-4}}$ , then by computing the same integral again with the substitution $x=-u$ .
e) Summarize the three substitutions of the set in a table: which root, which change of variable, which interval, and what the root turns into.

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Answers

a) $\ln|x+\sqrt{x^{2}-4}|+C$
b) $\sqrt{x^{2}-9}-3\arccos\frac{3}{x}+C$
c) $\sqrt{u^{2}}=|u|$ : $\tan\theta\leq 0$ on the negative branch
d) $\approx 0.3545$ by both computations
e) $a\sin\theta$ , $a\tan\theta$ , $a\sec\theta$ and their intervals

a)

$x=2\sec\theta$ , $dx=2\sec\theta\tan\theta\,d\theta$ and, for $x>2$ with $\theta\in\left[0, \frac{\pi}{2}\right)$ , $\sqrt{x^{2}-4}=2\tan\theta$ without an absolute value, since the tangent is positive there. $\int\frac{2\sec\theta\tan\theta}{2\tan\theta}d\theta=\int\sec\theta\,d\theta=\ln\left|\sec\theta+\tan\theta\right|+C$ . The triangle gives $\sec\theta=\frac{x}{2}$ and $\tan\theta=\frac{\sqrt{x^{2}-4}}{2}$ , hence $\ln\left|\frac{x+\sqrt{x^{2}-4}}{2}\right|+C=\ln\left|x+\sqrt{x^{2}-4}\right|+C'$ , the constant $-\ln 2$ being absorbed.

b)

$x=3\sec\theta$ : $\int\frac{3\tan\theta}{3\sec\theta}\cdot 3\sec\theta\tan\theta\,d\theta=3\int\tan^{2}\theta\,d\theta=3\int(\sec^{2}\theta-1)d\theta=3\tan\theta-3\theta+C$ . With $\tan\theta=\frac{\sqrt{x^{2}-9}}{3}$ and $\theta=\mathrm{arcsec}\frac{x}{3}=\arccos\frac{3}{x}$ : $\sqrt{x^{2}-9}-3\arccos\frac{3}{x}+C$ .

c)

Because $\sqrt{u^{2}}=\left|u\right|$ without exception, and the tangent CHANGES sign from one branch to the other. On the branch $x\geq a$ , take $\theta\in\left[0, \frac{\pi}{2}\right)$ , where $\sec\theta\geq 1$ and $\tan\theta\geq 0$ : the absolute value drops. On the branch $x\leq -a$ , take $\theta\in\left(\frac{\pi}{2}, \pi\right]$ , where $\sec\theta\leq -1$ , as needed to reach the negative values of $x$ , but where $\tan\theta\leq 0$ : the root is then $-a\tan\theta$ , and a minus sign runs through the whole computation. Forgetting this sign is the classic error, and it produces an antiderivative whose derivative is the negative of the integrand.

d)

With the antiderivative $\ln\left|x+\sqrt{x^{2}-4}\right|$ : at $x=-3$ , $\left|-3+\sqrt{5}\right|\approx 0.7639$ and at $x=-4$ , $\left|-4+\sqrt{12}\right|\approx 0.5359$ ; the difference of the logarithms is about $-0.2693+0.6238=0.3545$ . With the substitution $x=-u$ , the integral becomes $\int_{3}^{4}\frac{du}{\sqrt{u^{2}-4}}=\ln\left(4+\sqrt{12}\right)-\ln\left(3+\sqrt{5}\right)\approx 2.0101-1.6556=0.3545$ . The two values agree: the logarithmic form, unlike the inverse secant form, crosses both branches without any adjustment, which makes it the antiderivative to remember.

e)

Root $\sqrt{a^{2}-x^{2}}$ : set $x=a\sin\theta$ , with $\theta\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , the root becomes $a\cos\theta$ . Root $\sqrt{a^{2}+x^{2}}$ : set $x=a\tan\theta$ , with $\theta\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , the root becomes $a\sec\theta$ . Root $\sqrt{x^{2}-a^{2}}$ : set $x=a\sec\theta$ , with $\theta\in\left[0, \frac{\pi}{2}\right)$ if $x\geq a$ and $\theta\in\left(\frac{\pi}{2}, \pi\right]$ if $x\leq -a$ , the root becomes $a\left|\tan\theta\right|$ . In all three cases, the interval is chosen so that the inverse function exists AND the root is positive without an absolute value.

Exercise 9: Complete the square first

A root or a denominator of degree two almost never comes in the standard form $a^{2}\pm x^{2}$ . Completing the square is the step that brings it there, and on its own it decides which of the three substitutions, or which of the three inverse functions, will come out.

a) Compute $\int\frac{dx}{\sqrt{x^{2}+2x+5}}$ .
b) Compute $\int\frac{dx}{\sqrt{3-2x-x^{2}}}$ . Which detail of the sign of the $x^{2}$ term changes everything?
c) Compute $\int\frac{dx}{x^{2}+6x+13}$ .
d) Compute $\int\frac{x+1}{\sqrt{3-2x-x^{2}}}dx$ . Why is this computation much shorter than the one in b)?
e) Compute $\int\frac{2x+5}{x^{2}+6x+13}dx$ , then state the general rule for splitting a first-degree numerator.

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Answers

a) $\ln|x+1+\sqrt{x^{2}+2x+5}|+C$
b) $\arcsin\frac{x+1}{2}+C$ : the sign of $x^{2}$ decides
c) $\frac{1}{2}\arctan\frac{x+3}{2}+C$
d) $-\sqrt{3-2x-x^{2}}+C$ : the numerator is the derivative of the radicand
e) $\ln(x^{2}+6x+13)-\frac{1}{2}\arctan\frac{x+3}{2}+C$

a)

$x^{2}+2x+5=(x+1)^{2}+4$ : this is the form $u^{2}+a^{2}$ , so the root calls for the tangent, or directly for the formula $\int\frac{du}{\sqrt{u^{2}+a^{2}}}=\ln\left|u+\sqrt{u^{2}+a^{2}}\right|$ . With $u=x+1$ : $\ln\left|x+1+\sqrt{x^{2}+2x+5}\right|+C$ , which can also be written $\mathrm{arcsinh}\frac{x+1}{2}+C$ .

b)

The coefficient of $x^{2}$ is NEGATIVE: it must be factored out first, $3-2x-x^{2}=3-(x^{2}+2x)=3+1-(x+1)^{2}=4-(x+1)^{2}$ . This time you get the form $a^{2}-u^{2}$ , which calls for the sine and gives an inverse sine: $\arcsin\frac{x+1}{2}+C$ . The sign of the square term therefore decides everything: positive, you get a logarithm or an inverse hyperbolic sine; negative, an inverse sine. A sign error at this step leads to a function of an entirely different type.

c)

$x^{2}+6x+13=(x+3)^{2}+4$ , the form $u^{2}+a^{2}$ in a denominator with no root: this is the inverse tangent. $\int\frac{du}{u^{2}+4}=\frac{1}{2}\arctan\frac{u}{2}$ , that is $\frac{1}{2}\arctan\frac{x+3}{2}+C$ . The factor $\frac{1}{2}$ comes from the $a$ in the denominator and is very often forgotten.

d)

Because the numerator is proportional to the DERIVATIVE of the radicand: $\left(3-2x-x^{2}\right)'=-2-2x=-2(x+1)$ . The integral can therefore be written $-\frac{1}{2}\int\frac{(3-2x-x^{2})'}{\sqrt{3-2x-x^{2}}}dx=-\frac{1}{2}\cdot 2\sqrt{3-2x-x^{2}}=-\sqrt{3-2x-x^{2}}+C$ . No completing the square, no trigonometric substitution: having the right factor in the numerator short-circuits the whole machinery. So ALWAYS look at the numerator before completing the square.

e)

Split the numerator into two pieces: the one proportional to the derivative of the denominator, and a constant. Here $2x+5=(2x+6)-1$ , and $2x+6$ is exactly $\left(x^{2}+6x+13\right)'$ . Hence $\int\frac{2x+6}{x^{2}+6x+13}dx-\int\frac{dx}{x^{2}+6x+13}=\ln\left(x^{2}+6x+13\right)-\frac{1}{2}\arctan\frac{x+3}{2}+C$ ; the absolute value is unnecessary, the quadratic being always strictly positive. General rule: in front of $\frac{px+q}{\text{quadratic}}$ , write the numerator in the form $\alpha\times(\text{derivative of the denominator})+\beta$ . The first term gives a logarithm, the second an inverse tangent after completing the square. Under a root, the first gives a root and the second an inverse sine or a logarithm.

Exercise 10: When a trigonometric substitution is a bad idea

A root of the type $\sqrt{a^{2}\pm x^{2}}$ does not automatically call for a trigonometric substitution. In half the cases met on exams, a simple algebraic substitution is enough, and pulling out the triangle costs ten minutes.

a) Compute $\int\frac{x\,dx}{\sqrt{4-x^{2}}}$ , then compare with $\int\frac{dx}{\sqrt{4-x^{2}}}$ . What changes everything?
b) Compute $\int x\sqrt{x^{2}-9}\,dx$ .
c) Compute $\int\frac{x^{3}}{\sqrt{1+x^{2}}}dx$ . Is an ODD power in the numerator enough?
d) Compute $\int\frac{x\,dx}{x^{2}+4}$ and $\int\frac{dx}{x^{2}+4}$ : two neighbouring integrals, two answers of a different nature.
e) State the test that decides, in five seconds, between an algebraic substitution and a trigonometric substitution.

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Answers

a) $-\sqrt{4-x^{2}}+C$ versus $\arcsin\frac{x}{2}+C$
b) $\frac{(x^{2}-9)^{3/2}}{3}+C$
c) $\frac{(1+x^{2})^{3/2}}{3}-\sqrt{1+x^{2}}+C$
d) $\frac{1}{2}\ln(x^{2}+4)+C$ ; $\frac{1}{2}\arctan\frac{x}{2}+C$
e) The factor $x$ in the numerator decides

a)

The numerator $x$ is, up to the factor $-\frac{1}{2}$ , the derivative of $4-x^{2}$ . With $u=4-x^{2}$ and $du=-2x\,dx$ : $-\frac{1}{2}\int u^{-1/2}du=-\sqrt{u}=-\sqrt{4-x^{2}}+C$ . The second integral does NOT have this factor $x$ : no algebraic substitution works, you need $x=2\sin\theta$ , and you get $\arcsin\frac{x}{2}+C$ . A single factor $x$ separates an algebraic answer from an inverse sine.

b)

Same structure: $u=x^{2}-9$ , $du=2x\,dx$ , so $\frac{1}{2}\int u^{1/2}du=\frac{u^{3/2}}{3}=\frac{\left(x^{2}-9\right)^{3/2}}{3}+C$ . A trigonometric substitution would reach the same result after three times as many lines and a return through the triangle.

c)

Yes, provided you adjust. With $u=1+x^{2}$ , you have $x^{2}=u-1$ and $x\,dx=\frac{du}{2}$ , so $\int\frac{x^{2}\cdot x\,dx}{\sqrt{1+x^{2}}}=\frac{1}{2}\int\frac{u-1}{\sqrt{u}}du=\frac{1}{2}\int\left(u^{1/2}-u^{-1/2}\right)du=\frac{u^{3/2}}{3}-u^{1/2}$ , that is $\frac{\left(1+x^{2}\right)^{3/2}}{3}-\sqrt{1+x^{2}}+C$ . An odd power always lets you split off an $x\,dx$ and convert the rest, exactly like the odd powers of sine in Part A: it is the same idea in different clothing.

d)

$\int\frac{x\,dx}{x^{2}+4}=\frac{1}{2}\ln\left(x^{2}+4\right)+C$ , the form $\frac{f'}{f}$ , a LOGARITHMIC answer. $\int\frac{dx}{x^{2}+4}=\frac{1}{2}\arctan\frac{x}{2}+C$ , an INVERSE TANGENT answer. Two integrands that differ only by a factor $x$ , two families of functions with nothing in common. This is the example to keep in mind so that you never answer by reflex.

e)

Look at the NUMERATOR, and at it alone, before anything else. If it contains a factor proportional to the derivative of the expression under the root or in the denominator, that is, a lone $x$ when the expression is in $x^{2}$ , set $u=$ that expression and the computation is over in two lines. If it does not, or if it carries an EVEN power of $x$ , no algebraic substitution can work and the trigonometric substitution becomes necessary. In one sentence: the missing factor $x$ is exactly what costs you the triangle.

Exercise 11: The three basic antiderivatives and how to recognize them

Only three forms produce an inverse trigonometric function: $\int\frac{du}{\sqrt{a^{2}-u^{2}}}=\arcsin\frac{u}{a}$ , $\int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\arctan\frac{u}{a}$ and $\int\frac{du}{u\sqrt{u^{2}-a^{2}}}=\frac{1}{a}\mathrm{arcsec}\left|\frac{u}{a}\right|$ . The whole art is to bring the integrand back to one of the three.

a) Tell the three forms apart by their silhouette alone: root in the denominator or not, sign in front of $u^{2}$ , presence of a factor $u$ outside the root. Illustrate with $\int\frac{dx}{\sqrt{25-x^{2}}}$ , $\int\frac{dx}{25+x^{2}}$ and $\int\frac{dx}{x\sqrt{x^{2}-25}}$ .
b) Compute $\int\frac{dx}{\sqrt{8+2x-x^{2}}}$ .
c) Compute $\int\frac{dx}{(x+1)\sqrt{x^{2}+2x}}$ .
d) Compute $\int_{0}^{2}\frac{dx}{x^{2}-2x+5}$ and give a decimal value.
e) Compute $\int\frac{e^{x}}{1+e^{2x}}dx$ and $\int\frac{dx}{x\sqrt{1-\left(\ln x\right)^{2}}}$ . What common observation lets you handle both integrals at a glance?

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Answers

a) $\arcsin\frac{x}{5}$ , $\frac{1}{5}\arctan\frac{x}{5}$ , $\frac{1}{5}\mathrm{arcsec}\left|\frac{x}{5}\right|$
b) $\arcsin\frac{x-1}{3}+C$
c) $\mathrm{arcsec}|x+1|+C$
d) $\arctan\frac{1}{2}\approx 0.4636$
e) $\arctan(e^{x})+C$ ; $\arcsin(\ln x)+C$

a)

Root in the denominator with a MINUS SIGN in front of $u^{2}$ : inverse sine, here $\arcsin\frac{x}{5}+C$ . No root, PLUS sign: inverse tangent, here $\frac{1}{5}\arctan\frac{x}{5}+C$ . Root with a minus sign in front of $a^{2}$ AND a factor $u$ ALONE outside the root: inverse secant, here $\frac{1}{5}\mathrm{arcsec}\left|\frac{x}{5}\right|+C$ . The silhouette is enough, provided you check the three points in this order: the root, the sign, the outside factor.

b)

Complete the square under the root, factoring out the minus sign first: $8+2x-x^{2}=8-(x^{2}-2x)=8+1-(x-1)^{2}=9-(x-1)^{2}$ . With $u=x-1$ and $a=3$ , this is the first form: $\arcsin\frac{x-1}{3}+C$ .

c)

Here $x^{2}+2x=(x+1)^{2}-1$ , and the factor $x+1$ outside the root is exactly the $u$ of the third form: $\int\frac{du}{u\sqrt{u^{2}-1}}$ with $u=x+1$ and $a=1$ . Hence $\mathrm{arcsec}\left|x+1\right|+C$ . Without the factor $x+1$ in the denominator, the integral would have given a logarithm and not an inverse secant: that factor, which looks like a nuisance, is in fact what makes the formula applicable.

d)

$x^{2}-2x+5=(x-1)^{2}+4$ , the second form with $u=x-1$ and $a=2$ . The antiderivative is $\frac{1}{2}\arctan\frac{x-1}{2}$ , so $\left[\frac{1}{2}\arctan\frac{x-1}{2}\right]_{0}^{2}=\frac{1}{2}\left(\arctan\frac{1}{2}-\arctan\left(-\frac{1}{2}\right)\right)=\arctan\frac{1}{2}\approx 0.4636$ , the inverse tangent being odd.

e)

In both cases, an INNER substitution brings you back to a basic form. For the first, $u=e^{x}$ gives $du=e^{x}dx$ and $\int\frac{du}{1+u^{2}}=\arctan u=\arctan\left(e^{x}\right)+C$ . For the second, $u=\ln x$ gives $du=\frac{dx}{x}$ and $\int\frac{du}{\sqrt{1-u^{2}}}=\arcsin u=\arcsin\left(\ln x\right)+C$ . The common observation is that the factor that seems to clutter the integrand, $e^{x}$ on one side and $\frac{1}{x}$ on the other, is precisely the DERIVATIVE of the expression that plays the role of $u$ . The three basic forms almost never apply as they stand: they apply to a $u$ , and spotting that $u$ is all the work.

Exercise 12: The invisible factor 1: integrating an inverse function

An inverse trigonometric function has no obvious antiderivative, but it has a simple DERIVATIVE. That is exactly the situation where integration by parts with $dv=dx$ settles everything: differentiate the inverse function and integrate the factor 1, which had stayed invisible.

a) Compute $\int\arcsin x\,dx$ by setting $u=\arcsin x$ and $dv=dx$ .
b) Compute $\int\arccos x\,dx$ , then check that the two results are consistent using the identity $\arcsin x+\arccos x=\frac{\pi}{2}$ .
c) Compute $\int\arctan x\,dx$ and $\int\mathrm{arccot} x\,dx$ .
d) Compute $\int\mathrm{arcsec} x\,dx$ for $x>1$ .
e) Evaluate $\int_{0}^{1}\arctan x\,dx$ and give a decimal value.

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Answers

a) $x\arcsin x+\sqrt{1-x^{2}}+C$
b) $x\arccos x-\sqrt{1-x^{2}}+C$
c) $x\arctan x\mp\frac{1}{2}\ln(1+x^{2})+C$
d) $x\,\mathrm{arcsec}\,x-\ln|x+\sqrt{x^{2}-1}|+C$
e) $\frac{\pi}{4}-\frac{\ln 2}{2}\approx 0.4388$

a)

$u=\arcsin x$ gives $du=\frac{dx}{\sqrt{1-x^{2}}}$ , and $dv=dx$ gives $v=x$ . Hence $\int\arcsin x\,dx=x\arcsin x-\int\frac{x\,dx}{\sqrt{1-x^{2}}}$ . The second integral is algebraic, the numerator being proportional to the derivative of the radicand: it equals $-\sqrt{1-x^{2}}$ . Result: $x\arcsin x+\sqrt{1-x^{2}}+C$ .

b)

The same computation, with $du=-\frac{dx}{\sqrt{1-x^{2}}}$ , gives $x\arccos x-\sqrt{1-x^{2}}+C$ . Check: the sum of the two antiderivatives is $x\left(\arcsin x+\arccos x\right)=\frac{\pi x}{2}$ , the two roots cancelling. And that is indeed an antiderivative of $\arcsin x+\arccos x=\frac{\pi}{2}$ , a constant function. The two computations therefore check each other, and the fact that the roots cancel exactly is no accident: it is the identity, translated.

c)

$\int\arctan x\,dx=x\arctan x-\int\frac{x\,dx}{1+x^{2}}=x\arctan x-\frac{1}{2}\ln\left(1+x^{2}\right)+C$ . For the inverse cotangent, the derivative changes sign: $\int\mathrm{arccot}x\,dx=x\mathrm{arccot}x+\frac{1}{2}\ln\left(1+x^{2}\right)+C$ . Here again, the sum of the two antiderivatives is $\frac{\pi x}{2}$ , since $\arctan x+\mathrm{arccot}x=\frac{\pi}{2}$ .

d)

$u=\mathrm{arcsec}x$ gives $du=\frac{dx}{\left|x\right|\sqrt{x^{2}-1}}$ , which equals $\frac{dx}{x\sqrt{x^{2}-1}}$ since $x>1$ . Hence $\int\mathrm{arcsec}x\,dx=x\mathrm{arcsec}x-\int\frac{x\,dx}{x\sqrt{x^{2}-1}}=x\mathrm{arcsec}x-\int\frac{dx}{\sqrt{x^{2}-1}}$ . The factor $x$ cancels, and what remains is the integral from Part B: $x\mathrm{arcsec}x-\ln\left|x+\sqrt{x^{2}-1}\right|+C$ .

e)

$\left[x\arctan x-\frac{1}{2}\ln\left(1+x^{2}\right)\right]_{0}^{1}=\frac{\pi}{4}-\frac{\ln 2}{2}-0=\frac{\pi}{4}-\frac{\ln 2}{2}\approx 0.7854-0.3466\approx 0.4388$ .

Exercise 13: Products with x: the division that follows integration by parts

When an inverse function is multiplied by a power of $x$ , you still integrate by parts, but the remaining integral is no longer immediate: it requires a polynomial division or an identity. That second step is what separates a careful computation from a sloppy one.

a) Compute $\int x\arctan x\,dx$ , showing in detail the division that follows the integration by parts.
b) Compute $\int x\arcsin x\,dx$ .
c) Compute $\int x^{2}\arctan x\,dx$ .
d) Evaluate $\int_{0}^{1}x\arctan x\,dx$ and give a decimal value.
e) Why is the inverse function always chosen as $u$ , and never as $dv$ ? Connect your answer to the mnemonic rule for choosing $u$ .

Type your answers, the page tells you right or wrong 0/7

a)

Antiderivative

b)

Coefficient of

x\sqrt{1-x^{2}}

Coefficient of

\arcsin x

c)

Coefficient of

x^{2}

Coefficient of

\ln(1+x^{2})

d)

Value

e)

The inverse function is taken as u because

Show the solution

Answers

a) $\frac{x^{2}+1}{2}\arctan x-\frac{x}{2}+C$
b) $\frac{2x^{2}-1}{4}\arcsin x+\frac{x\sqrt{1-x^{2}}}{4}+C$
c) $\frac{x^{3}}{3}\arctan x-\frac{x^{2}}{6}+\frac{1}{6}\ln(1+x^{2})+C$
d) $\frac{\pi}{4}-\frac{1}{2}\approx 0.2854$
e) Differentiate the inverse function: the I of LIATE

a)

$u=\arctan x$ , $dv=x\,dx$ , so $v=\frac{x^{2}}{2}$ : $\frac{x^{2}}{2}\arctan x-\frac{1}{2}\int\frac{x^{2}}{1+x^{2}}dx$ . The degree of the numerator equals that of the denominator: the division is COMPULSORY before anything else, and it gives $\frac{x^{2}}{1+x^{2}}=1-\frac{1}{1+x^{2}}$ . Hence $\frac{x^{2}}{2}\arctan x-\frac{1}{2}\left(x-\arctan x\right)=\frac{x^{2}+1}{2}\arctan x-\frac{x}{2}+C$ .

b)

$u=\arcsin x$ , $v=\frac{x^{2}}{2}$ : $\frac{x^{2}}{2}\arcsin x-\frac{1}{2}\int\frac{x^{2}}{\sqrt{1-x^{2}}}dx$ . The remaining integral is the one from Part B, with $a=1$ : $\int\frac{x^{2}}{\sqrt{1-x^{2}}}dx=\frac{1}{2}\arcsin x-\frac{x\sqrt{1-x^{2}}}{2}$ . Hence $\frac{x^{2}}{2}\arcsin x-\frac{1}{4}\arcsin x+\frac{x\sqrt{1-x^{2}}}{4}=\frac{2x^{2}-1}{4}\arcsin x+\frac{x\sqrt{1-x^{2}}}{4}+C$ .

c)

$u=\arctan x$ , $v=\frac{x^{3}}{3}$ : $\frac{x^{3}}{3}\arctan x-\frac{1}{3}\int\frac{x^{3}}{1+x^{2}}dx$ . Division: $\frac{x^{3}}{1+x^{2}}=x-\frac{x}{1+x^{2}}$ , whose integral is $\frac{x^{2}}{2}-\frac{1}{2}\ln\left(1+x^{2}\right)$ . Result: $\frac{x^{3}}{3}\arctan x-\frac{x^{2}}{6}+\frac{1}{6}\ln\left(1+x^{2}\right)+C$ .

d)

$\left[\frac{x^{2}+1}{2}\arctan x-\frac{x}{2}\right]_{0}^{1}=\left(1\cdot\frac{\pi}{4}-\frac{1}{2}\right)-0=\frac{\pi}{4}-\frac{1}{2}\approx 0.2854$ .

e)

Because integration by parts only pays off if you DIFFERENTIATE the difficult factor. An inverse trigonometric function has a very simple algebraic derivative, a rational function or a root, whereas none of its antiderivatives is known when you start: taking it as $dv$ would force you to integrate what you cannot integrate, and the computation would stop at the first line. This is exactly the order of priority summed up by the acronym LIATE, where the I of Inverse functions comes right after the L of Logarithms: those two families are the ones you always differentiate, and never the ones you integrate.

Exercise 14: The inverse secant and the absolute value

The inverse secant is the only one of the six inverse trigonometric functions whose derivative carries an absolute value: $\left(\mathrm{arcsec}x\right)'=\frac{1}{\left|x\right|\sqrt{x^{2}-1}}$ . This absolute value is not a matter of careful notation, it is what makes the formula true on both branches of the domain.

a) The domain of $\mathrm{arcsec}$ is $\left|x\right|\geq 1$ , so it comes in two pieces. Justify the presence of $\left|x\right|$ by differentiating $\mathrm{arcsec}x=\arccos\frac{1}{x}$ in both sign cases.
b) Compute $\int\frac{dx}{x\sqrt{x^{2}-9}}$ and write the antiderivative in a form valid on BOTH branches.
c) A student suggests $\frac{1}{3}\mathrm{arcsec}\frac{x}{3}$ , without an absolute value. Test the suggestion at $x=-4$ by differentiating, and say exactly what goes wrong.
d) Evaluate $\int_{-6}^{-4}\frac{dx}{x\sqrt{x^{2}-9}}$ . Could the sign of your answer be predicted without computing?
e) Some textbooks define the inverse secant on $\left[0, \frac{\pi}{2}\right)\cup\left[\pi, \frac{3\pi}{2}\right)$ so as to write the derivative without an absolute value. Explain the mechanism and say what drawback this convention brings.

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Answers

a) $\sqrt{x^{2}}=|x|$ : derivative positive everywhere
b) $\frac{1}{3}\mathrm{arcsec}\left|\frac{x}{3}\right|+C$
c) At $x=-4$ : $+0.0945$ versus $-0.0945$
d) $\approx -0.108$ , negative like the integrand
e) Derivative without absolute value, but $\mathrm{arcsec}\,x\neq\arccos\frac{1}{x}$

a)

Differentiating $\arccos\frac{1}{x}$ : $\left(\arccos\frac{1}{x}\right)'=-\frac{1}{\sqrt{1-\frac{1}{x^{2}}}}\cdot\left(-\frac{1}{x^{2}}\right)=\frac{1}{x^{2}\sqrt{\frac{x^{2}-1}{x^{2}}}}=\frac{1}{x^{2}\cdot\frac{\sqrt{x^{2}-1}}{\left|x\right|}}=\frac{1}{\left|x\right|\sqrt{x^{2}-1}}$ . It is the simplification $\sqrt{x^{2}}=\left|x\right|$ that produces the absolute value, and it cannot be avoided: for $x>1$ it gives $\frac{1}{x\sqrt{x^{2}-1}}$ , but for $x<-1$ it gives $-\frac{1}{x\sqrt{x^{2}-1}}$ , a POSITIVE quantity even though $x$ is negative. The derivative of the inverse secant is therefore positive everywhere, which is consistent with a function that increases on each of its branches.

b)

With $u=\frac{x}{3}$ , the third basic form gives $\frac{1}{3}\mathrm{arcsec}\left|\frac{x}{3}\right|+C$ , also written $\frac{1}{3}\arccos\frac{3}{\left|x\right|}+C$ . The absolute value INSIDE is what makes the expression valid for $x>3$ as well as for $x<-3$ , without changing formula from one branch to the other.

c)

Without the absolute value, the suggestion reads $\frac{1}{3}\arccos\frac{3}{x}$ . At $x=-4$ , its derivative is $\frac{1}{3}\cdot\frac{3}{\left|x\right|\sqrt{x^{2}-9}}=\frac{1}{\left|x\right|\sqrt{x^{2}-9}}=\frac{1}{4\sqrt{7}}\approx +0.0945$ , whereas the integrand there is $\frac{1}{x\sqrt{x^{2}-9}}=\frac{1}{-4\sqrt{7}}\approx -0.0945$ . The SIGN is wrong: the suggestion is an antiderivative of the negative of the integrand on the whole negative branch. It remains perfectly correct for $x>3$ , which makes it all the more dangerous: it passes the check if you only test a positive point.

d)

With the correct antiderivative: at $x=-4$ , $\frac{1}{3}\arccos\frac{3}{4}\approx\frac{0.7227}{3}\approx 0.2409$ ; at $x=-6$ , $\frac{1}{3}\arccos\frac{1}{2}=\frac{\pi}{9}\approx 0.3491$ . The integral is $0.2409-0.3491\approx -0.108$ . The negative sign was entirely predictable: on $\left[-6, -4\right]$ , the root is positive but $x$ is negative, so the integrand is negative on the whole interval, and the integral of a negative function can only be negative. This is the check to make every time before accepting an answer.

e)

The mechanism is as follows: on $\left[\pi, \frac{3\pi}{2}\right)$ , the tangent is POSITIVE while the secant is negative, so the relation $\sqrt{\sec^{2}\theta-1}=\tan\theta$ holds without an absolute value and the derivative becomes $\frac{1}{x\sqrt{x^{2}-1}}$ on both branches. The notation gains from it, and the trigonometric substitutions of Part B become lighter. The drawback is real: this convention breaks the relation $\mathrm{arcsec}x=\arccos\frac{1}{x}$ , which is the one calculators and computer algebra systems implement. A result obtained with one convention and checked with the other will look wrong when it is not, and that is exactly why, in a written solution, you must state the convention you use or, more simply, prefer the logarithmic form $\ln\left|x+\sqrt{x^{2}-a^{2}}\right|$ whenever it is available.

Exercise 15: Compositions and identities: two antiderivatives for one integral

An inverse function composed with something else is handled by substituting first, then integrating by parts. And since inverse functions obey identities, the same integral often has two answers that look incompatible.

a) Compute $\int\arctan\sqrt{x}\,dx$ by setting $t=\sqrt{x}$ .
b) Compute $\int e^{x}\arcsin\left(e^{x}\right)dx$ .
c) Compute $\int\frac{\arctan x}{1+x^{2}}dx$ and $\int\frac{dx}{\left(1+x^{2}\right)\arctan x}$ . What structure do they share?
d) Given that $\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$ for $x>0$ , compute $\int\arctan\frac{1}{x}\,dx$ without any integration by parts.
e) Two students compute $\int\frac{dx}{x\sqrt{x^{2}-1}}$ for $x>1$ . One answers $\mathrm{arcsec}x$ , the other $\arctan\sqrt{x^{2}-1}$ . Settle the matter.

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Answers

a) $(x+1)\arctan\sqrt{x}-\sqrt{x}+C$
b) $e^{x}\arcsin(e^{x})+\sqrt{1-e^{2x}}+C$
c) $\frac{(\arctan x)^{2}}{2}+C$ ; $\ln|\arctan x|+C$
d) $\frac{\pi x}{2}-x\arctan x+\frac{1}{2}\ln(1+x^{2})+C$
e) Both: the same function on $(1, +\infty)$

a)

$t=\sqrt{x}$ gives $x=t^{2}$ and $dx=2t\,dt$ : $\int\arctan t\cdot 2t\,dt=2\int t\arctan t\,dt$ . This is the integral of exercise 13: $2\left(\frac{t^{2}+1}{2}\arctan t-\frac{t}{2}\right)=\left(t^{2}+1\right)\arctan t-t$ . Going back to $x$ : $\left(x+1\right)\arctan\sqrt{x}-\sqrt{x}+C$ . Substituting first turned a composition into a product already handled.

b)

The factor $e^{x}$ is the derivative of the argument: that is the signal to set $u=e^{x}$ , hence $\int\arcsin u\,du=u\arcsin u+\sqrt{1-u^{2}}$ , that is $e^{x}\arcsin\left(e^{x}\right)+\sqrt{1-e^{2x}}+C$ , valid for $e^{x}\leq 1$ , that is, $x\leq 0$ .

c)

In both cases, $\frac{dx}{1+x^{2}}$ is exactly $d\left(\arctan x\right)$ : setting $u=\arctan x$ , the first becomes $\int u\,du=\frac{u^{2}}{2}$ , that is $\frac{\left(\arctan x\right)^{2}}{2}+C$ , and the second $\int\frac{du}{u}=\ln\left|u\right|$ , that is $\ln\left|\arctan x\right|+C$ . The shared structure is that the inverse tangent appears TWICE, once as a function and once, in disguise, as a differential. Spotting $\frac{dx}{1+x^{2}}$ as a ready-made $du$ brings these integrals down in one line.

d)

$\int\arctan\frac{1}{x}dx=\int\left(\frac{\pi}{2}-\arctan x\right)dx=\frac{\pi x}{2}-x\arctan x+\frac{1}{2}\ln\left(1+x^{2}\right)+C$ , reusing the antiderivative of the inverse tangent. The identity has saved a whole integration by parts, and it is a reminder that $\arctan\frac{1}{x}$ is not a new function but the inverse tangent turned around. Be careful, though: for $x<0$ , the identity becomes $\arctan x+\arctan\frac{1}{x}=-\frac{\pi}{2}$ , and the constant changes from one branch to the other.

e)

Both are right. Differentiating the second suggestion: $\left(\arctan\sqrt{x^{2}-1}\right)'=\frac{1}{1+\left(x^{2}-1\right)}\cdot\frac{x}{\sqrt{x^{2}-1}}=\frac{1}{x\sqrt{x^{2}-1}}$ , which is indeed the integrand. The two functions are therefore two antiderivatives of the same function on the interval $\left(1, +\infty\right)$ , which is connected: they differ by a constant, and since both are zero at $x=1$ , that constant is ZERO, and they are exactly equal. It is in fact only a disguised identity: if $\theta=\mathrm{arcsec}x$ , then $\sec\theta=x$ and $\tan\theta=\sqrt{x^{2}-1}$ , so $\theta=\arctan\sqrt{x^{2}-1}$ . The lesson holds for the whole chapter: two answers that look very different are often the same, and the only judge is differentiation.

Exercise 16: The bounds you forget to change

In a DEFINITE integral, a change of variable must come with a change of bounds. Two methods are correct: transform the bounds and never come back to $x$ , or keep the bounds in $x$ and return to the original variable before evaluating. Mixing the two is the most costly error of the chapter.

a) Compute $\int_{0}^{1}\frac{dx}{\sqrt{4-x^{2}}}$ by trigonometric substitution, transforming the bounds.
b) Compute $\int_{0}^{2}x^{2}\sqrt{4-x^{2}}\,dx$ .
c) Compute $\int_{0}^{\sqrt{3}}\frac{dx}{1+x^{2}}$ and interpret the angle obtained geometrically.
d) A student writes $\int_{0}^{1}\frac{dx}{\sqrt{4-x^{2}}}=\int_{0}^{1}d\theta=1$ . Diagnose the error precisely and assess its size.
e) State the two correct methods and say which one you would recommend on an exam, with your reason.

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Answers

a) $\frac{\pi}{6}\approx 0.5236$
b) $\pi$
c) $\frac{\pi}{3}$ , that is $60°$
d) Bounds not changed: $1$ instead of $\frac{\pi}{6}$ , a $91$ % error
e) Transform the bounds and stay in $\theta$

a)

Set $x=2\sin\theta$ , so $dx=2\cos\theta\,d\theta$ and $\sqrt{4-x^{2}}=2\cos\theta$ . The bounds follow: $x=0$ gives $\theta=0$ , and $x=1$ gives $\sin\theta=\frac{1}{2}$ , that is $\theta=\frac{\pi}{6}$ . The integral becomes $\int_{0}^{\pi/6}d\theta=\frac{\pi}{6}\approx 0.5236$ . Direct check: $\left[\arcsin\frac{x}{2}\right]_{0}^{1}=\arcsin\frac{1}{2}=\frac{\pi}{6}$ .

b)

Same substitution and bounds, except that $x=2$ gives $\theta=\frac{\pi}{2}$ : $\int_{0}^{\pi/2}4\sin^{2}\theta\cdot 2\cos\theta\cdot 2\cos\theta\,d\theta=16\int_{0}^{\pi/2}\sin^{2}\theta\cos^{2}\theta\,d\theta$ . Part A showed that $\sin^{2}\theta\cos^{2}\theta=\frac{1-\cos 4\theta}{8}$ , hence $2\left[\theta-\frac{\sin 4\theta}{4}\right]_{0}^{\pi/2}=2\cdot\frac{\pi}{2}=\pi\approx 3.1416$ .

c)

$\left[\arctan x\right]_{0}^{\sqrt{3}}=\arctan\sqrt{3}=\frac{\pi}{3}\approx 1.0472$ . Geometrically, the substitution $x=\tan\theta$ makes $\theta$ the angle whose tangent is $x$ : integrating $\frac{dx}{1+x^{2}}$ amounts to accumulating $d\theta$ , so the integral measures exactly the angle swept as $x$ goes from 0 to $\sqrt{3}$ , that is, 60 degrees. This is also what makes $\arctan$ the most convenient tool for computing $\pi$ numerically.

d)

The student did substitute $dx$ and the root, but kept the bounds in $x$ while writing the integral in $\theta$ : the upper bound $x=1$ was copied as $\theta=1$ . Yet $\theta=1$ radian corresponds to $x=2\sin 1\approx 1.68$ , which is not the requested bound. The result, 1, exceeds the true value $\frac{\pi}{6}\approx 0.524$ by 91%. The symptom is easy to spot: an answer without $\pi$ for an integral that should have produced one.

e)

First method: transform the bounds along with the variable, then evaluate directly in $\theta$ without ever going back to $x$ . Second method: first work out the INDEFINITE integral, go back to $x$ through the triangle, then evaluate between the original bounds. On an exam, I recommend the first: it avoids the return through the triangle entirely, and that return is the longest and most error-prone step of any trigonometric substitution. It also has the advantage of producing simple numerical bounds, often $0$ , $\frac{\pi}{6}$ , $\frac{\pi}{4}$ or $\frac{\pi}{2}$ , which make the evaluation immediate. The only absolute rule is never to leave on the same line a variable and bounds that do not match.

Exercise 17: Two antiderivatives that look different

The same integral can receive answers that look incompatible depending on the method used. The theorem that settles the question is well known, but it carries a hypothesis that the chapter on inverse functions specifically violates.

a) Compute $\int\sin x\cos x\,dx$ in THREE ways: with $u=\sin x$ , with $u=\cos x$ , then with the double angle.
b) Show that the three answers differ pairwise by a constant, and compute these constants.
c) State the theorem that guarantees this result, stressing its hypothesis.
d) On $\mathbb{R}\setminus\left\{0\right\}$ , $\ln\left|x\right|+3$ and the function equal to $\ln\left|x\right|$ for $x>0$ and to $\ln\left|x\right|+7$ for $x<0$ are both antiderivatives of $\frac{1}{x}$ . Is the hypothesis of c) satisfied?
e) Connect this phenomenon to the question about the inverse secant handled earlier, and say what you must write in a solution to be beyond reproach.

Type your answers, the page tells you right or wrong 0/7

a)

With

u=\cos x

With the double angle

b)

\frac{\sin^{2}x}{2}-\left(-\frac{\cos^{2}x}{2}\right)

\frac{\sin^{2}x}{2}-\left(-\frac{\cos 2x}{4}\right)

c)

Hypothesis of the theorem

d)

Is the hypothesis satisfied?

e)

On a domain made of two pieces, the antiderivative carries

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Answers

a) $\frac{\sin^{2}x}{2}$ , $-\frac{\cos^{2}x}{2}$ , $-\frac{\cos 2x}{4}$
b) Differences $\frac{1}{2}$ , $\frac{1}{4}$ , $-\frac{1}{4}$
c) $F-G$ constant on an INTERVAL
d) No: one constant per piece
e) State the working interval

a)

With $u=\sin x$ : $\int u\,du=\frac{\sin^{2}x}{2}+C$ . With $u=\cos x$ , where $du=-\sin x\,dx$ : $-\int u\,du=-\frac{\cos^{2}x}{2}+C$ . With the double angle, $\sin x\cos x=\frac{\sin 2x}{2}$ : $\frac{1}{2}\int\sin 2x\,dx=-\frac{\cos 2x}{4}+C$ .

b)

$\frac{\sin^{2}x}{2}-\left(-\frac{\cos^{2}x}{2}\right)=\frac{\sin^{2}x+\cos^{2}x}{2}=\frac{1}{2}$ , a constant. $\frac{\sin^{2}x}{2}-\left(-\frac{\cos 2x}{4}\right)=\frac{\sin^{2}x}{2}+\frac{1-2\sin^{2}x}{4}=\frac{1}{4}$ , a constant as well. And by subtracting the previous two, the third pair differs by $-\frac{1}{4}$ . The three answers are therefore the same antiderivative up to a constant, which is exactly what the $+C$ allows: three correct methods cannot give essentially different functions.

c)

Theorem: if $F$ and $G$ are two antiderivatives of the same function on an INTERVAL, then $F-G$ is constant on that interval. The proof takes one line, $(F-G)'=0$ , but it relies on the Mean Value Theorem, which requires an interval, that is, a CONNECTED domain, all in one piece. This is the hypothesis people forget to check, because it is almost always satisfied.

d)

No, it is not: $\mathbb{R}\setminus\left\{0\right\}$ is not an interval but the union of two disjoint intervals. The difference of the two proposed functions is $-3$ on the positive half-line and $+4$ on the negative one: it is LOCALLY constant without being constant. Yet both functions really are antiderivatives of $\frac{1}{x}$ on each of the two pieces. The correct conclusion is therefore that on a domain made of several pieces, an antiderivative carries one constant PER PIECE, and the notation $\ln\left|x\right|+C$ is a convenient abuse that you must know how to defuse when you work on both sides of zero.

e)

It is exactly the same situation. The domain of $\frac{1}{x\sqrt{x^{2}-9}}$ is $\left(-\infty, -3\right)\cup\left(3, +\infty\right)$ , also made of two pieces: nothing forces the antiderivative to be written in one piece, and that is why a formula that is correct on one branch can be wrong on the other, like the version of the inverse secant without an absolute value. To be beyond reproach in a solution, do three things: state the working interval before computing, check that the proposed antiderivative is defined and differentiable on it, and, when the domain comes in several pieces, say so explicitly rather than writing a global $+C$ that hides the problem. A definite integral is also never computed across a discontinuity: it must be split, or recognized as improper.

Exercise 18: Improper integrals with a trigonometric answer

The inverse trigonometric functions are bounded: that is what lets some integrals over an infinite interval converge, and it is also what makes $\pi$ appear in results where nothing announced it.

a) Compute $\int_{0}^{+\infty}\frac{dx}{1+x^{2}}$ by going back to the definition with a limit.
b) Compute $\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}$ . In what sense is this integral improper, even though its bounds are finite?
c) Compute $\int_{-\infty}^{+\infty}\frac{dx}{x^{2}+2x+5}$ .
d) Compute $\int_{0}^{+\infty}\frac{dx}{\left(1+x^{2}\right)^{2}}$ .
e) Compute $\int_{1}^{+\infty}\frac{dx}{x\sqrt{x^{2}-1}}$ and explain why this integral, improper at both ends, still converges.

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Answers

a) $\frac{\pi}{2}$
b) $\frac{\pi}{2}$ , integrand infinite at $1$
c) $\frac{\pi}{2}$
d) $\frac{\pi}{4}$
e) $\frac{\pi}{2}$ : singularity like $\frac{1}{\sqrt{x-1}}$ , decrease like $\frac{1}{x^{2}}$

a)

By definition, $\int_{0}^{+\infty}\frac{dx}{1+x^{2}}=\lim_{b\to+\infty}\int_{0}^{b}\frac{dx}{1+x^{2}}=\lim_{b\to+\infty}\arctan b=\frac{\pi}{2}\approx 1.5708$ . Convergence rests entirely on the inverse tangent being BOUNDED: the area keeps accumulating but stays capped, the integrand decreasing like $\frac{1}{x^{2}}$ .

b)

Its bounds are finite, but the INTEGRAND is not: $\frac{1}{\sqrt{1-x^{2}}}$ tends to infinity as $x\to 1^{-}$ . It is an improper integral of the second kind, and you must write $\lim_{b\to 1^{-}}\left[\arcsin x\right]_{0}^{b}=\lim_{b\to 1^{-}}\arcsin b=\frac{\pi}{2}$ . An infinitely tall yet finite area: the graph shoots up vertically at $x=1$ while enclosing only $\frac{\pi}{2}$ of area. Notice the unsettling result: it is the same value as in a), obtained on a bounded interval.

c)

Complete the square: $x^{2}+2x+5=(x+1)^{2}+4$ , so the antiderivative is $\frac{1}{2}\arctan\frac{x+1}{2}$ . The integral, improper at both bounds, is split in two and equals $\frac{1}{2}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=\frac{\pi}{2}\approx 1.5708$ . Each of the two halves converges separately, an essential condition for being allowed to conclude.

d)

The substitution $x=\tan\theta$ turns the bounds $0$ and $+\infty$ into $0$ and $\frac{\pi}{2}$ , which makes the integral PROPER: $\int_{0}^{\pi/2}\frac{\sec^{2}\theta}{\sec^{4}\theta}d\theta=\int_{0}^{\pi/2}\cos^{2}\theta\,d\theta=\frac{\pi}{4}\approx 0.7854$ . This is one of the great services of the trigonometric substitution: it brings an infinite interval down to a bounded one, and the question of convergence settles itself.

e)

The integral is improper at $1$ , where the denominator vanishes, and at $+\infty$ . The antiderivative is $\mathrm{arcsec}x$ , so the value is $\lim_{b\to+\infty}\mathrm{arcsec}b-\lim_{a\to 1^{+}}\mathrm{arcsec}a=\frac{\pi}{2}-0=\frac{\pi}{2}$ . It converges at both ends for two distinct reasons: at $1$ , the singularity behaves like $\frac{1}{\sqrt{x-1}}$ , with exponent $\frac{1}{2}<1$ , hence integrable; at infinity, the integrand decreases like $\frac{1}{x^{2}}$ , with exponent $2>1$ , hence integrable as well. The fact that the inverse secant is bounded by $\frac{\pi}{2}$ sums up both at once, and the same value comes back for the fourth time, after a), b) and c): this is no coincidence, the four integrands are images of one another under the substitutions of the chapter.

Exercise 19: Applications: area, arc length and volume

The integrals of this chapter are not algebra drills: they come up naturally in the three great applications of integral calculus, and that is where you meet them, in physics as well as in geometry.

a) Prove the formula for the area of a disk by computing $4\int_{0}^{a}\sqrt{a^{2}-x^{2}}\,dx$ .
b) Compute the length of the curve $y=\ln\left(\cos x\right)$ for $x\in\left[0, \frac{\pi}{4}\right]$ . Which integral from Part A do you recognize?
c) Compute the length of the parabola $y=\frac{x^{2}}{2}$ for $x\in\left[0, 1\right]$ .
d) Compute the volume generated by rotating about the $x$ -axis the region under $y=\frac{1}{\sqrt{1+x^{2}}}$ for $x\in\left[0, 1\right]$ .
e) A part has the shape of the region under $y=\frac{1}{1+x^{2}}$ for $x\geq 0$ . Is its area finite? And the volume generated by rotating it about the $x$ -axis?

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Answers

a) $4\cdot\frac{\pi a^{2}}{4}=\pi a^{2}$
b) $\int_{0}^{\pi/4}\sec x\,dx\approx 0.8814$
c) $\frac{\sqrt{2}}{2}+\frac{1}{2}\ln(1+\sqrt{2})\approx 1.148$
d) $\frac{\pi^{2}}{4}\approx 2.467$
e) Area $\frac{\pi}{2}$ , volume $\frac{\pi^{2}}{4}$ : both finite

a)

By symmetry, the quarter disk in the first quadrant has area $\int_{0}^{a}\sqrt{a^{2}-x^{2}}\,dx$ . With $x=a\sin\theta$ , the bounds become $0$ and $\frac{\pi}{2}$ : $\int_{0}^{\pi/2}a\cos\theta\cdot a\cos\theta\,d\theta=a^{2}\int_{0}^{\pi/2}\cos^{2}\theta\,d\theta=a^{2}\cdot\frac{\pi}{4}$ . Multiplying by 4: $\pi a^{2}$ . The formula for the area of a disk, learned in high school without proof, thus comes out of a trigonometric substitution and the power reduction of $\cos^{2}$ .

b)

$y'=-\tan x$ , so $\sqrt{1+y'^{2}}=\sqrt{1+\tan^{2}x}=\sec x$ , the secant being positive on $\left[0, \frac{\pi}{4}\right]$ . The length is $\int_{0}^{\pi/4}\sec x\,dx=\ln\left(\sqrt{2}+1\right)\approx 0.8814$ . This is exactly the integral of the secant computed in Part A, and the coincidence is no coincidence: the identity $1+\tan^{2}=\sec^{2}$ brings out the secant in every arc length of a curve whose slope is a tangent.

c)

$y'=x$ , so the length is $\int_{0}^{1}\sqrt{1+x^{2}}\,dx$ , the integral from Part B with $a=1$ : $\left[\frac{x\sqrt{1+x^{2}}}{2}+\frac{1}{2}\ln\left(x+\sqrt{1+x^{2}}\right)\right]_{0}^{1}=\frac{\sqrt{2}}{2}+\frac{1}{2}\ln\left(1+\sqrt{2}\right)\approx 0.7071+0.4407\approx 1.148$ . Compare with the chord, of length $\sqrt{1+0.25}\approx 1.118$ : the parabola is only 3% longer than the segment joining its endpoints.

d)

By the disk method, $V=\pi\int_{0}^{1}\frac{dx}{1+x^{2}}=\pi\left[\arctan x\right]_{0}^{1}=\frac{\pi^{2}}{4}\approx 2.467$ . A volume equal to $\frac{\pi^{2}}{4}$ : the square of $\pi$ appears because $\pi$ comes once from the rotation and once from the inverse tangent, two completely independent origins.

e)

The area is $\int_{0}^{+\infty}\frac{dx}{1+x^{2}}=\frac{\pi}{2}$ , so it is FINITE. The volume of revolution is $\pi\int_{0}^{+\infty}\frac{dx}{\left(1+x^{2}\right)^{2}}=\pi\cdot\frac{\pi}{4}=\frac{\pi^{2}}{4}\approx 2.467$ , finite as well. Both converge because the integrand decreases fast enough, like $\frac{1}{x^{2}}$ for the area and like $\frac{1}{x^{4}}$ for the volume: squaring speeds up the decrease and makes convergence even more comfortable. This is not always the case, and the comparison is worth making with Gabriel's horn, generated by $y=\frac{1}{x}$ on $\left[1, +\infty\right)$ , whose volume is finite while the area under the curve is infinite.

Exercise 20: Synthesis problem: eight integrals, one decision tree

Here are eight mixed integrals, with no hint about the method, as on an exam. For each one, the difficulty is not the computation but the DIAGNOSIS: identifying in a few seconds which family it belongs to.

a) Compute $\int\sin^{3}x\cos^{4}x\,dx$ and $\int\sec^{4}x\tan^{2}x\,dx$ .
b) Compute $\int\frac{dx}{\sqrt{x^{2}-16}}$ and $\int\frac{x\,dx}{\sqrt{x^{2}-16}}$ .
c) Compute $\int\frac{dx}{x^{2}+4x+8}$ and $\int\frac{dx}{x^{2}\sqrt{x^{2}+1}}$ .
d) Compute $\int\arctan\left(3x\right)dx$ and $\int x\arcsin\left(x^{2}\right)dx$ .
e) Write out the complete decision tree of the set: in front of a trigonometric integral or one with a quadratic under a root, which questions should you ask, and in what order?

Type your answers, the page tells you right or wrong 0/8

a)

\int\sin^{3}x\cos^{4}x\,dx

\int\sec^{4}x\tan^{2}x\,dx

b)

\int\frac{dx}{\sqrt{x^{2}-16}}

\int\frac{x\,dx}{\sqrt{x^{2}-16}}

c)

Coefficient of

\arctan\frac{x+2}{2}

\int\frac{dx}{x^{2}\sqrt{x^{2}+1}}

d)

Coefficient of

\ln(1+9x^{2})

e)

First reflex

Show the solution

Answers

a) $-\frac{\cos^{5}x}{5}+\frac{\cos^{7}x}{7}+C$ ; $\frac{\tan^{3}x}{3}+\frac{\tan^{5}x}{5}+C$
b) $\ln|x+\sqrt{x^{2}-16}|+C$ ; $\sqrt{x^{2}-16}+C$
c) $\frac{1}{2}\arctan\frac{x+2}{2}+C$ ; $-\frac{\sqrt{x^{2}+1}}{x}+C$
d) $x\arctan 3x-\frac{1}{6}\ln(1+9x^{2})+C$ ; $\frac{x^{2}\arcsin(x^{2})+\sqrt{1-x^{4}}}{2}+C$
e) Derivative in the numerator, inverse function, parity, square, triangle

a)

Odd power of sine: split off $\sin x$ and set $u=\cos x$ , hence $-\int(1-u^{2})u^{4}du=-\frac{\cos^{5}x}{5}+\frac{\cos^{7}x}{7}+C$ . For the second, EVEN power of secant: split off $\sec^{2}x$ and set $u=\tan x$ , hence $\int(1+u^{2})u^{2}du=\frac{\tan^{3}x}{3}+\frac{\tan^{5}x}{5}+C$ .

b)

The first has no factor $x$ in the numerator: it is the substitution $x=4\sec\theta$ , or the formula directly, and you get $\ln\left|x+\sqrt{x^{2}-16}\right|+C$ . The second has one, and it is proportional to the derivative of the radicand: algebraic substitution $u=x^{2}-16$ , hence $\sqrt{x^{2}-16}+C$ . Two neighbouring integrals, two opposite families, and the factor $x$ is the only judge.

c)

$x^{2}+4x+8=(x+2)^{2}+4$ : complete the square, then inverse tangent, $\frac{1}{2}\arctan\frac{x+2}{2}+C$ . For the second, a root of the form $\sqrt{x^{2}+a^{2}}$ with no useful factor in the numerator: substitution $x=\tan\theta$ , which gives $-\frac{\sqrt{x^{2}+1}}{x}+C$ .

d)

An inverse function alone: integration by parts with the invisible factor 1. $\int\arctan 3x\,dx=x\arctan 3x-\int\frac{3x\,dx}{1+9x^{2}}=x\arctan 3x-\frac{1}{6}\ln\left(1+9x^{2}\right)+C$ . For the second, the factor $x$ is the derivative of $x^{2}$ : first substitute $u=x^{2}$ , which gives $\frac{1}{2}\int\arcsin u\,du=\frac{1}{2}\left(u\arcsin u+\sqrt{1-u^{2}}\right)$ , that is $\frac{x^{2}\arcsin\left(x^{2}\right)+\sqrt{1-x^{4}}}{2}+C$ .

e)

First question: is there, in the numerator, a factor proportional to the DERIVATIVE of what is under the root or in the denominator? If so, algebraic substitution, and the work is done. Second question: is the integrand an inverse trigonometric function, alone or multiplied by a power of $x$ ? If so, integration by parts, differentiating the inverse function. Third question: is it a product of trigonometric powers? If so, apply the parity rule, splitting off one factor from the odd power or reducing the powers if both are even. Fourth question: is the denominator or the radicand a quadratic? If so, complete the square, which brings you back to one of the three standard forms. Only then, fifth question: is there still a root of the type $\sqrt{a^{2}-x^{2}}$ , $\sqrt{a^{2}+x^{2}}$ or $\sqrt{x^{2}-a^{2}}$ ? Then, and only then, bring out the trigonometric substitution and its triangle. The order matters as much as the list: the trigonometric substitution is the last resort, not the first reflex, and half the marks lost on exams come from drawing it too early.

Corrected exercises: trigonometric integrals (201-SN3-RE)

Course recap

Part A: Powers and products of trigonometric functions (/50)

Exercise 1: Powers of sine and cosine: the three cases

Exercise 2: Secant and cosecant: the two antiderivatives you cannot guess

Exercise 3: Tangent and secant: the two cases and the stubborn case

Exercise 4: Cotangent and cosecant: the mirror image, and its signs

Exercise 5: The identities that replace a technique

Part B: Trigonometric substitutions (/50)

Exercise 6: The root of a squared minus x squared: the sine

Exercise 7: The root of a squared plus x squared: the tangent

Exercise 8: The root of x squared minus a squared: the secant and its two branches

Exercise 9: Complete the square first

Exercise 10: When a trigonometric substitution is a bad idea

Part C: Inverse trigonometric functions (/50)

Exercise 11: The three basic antiderivatives and how to recognize them

Exercise 12: The invisible factor 1: integrating an inverse function

Exercise 13: Products with x: the division that follows integration by parts

Exercise 14: The inverse secant and the absolute value

Exercise 15: Compositions and identities: two antiderivatives for one integral

Part D: Definite integrals, pitfalls and synthesis (/50)

Exercise 16: The bounds you forget to change

Exercise 17: Two antiderivatives that look different

Exercise 18: Improper integrals with a trigonometric answer

Exercise 19: Applications: area, arc length and volume

Exercise 20: Synthesis problem: eight integrals, one decision tree

See also

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