Integral Calculus 201-SN3-RE • CEGEP, Montreal

Revision sheet: trigonometric integrals (201-SN3-RE)

Q: How do you choose the trigonometric substitution?

Look at the form under the root. A difference of the type constant minus x squared calls for a sine. A sum calls for a tangent. A difference of the type x squared minus a constant calls for a secant. If the expression is a quadratic, first complete the square to reveal which of the three forms you really have.

Q: What do you do when both exponents of sine and cosine are even?

Reduce the powers, that is, replace each square by its double-angle formula. No substitution can work in this case, because the factor you would split off would not be the derivative of the function you set. Power reduction turns the product into a sum of cosines of multiple angles, which integrates at once.

Q: Why do you have to come back to x after a trigonometric substitution?

Because the angle you introduced is a computing tool that does not exist in the question. An antiderivative must be expressed in the variable of the question. The return is made with the reference right triangle, on which you read the sine, cosine and tangent of the angle in terms of x. For a definite integral, you can also change the bounds and avoid this return.

Q: When is the absolute value necessary?

On the square root of x squared minus a squared, which becomes a times the absolute value of the tangent, and on the antiderivative involving the inverse secant. In both cases, the root is positive by definition while the trigonometric function changes sign depending on the branch. Stating the branch you are working on often lets you drop the absolute value.

Q: How do you know whether an antiderivative gives an inverse sine or an inverse tangent?

Check whether there is a root in the denominator. The root of a difference gives an inverse sine, with no factor in front. A sum with no root gives an inverse tangent, with a factor equal to the reciprocal of the constant. That is the only criterion to remember, and it settles the question with no ambiguity.

Q: Can two different antiderivatives both be right?

Yes, if they differ by a constant. An antiderivative is never unique. The only valid check is to differentiate both expressions and compare the derivatives with the integrand. Two answers that seem far apart can be turned into one another by a trigonometric identity.

In Integral Calculus, 201-SN3-RE (formerly 201-NYB), this chapter splits into three families: powers of trigonometric functions, handled by the parity of the exponents; trigonometric substitutions, called for by a square root; and inverse functions, which show up as antiderivatives. Each one has its recognition criterion, and it is that criterion, more than the technique, that you need to make automatic.

This sheet gives the three reference triangles, the decision table by parity, and the list of final steps that exam papers skip: the return to $x$ , the change of bounds and the absolute value.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A trigonometric substitution is not finished when the integral in $\theta$ has been computed: you still have to come BACK to $x$ , and that always goes through the right triangle. It is this last step that exam papers leave out.

This chapter is part of 201-SN3-RE, Integral Calculus

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Powers of sine and cosine
Power-reducing formulas
Tangent-secant family
Trigonometric substitution
Completing the square
Inverse trigonometric functions
Inverse sine
Inverse tangent
Reference triangle
Changing the bounds of integration

The essentials

Powers: everything is decided by PARITY

• $\displaystyle\int\sin^{m}x\cos^{n}x\,dx$ : if one exponent is ODD, split off one factor from it and substitute the OTHER function.
• If BOTH exponents are even, no substitution works: you must REDUCE THE POWERS with $\sin^{2}x=\dfrac{1-\cos 2x}{2}$ and $\cos^{2}x=\dfrac{1+\cos 2x}{2}$ .
• A useful shortcut: $\sin x\cos x=\dfrac{\sin 2x}{2}$ , which saves a whole substitution.
• Tangent-secant family: if the power of $\sec$ is EVEN, split off $\sec^{2}x$ and set $u=\tan x$ ; if the power of $\tan$ is ODD, split off $\sec x\tan x$ and set $u=\sec x$ .
• Four antiderivatives to know: $\displaystyle\int\tan x\,dx=-\ln\left|\cos x\right|$ , $\displaystyle\int\cot x\,dx=\ln\left|\sin x\right|$ , $\displaystyle\int\sec x\,dx=\ln\left|\sec x+\tan x\right|$ , $\displaystyle\int\csc x\,dx=-\ln\left|\csc x+\cot x\right|$ .

\sin^{2}x

and

\cos^{2}x

oscillate around

\frac{1}{2}

at twice the frequency: that is what

\frac{1\mp\cos 2x}{2}

says, and it is why even powers are reduced.

The cotangent-cosecant family is the exact mirror image of the tangent-secant family, up to sign: every name that starts with “co” carries a minus sign in its derivative, and therefore in its antiderivative.

The three substitutions and their triangles

• $\sqrt{a^{2}-x^{2}}$ calls for $x=a\sin\theta$ , with $dx=a\cos\theta\,d\theta$ and $\sqrt{a^{2}-x^{2}}=a\cos\theta$ .
• $\sqrt{a^{2}+x^{2}}$ calls for $x=a\tan\theta$ , with $dx=a\sec^{2}\theta\,d\theta$ and $\sqrt{a^{2}+x^{2}}=a\sec\theta$ .
• $\sqrt{x^{2}-a^{2}}$ calls for $x=a\sec\theta$ , with $dx=a\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^{2}-a^{2}}=a\left|\tan\theta\right|$ , the absolute value being mandatory.
• The reference TRIANGLE is what gets you back: on it you read $\sin\theta$ , $\cos\theta$ and $\tan\theta$ in terms of $x$ and $a$ .
• A quadratic always comes down to one of the three forms by COMPLETING THE SQUARE: $x^{2}-6x+13=(x-3)^{2}+4$ .

Three roots, three triangles: each time, the quantity under the root takes the side that is left once

x

and

a

are placed. This drawing is what gives the return to

x

at the end of the computation.

Draw the triangle once and for all at the start of the question: it gives at the same time the substitution, the simplification of the root and the return to $x$ . Skipping it costs you all three.

Inverse functions, as antiderivatives and as integrands

• $\displaystyle\int\frac{du}{\sqrt{a^{2}-u^{2}}}=\arcsin\frac{u}{a}$ ; $\displaystyle\int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\arctan\frac{u}{a}$ ; $\displaystyle\int\frac{du}{u\sqrt{u^{2}-a^{2}}}=\frac{1}{a}\,\mathrm{arcsec}\left|\frac{u}{a}\right|$ .
• The absolute value in the third is not decoration: without it, the antiderivative is wrong on the negative branch.
• The square root of a DIFFERENCE (a constant minus a square) in the DENOMINATOR gives an inverse sine; a sum with no root gives an inverse tangent. That is the only criterion to remember.
• Integrating an inverse function is done by PARTS with the invisible factor $1$ : $\displaystyle\int\arcsin x\,dx=x\arcsin x+\sqrt{1-x^{2}}$ and $\displaystyle\int\arctan x\,dx=x\arctan x-\tfrac{1}{2}\ln(1+x^{2})$ .
• The $\arccos$ and $\mathrm{arccot}$ versions differ only by a sign: the sum of the two antiderivatives of a pair is always $\dfrac{\pi x}{2}$ .

\arcsin

exists only on

\left[-1, 1\right]

;

\arctan

is defined everywhere. Both stay between

-\frac{\pi}{2}

and

\frac{\pi}{2}

Two antiderivatives that look different can both be right: they then differ by a CONSTANT. The check is to differentiate them, not to compare them term by term.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Not coming back to $x$ after a trigonometric substitution

3 marks, and an answer in a variable that does not exist in the question

What not to write

“ $\displaystyle\int\frac{dx}{\sqrt{16-x^{2}}}=\int d\theta=\theta+C$ .”

What to write

“You must come back to the original variable: $x=4\sin\theta$ gives $\sin\theta=\dfrac{x}{4}$ , so $\theta=\arcsin\dfrac{x}{4}$ and the antiderivative is $\arcsin\dfrac{x}{4}+C$ .”

Why: The variable $\theta$ is a computing tool that we introduced: it has no meaning in the question. An antiderivative must be expressed in the variable of the question.

2. Keeping the bounds in $x$ after switching to $\theta$

the whole question, 5 marks

What not to write

“ $\displaystyle\int_{0}^{2}\frac{x^{2}dx}{\sqrt{16-x^{2}}}=\int_{0}^{2}16\sin^{2}\theta\,d\theta$ .”

What to write

“The bounds follow the variable: $x=0$ gives $\theta=0$ and $x=2$ gives $\theta=\dfrac{\pi}{6}$ , so $\displaystyle\int_{0}^{\pi/6}16\sin^{2}\theta\,d\theta$ .”

Why: Evaluating an antiderivative in $\theta$ at bounds expressed in $x$ makes no sense. The alternative is to come back to $x$ before evaluating, but you have to choose one of the two routes.

3. Writing $\sqrt{x^{2}-a^{2}}=a\tan\theta$ without an absolute value

3 marks, and an antiderivative that is wrong on half the domain

What not to write

“With $x=a\sec\theta$ : $\sqrt{x^{2}-a^{2}}=a\tan\theta$ .”

What to write

“ $\sqrt{x^{2}-a^{2}}=a\left|\tan\theta\right|$ : the root is always positive, whereas $\tan\theta$ changes sign depending on the branch.”

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

lives on TWO branches: on the left,

x=a\sec\theta

gives

\tan\theta<0

, hence the absolute value.

Why: The function $x=a\sec\theta$ has two branches, $x\ge a$ and $x\le -a$ , and $\tan\theta$ has opposite signs on them. The case $x\ge a$ is the only one where the absolute value can be dropped, and you have to say so.

4. Attacking two even exponents with a substitution

the whole question, 5 marks, and ten minutes in a dead end

What not to write

“ $\displaystyle\int\sin^{2}x\cos^{2}x\,dx$ : I split off a $\cos x$ and set $u=\sin x$ .”

What to write

“Splitting off a cosine leaves $\cos^{1}x$ , which you can do nothing with: you must REDUCE THE POWERS. With $\sin x\cos x=\dfrac{\sin 2x}{2}$ , the integral becomes $\displaystyle\int\frac{\sin^{2}2x}{4}dx=\frac{x}{8}-\frac{\sin 4x}{32}$ .”

Why: A substitution only works if the factor split off really is the derivative of the variable you set. With two even exponents, that factor does not exist, and no manipulation will make it appear.

5. Substituting on a quadratic without completing the square

the whole question, 5 marks

What not to write

“ $\displaystyle\int\frac{dx}{\sqrt{x^{2}-6x+13}}$ : the form is $x^{2}-a^{2}$ , so I set $x=a\sec\theta$ .”

What to write

“First complete the square: $x^{2}-6x+13=(x-3)^{2}+4$ . So this is the form $u^{2}+a^{2}$ with $u=x-3$ and $a=2$ , which calls for $u=2\tan\theta$ .”

The quadratic never drops below

4

: it never vanishes, so it does not factor. Its real form is a SQUARE PLUS a constant, which forces the tangent substitution.

Why: The quadratic has a negative discriminant: it does not factor, so it cannot be of the form $x^{2}-a^{2}$ . Completing the square reveals its real form, and with it the right substitution.

6. Forgetting the $dx$ of the trigonometric substitution

the whole question, 5 marks

What not to write

“ $x=4\sin\theta$ , so $\displaystyle\int\frac{x^{2}dx}{\sqrt{16-x^{2}}}=\int\frac{16\sin^{2}\theta}{4\cos\theta}d\theta$ .”

What to write

“ $dx=4\cos\theta\,d\theta$ : the integral becomes $\displaystyle\int\frac{16\sin^{2}\theta}{4\cos\theta}\cdot 4\cos\theta\,d\theta=\int 16\sin^{2}\theta\,d\theta$ .”

Why: It is precisely the $dx$ that cancels the cosine in the denominator, and therefore the whole point of the substitution. Without it, the integral you get is harder than the one you started with.

7. Mixing up the criterion for the inverse sine and the one for the inverse tangent

3 marks

What not to write

“ $\displaystyle\int\frac{dx}{9+x^{2}}=\arcsin\frac{x}{3}$ .”

What to write

“With no root in the denominator, it is an inverse tangent: $\displaystyle\int\frac{dx}{9+x^{2}}=\frac{1}{3}\arctan\frac{x}{3}$ .”

With a root,

\frac{1}{\sqrt{9-x^{2}}}

exists only between

-3

and

3

: inverse sine. Without a root,

\frac{1}{9+x^{2}}

is defined everywhere: inverse tangent.

Why: The only criterion is the presence of a root: $\dfrac{1}{\sqrt{a^{2}-u^{2}}}$ gives an inverse sine, $\dfrac{1}{a^{2}+u^{2}}$ an inverse tangent. The factor $\dfrac{1}{a}$ , by the way, appears only in the second.

8. Believing that two antiderivatives that look different cannot both be right

no marks lost, but time and confidence

What not to write

“The answer key gives $-\dfrac{\cos 2x}{4}$ and I found $\dfrac{\sin^{2}x}{2}$ : one of the two is wrong.”

What to write

“ $\dfrac{\sin^{2}x}{2}=\dfrac{1-\cos 2x}{4}=-\dfrac{\cos 2x}{4}+\dfrac{1}{4}$ : the two differ by a CONSTANT, so both are right.”

\frac{\sin^{2}x}{2}

and

-\frac{\cos 2x}{4}

have the same shape, shifted by

\frac{1}{4}

: two correct antiderivatives of

\sin x\cos x

Why: An antiderivative is only defined up to a constant. The only valid check is to differentiate both expressions: if the derivatives agree, both answers are correct.

9. Using a trigonometric substitution when a simple substitution is enough

no marks, but five to ten minutes lost on the exam

What not to write

“ $\displaystyle\int\frac{x\,dx}{\sqrt{16-x^{2}}}$ : I set $x=4\sin\theta$ .”

What to write

“The derivative of $16-x^{2}$ is $-2x$ , which is already there: $u=16-x^{2}$ gives $-\sqrt{16-x^{2}}+C$ in three lines.”

Why: Trigonometric substitution is the last resort, not the first reflex. A factor $x$ in front of the root almost always signals an ordinary substitution.

10. Getting the sign wrong in the cotangent-cosecant family

2 marks, but the error repeats over a whole question

What not to write

“ $\displaystyle\int\csc^{2}x\,dx=\cot x$ .”

What to write

“ $\displaystyle\int\csc^{2}x\,dx=-\cot x$ : everything that starts with “co” carries a minus sign in its derivative, and therefore in its antiderivative.”

Why: The “co” rule is systematic: $(\cos)'=-\sin$ , $(\cot)'=-\csc^{2}$ , $(\csc)'=-\csc\cot$ . It turns every formula of the tangent-secant family into its twin with a sign.

Which method to choose

Which technique for which trigonometric form

What you see under the integral sign

If an ODD exponent on $\sin$ or $\cos$ → split off one factor from the function with the odd exponent and substitute the other one

Example: $\displaystyle\int\sin^{3}x\cos^{2}x\,dx$ with $u=\cos x$
If BOTH exponents even → REDUCE THE POWERS with the double-angle formulas; no substitution works

Example: $\displaystyle\int\sin^{2}x\cos^{2}x\,dx=\frac{x}{8}-\frac{\sin 4x}{32}$
If an EVEN power of $\sec$ → split off $\sec^{2}x$ and set $u=\tan x$

Example: $\displaystyle\int\tan^{3}x\sec^{4}x\,dx$
If an ODD power of $\tan$ → split off $\sec x\tan x$ and set $u=\sec x$

Example: $\displaystyle\int\tan^{3}x\sec^{3}x\,dx$
If neither of the two cases above → convert with $\tan^{2}x=\sec^{2}x-1$ and start again

this is the road that leads to $\int\sec^{3}x\,dx$ , the cyclic case of the family

Parity is read on the exponent, not on the function: it is the only criterion, and it separates the three cases with no ambiguity at all.

Which substitution for which root

The form of the expression under the root, AFTER completing the square if it is a quadratic

If $\sqrt{a^{2}-u^{2}}$ → $u=a\sin\theta$ , the root becomes $a\cos\theta$

Example: $\sqrt{16-x^{2}}$ with $x=4\sin\theta$
If $\sqrt{a^{2}+u^{2}}$ → $u=a\tan\theta$ , the root becomes $a\sec\theta$

Example: $\sqrt{(x-3)^{2}+4}$ with $x-3=2\tan\theta$
If $\sqrt{u^{2}-a^{2}}$ → $u=a\sec\theta$ , the root becomes $a\left|\tan\theta\right|$

Example: $\sqrt{x^{2}-9}$ with $x=3\sec\theta$
If a factor $u$ appears in front of the root → ORDINARY substitution $t=$ what is under the root, much faster

Example: $\displaystyle\int x\sqrt{16-x^{2}}\,dx$
If the expression under the root is a quadratic → complete the square FIRST, then apply one of the three rows above

after completing the square, a negative squared term gives the sine form; with a positive squared term, the sign of the constant decides between the tangent form and the secant form

Before any trigonometric substitution, check whether the derivative of what is under the root is already there: in that case an ordinary substitution does the job in three lines instead of fifteen.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Carrying a trigonometric substitution through to the end

When to use it: The integrand contains a root of the form $\sqrt{a^{2}\pm u^{2}}$ or $\sqrt{u^{2}-a^{2}}$ , and the derivative of what is under the root does not appear.

1 Identify the form and set the matching substitution, writing $x$ , $dx$ AND the simplification of the root on three separate lines.
2 DRAW the reference triangle right away: it will be used for the return, and it saves you from hunting for identities later.
3 If the integral is definite, compute the new bounds in $\theta$ and say so; otherwise, plan the return to $x$ .
4 Rewrite the integral entirely in $\theta$ , check that no $x$ is left, then compute it.
5 Come back to $x$ through the triangle, or evaluate directly at the bounds in $\theta$ if they were changed.
6 Check by differentiating, or by testing a numerical value.

Concluding sentence

“Let $x=4\sin\theta$ , so $dx=4\cos\theta\,d\theta$ and $\sqrt{16-x^{2}}=4\cos\theta$ ; the bounds $x=0$ and $x=2$ become $\theta=0$ and $\theta=\dfrac{\pi}{6}$ , and the integral reduces to $\displaystyle\int_{0}^{\pi/6}16\sin^{2}\theta\,d\theta$ .”

The trap: Skipping the triangle on the grounds that you will recall the identities from memory. That works for $\sin\theta$ , not for $\tan\theta$ and $\sec\theta$ , and that is where exam papers stop.

Marking: Typically 2 marks for the complete substitution, 1 for the bounds or the triangle, 3 for the computation in $\theta$ , 2 for the return or the evaluation, 1 for the conclusion.

Handling a power of sine and cosine

When to use it: The integrand is a product of powers of $\sin$ and $\cos$ .

1 Note the two exponents and determine their parity, in writing.
2 If one is odd, split off ONE factor of that function and convert the rest with $\sin^{2}+\cos^{2}=1$ .
3 Set the substitution on the OTHER function, and check that the factor split off really is its derivative, up to sign.
4 If both are even, replace each square by its double-angle formula and expand, starting over once if needed.
5 Integrate term by term, then come back to $x$ if a substitution was made.

Concluding sentence

“The exponent of the sine is odd: split off $\sin x$ and write $\sin^{2}x=1-\cos^{2}x$ ; the substitution $u=\cos x$ , whose derivative $-\sin x$ is exactly the factor split off, gives a polynomial integral in $u$ .”

The trap: Splitting off a factor from the function with the EVEN exponent. What is left then contains an odd power of the other function, which cannot be converted into squares, and the substitution fails.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

No $\theta$ in the final answer

Reread the antiderivative: if it still contains $\theta$ , the return to $x$ has not been made. It is the fastest check in the whole chapter.

An answer “ $\theta+C$ ”: you had to write $\arcsin\dfrac{x}{4}+C$ .
Bounds consistent with the variable

Check that the bounds of the final integral are in the same variable as the integrand. Bounds in $x$ on an integral in $\theta$ are a visible contradiction.

An integral $\displaystyle\int_{0}^{2}\sin^{2}\theta\,d\theta$ : the bound $2$ is not an angle of the problem.
The derivative of the result

Differentiate the antiderivative you found and compare it with the integrand. This validates the substitution, the return and the constants all at once.

A derivative equal to the expected integrand divided by $4\cos\theta$ : the $dx$ of the substitution was forgotten.
The absolute value where it is required

On $\sqrt{x^{2}-a^{2}}$ and on the inverse secant, check that the absolute value is there and state which branch you are working on.

An antiderivative without an absolute value, correct for $x>a$ and wrong for $x<-a$ .
The order of magnitude of a definite integral

Compare the result with the area of a rough rectangle built on the interval and on an average value of the integrand.

A result of $0.72$ for an integrand that rises from $0$ to $1.15$ on $[0, 2]$ : plausible.

The typical problem, taken apart

A definite integral by trigonometric substitution

Compute $\displaystyle\int_{0}^{2}\frac{x^{2}}{\sqrt{16-x^{2}}}\,dx$ .

Also give the antiderivative expressed in $x$ , so as to check the result by a second route.

The triangle carries the whole substitution: the hypotenuse is

4

, the opposite side is

x

, and the adjacent side is exactly the root you want to simplify. It is what will give the return to

x

at the last step.

Step 1

The form $\sqrt{16-x^{2}}$ is of type $\sqrt{a^{2}-x^{2}}$ with $a=4$ : set $x=4\sin\theta$ , hence $dx=4\cos\theta\,d\theta$ and $\sqrt{16-x^{2}}=4\cos\theta$ .

Why

The three lines are written together. No factor $x$ stands in front of the root, so an ordinary substitution is ruled out and the trigonometric substitution is indeed the right one.

Step 2

New bounds: $x=0$ gives $\sin\theta=0$ , so $\theta=0$ ; $x=2$ gives $\sin\theta=\dfrac{1}{2}$ , so $\theta=\dfrac{\pi}{6}$ .

Why

Change the bounds right away, which will spare you the return to $x$ for the numerical part. The substitution is increasing on $\left[0, \dfrac{\pi}{6}\right]$ , so it is legitimate.

Step 3

Complete substitution: $\displaystyle\int_{0}^{\pi/6}\frac{16\sin^{2}\theta}{4\cos\theta}\cdot 4\cos\theta\,d\theta=\int_{0}^{\pi/6}16\sin^{2}\theta\,d\theta$ .

Why

It is $dx=4\cos\theta\,d\theta$ that cancels the cosine in the denominator: without it, a fraction would remain, and the whole point of the substitution would vanish.

Step 4

Both exponents are even: reduce the power. $16\sin^{2}\theta=8(1-\cos 2\theta)$ , so the integral equals $\left[8\theta-4\sin 2\theta\right]_{0}^{\pi/6}$ .

Why

Two techniques of the chapter are chained: the substitution has brought the problem down to an even power, which calls for power reduction. It is the most frequent case on final exams.

Step 5

Evaluation: $8\cdot\dfrac{\pi}{6}-4\sin\dfrac{\pi}{3}=\dfrac{4\pi}{3}-4\cdot\dfrac{\sqrt{3}}{2}=\dfrac{4\pi}{3}-2\sqrt{3}\approx 0.7247$ .

Why

The lower bound makes both terms vanish, which makes the evaluation immediate. Keep the exact form before giving the decimal value.

Step 6

Antiderivative in $x$ , through the triangle: $\sin\theta=\dfrac{x}{4}$ and $\cos\theta=\dfrac{\sqrt{16-x^{2}}}{4}$ , so $8\theta-8\sin\theta\cos\theta=8\arcsin\dfrac{x}{4}-\dfrac{x\sqrt{16-x^{2}}}{2}$ .

Why

First write $\sin 2\theta=2\sin\theta\cos\theta$ so that the triangle is enough: without this identity, you would have to express a double angle, which cannot be read on the drawing.

Step 7

Check with the antiderivative: at $x=2$ , $8\arcsin\dfrac{1}{2}-\dfrac{2\sqrt{12}}{2}=\dfrac{4\pi}{3}-2\sqrt{3}$ ; at $x=0$ , everything vanishes. The two routes agree.

Why

Two independent paths, one by changing the bounds and the other by coming back to $x$ , give the same number: it is the best possible validation of a trigonometric substitution.

Step 8

Numerical check: the integrand is $0$ at $x=0$ and $1.15$ at $x=2$ ; an average rectangle of height $0.36$ on a base of $2$ gives $0.72$ .

Why

The order of magnitude confirms the result without any symbolic computation. A negative result, or one greater than $2.3$ , would have been rejected at once.

The conclusion, written out

“ $\displaystyle\int_{0}^{2}\frac{x^{2}}{\sqrt{16-x^{2}}}\,dx=\frac{4\pi}{3}-2\sqrt{3}\approx 0.725$ , and the antiderivative is $8\arcsin\dfrac{x}{4}-\dfrac{x\sqrt{16-x^{2}}}{2}+C$ .”

The classic mistake on this problem: Stopping at $\left[8\theta-4\sin 2\theta\right]$ and evaluating between $0$ and $2$ . The bounds are then values of $x$ used as angles, and the result obtained, $16-4\sin 4\approx 19.0$ , has nothing to do with the area sought.

Learn by heart

• ODD exponent: split off one factor and substitute the other function. BOTH even: reduce the powers.
• $\sec$ to an even power: $u=\tan x$ . $\tan$ to an odd power: $u=\sec x$ .
• $\displaystyle\int\sec x\,dx=\ln\left|\sec x+\tan x\right|$ and $\displaystyle\int\tan x\,dx=-\ln\left|\cos x\right|$ .
• $\sqrt{a^{2}-x^{2}}$ : $x=a\sin\theta$ . $\sqrt{a^{2}+x^{2}}$ : $x=a\tan\theta$ . $\sqrt{x^{2}-a^{2}}$ : $x=a\sec\theta$ .
• $\sqrt{x^{2}-a^{2}}=a\left|\tan\theta\right|$ : the absolute value is mandatory.
• A quadratic under a root: COMPLETE THE SQUARE before choosing the substitution.
• Root of a constant minus a square in the denominator: inverse sine. A sum with no root: inverse tangent, with a factor $\dfrac{1}{a}$ .
• Always finish with the RETURN to $x$ , through the triangle, or by changing the bounds.

Frequently asked questions

How do you choose the trigonometric substitution?

Look at the form under the root. A difference of the type constant minus x squared calls for a sine. A sum calls for a tangent. A difference of the type x squared minus a constant calls for a secant. If the expression is a quadratic, first complete the square to reveal which of the three forms you really have.

What do you do when both exponents of sine and cosine are even?

Reduce the powers, that is, replace each square by its double-angle formula. No substitution can work in this case, because the factor you would split off would not be the derivative of the function you set. Power reduction turns the product into a sum of cosines of multiple angles, which integrates at once.

Why do you have to come back to x after a trigonometric substitution?

Because the angle you introduced is a computing tool that does not exist in the question. An antiderivative must be expressed in the variable of the question. The return is made with the reference right triangle, on which you read the sine, cosine and tangent of the angle in terms of x. For a definite integral, you can also change the bounds and avoid this return.

When is the absolute value necessary?

On the square root of x squared minus a squared, which becomes a times the absolute value of the tangent, and on the antiderivative involving the inverse secant. In both cases, the root is positive by definition while the trigonometric function changes sign depending on the branch. Stating the branch you are working on often lets you drop the absolute value.

How do you know whether an antiderivative gives an inverse sine or an inverse tangent?

Check whether there is a root in the denominator. The root of a difference gives an inverse sine, with no factor in front. A sum with no root gives an inverse tangent, with a factor equal to the reciprocal of the constant. That is the only criterion to remember, and it settles the question with no ambiguity.

Can two different antiderivatives both be right?

Yes, if they differ by a constant. An antiderivative is never unique. The only valid check is to differentiate both expressions and compare the derivatives with the integrand. Two answers that seem far apart can be turned into one another by a trigonometric identity.

Practise it

Corrected exercises: Trigonometric integrals: 20 exercises, 201-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

20 corrected exercises
200 points
300 minutes

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Revision sheet: trigonometric integrals (201-SN3-RE)

The essentials

Powers: everything is decided by PARITY

The three substitutions and their triangles

Inverse functions, as antiderivatives and as integrands

The mistakes that cost marks

1. Not coming back to xxx after a trigonometric substitution

2. Keeping the bounds in xxx after switching to θ\thetaθ

3. Writing x2−a2=atan⁡θ\sqrt{x^{2}-a^{2}}=a\tan\thetax2−a2​=atanθ without an absolute value

4. Attacking two even exponents with a substitution

5. Substituting on a quadratic without completing the square

6. Forgetting the dxdxdx of the trigonometric substitution

7. Mixing up the criterion for the inverse sine and the one for the inverse tangent

8. Believing that two antiderivatives that look different cannot both be right

9. Using a trigonometric substitution when a simple substitution is enough

10. Getting the sign wrong in the cotangent-cosecant family

Which method to choose

Which technique for which trigonometric form

Which substitution for which root

How the answer is expected to be written

Carrying a trigonometric substitution through to the end

Handling a power of sine and cosine

Check before you hand in

The typical problem, taken apart

A definite integral by trigonometric substitution

Learn by heart

Frequently asked questions

How do you choose the trigonometric substitution?

What do you do when both exponents of sine and cosine are even?

Why do you have to come back to x after a trigonometric substitution?

When is the absolute value necessary?

How do you know whether an antiderivative gives an inverse sine or an inverse tangent?

Can two different antiderivatives both be right?

Corrected exercises: Trigonometric integrals: 20 exercises, 201-SN3-RE

See also

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1. Not coming back to $x$ after a trigonometric substitution

2. Keeping the bounds in $x$ after switching to $\theta$

3. Writing $\sqrt{x^{2}-a^{2}}=a\tan\theta$ without an absolute value

6. Forgetting the $dx$ of the trigonometric substitution