Exercise 0.1: Integrating by parts: ten direct calculations

For each definite integral, integrate by parts, once or several times, then give its exact value and its value to three decimal places.

One method for all ten lines: choose $u$ in the LIATE order (logarithmic, inverse trigonometric, algebraic, trigonometric, exponential) and take everything else as $dv$ , $dx$ included; compute $du$ and $v$ ; write $\int_{a}^{b}u\,dv=\left[uv\right]_{a}^{b}-\int_{a}^{b}v\,du$ and evaluate the boundary term right away. If the new integral still resists, start again with the SAME kind of choice.

a) $\int_{0}^{1}x\,e^{2x}\,dx$ .
b) $\int_{0}^{\pi/2}x\sin(2x)\,dx$ .
c) $\int_{1}^{e}x^{3}\ln x\,dx$ .
d) $\int_{1}^{e^{2}}\frac{\ln x}{\sqrt{x}}\,dx$ .
e) $\int_{0}^{1}\arcsin\left(\frac{x}{2}\right)dx$ .
f) $\int_{0}^{\pi/2}x^{2}\cos x\,dx$ .
g) $\int_{0}^{1}x^{3}e^{x}\,dx$ .
h) $\int_{1}^{e}x\left(\ln x\right)^{2}dx$ .
i) $\int_{0}^{\pi/4}x\sec^{2}x\,dx$ .
j) $\int_{0}^{\pi}e^{-x}\sin(2x)\,dx$ .

Type your answers, the page tells you right or wrong 0/10

a)

value

b)

value

c)

value

d)

value

e)

value

f)

value

g)

value

h)

value

i)

value

j)

value

Show the solution

Answers

a) $\left[\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\right]_{0}^{1}=\frac{e^{2}+1}{4}\approx 2.097$
b) $\left[-\frac{x\cos(2x)}{2}+\frac{\sin(2x)}{4}\right]_{0}^{\pi/2}=\frac{\pi}{4}\approx 0.785$
c) $\left[\frac{x^{4}\ln x}{4}-\frac{x^{4}}{16}\right]_{1}^{e}=\frac{3e^{4}+1}{16}\approx 10.300$
d) $\left[2\sqrt{x}\ln x-4\sqrt{x}\right]_{1}^{e^{2}}=4$
e) $\left[x\arcsin\frac{x}{2}+\sqrt{4-x^{2}}\right]_{0}^{1}=\frac{\pi}{6}+\sqrt{3}-2\approx 0.256$
f) $\left[x^{2}\sin x+2x\cos x-2\sin x\right]_{0}^{\pi/2}=\frac{\pi^{2}}{4}-2\approx 0.467$
g) $\left[e^{x}\left(x^{3}-3x^{2}+6x-6\right)\right]_{0}^{1}=6-2e\approx 0.563$
h) $\left[\frac{x^{2}(\ln x)^{2}}{2}-\frac{x^{2}\ln x}{2}+\frac{x^{2}}{4}\right]_{1}^{e}=\frac{e^{2}-1}{4}\approx 1.597$
i) $\left[x\tan x+\ln|\cos x|\right]_{0}^{\pi/4}=\frac{\pi}{4}-\frac{\ln 2}{2}\approx 0.439$
j) $I=\frac{2\left(1-e^{-\pi}\right)}{5}\approx 0.383$

The move that decides everything: the choice of $u$ . You differentiate what SIMPLIFIES when differentiated (a logarithm, an arcsine, a power of $x$ ), and you integrate what does not get more complicated (an exponential, a sine, a $\sec^{2}$ ). LIATE is only the order of that preference, and a bad choice does not give a wrong result: it gives an integral worse than the one you started with.

a)

$u=x$ , $dv=e^{2x}dx$ , so $du=dx$ and $v=\frac{e^{2x}}{2}$ : $\int_{0}^{1}xe^{2x}dx=\left[\frac{xe^{2x}}{2}\right]_{0}^{1}-\frac{1}{2}\int_{0}^{1}e^{2x}dx=\frac{e^{2}}{2}-\frac{e^{2}-1}{4}=\frac{e^{2}+1}{4}\approx 2.097$ . Pitfall: taking $v=e^{2x}$ ; an antiderivative of $e^{2x}$ carries the factor $\frac{1}{2}$ .

b)

$u=x$ , $dv=\sin(2x)dx$ , $v=-\frac{\cos(2x)}{2}$ : $\left[-\frac{x\cos(2x)}{2}\right]_{0}^{\pi/2}+\frac{1}{2}\int_{0}^{\pi/2}\cos(2x)dx=-\frac{\frac{\pi}{2}\cos\pi}{2}+\frac{1}{2}\left[\frac{\sin(2x)}{2}\right]_{0}^{\pi/2}=\frac{\pi}{4}+0\approx 0.785$ . Pitfall: the double minus sign, the one in $v$ and the one in the formula, which gives a PLUS in front of the remaining integral.

c)

The logarithm comes first: $u=\ln x$ , $dv=x^{3}dx$ , $du=\frac{dx}{x}$ , $v=\frac{x^{4}}{4}$ . So $\left[\frac{x^{4}\ln x}{4}\right]_{1}^{e}-\frac{1}{4}\int_{1}^{e}x^{3}dx=\frac{e^{4}}{4}-\frac{e^{4}-1}{16}=\frac{3e^{4}+1}{16}\approx 10.300$ . Pitfall: $u=x^{3}$ would force you to integrate $\ln x$ , and the integral would stay just as hard.

d)

$u=\ln x$ , $dv=x^{-\frac{1}{2}}dx$ , $v=2\sqrt{x}$ : $\left[2\sqrt{x}\ln x\right]_{1}^{e^{2}}-\int_{1}^{e^{2}}\frac{2}{\sqrt{x}}dx$ . The boundary term equals $2e\times 2-0=4e$ , since $\sqrt{e^{2}}=e$ and $\ln 1=0$ ; the remaining integral equals $\left[4\sqrt{x}\right]_{1}^{e^{2}}=4e-4$ . So the value is $4e-(4e-4)=4$ . Pitfall: forgetting the boundary term, which gives $-(4e-4)\approx-6.873$ , a negative integral of a positive function.

e)

The invisible factor $1$ : $u=\arcsin\frac{x}{2}$ , $dv=dx$ , $v=x$ , and $du=\frac{\frac{1}{2}}{\sqrt{1-\frac{x^{2}}{4}}}dx=\frac{dx}{\sqrt{4-x^{2}}}$ . So $\left[x\arcsin\frac{x}{2}\right]_{0}^{1}-\int_{0}^{1}\frac{x\,dx}{\sqrt{4-x^{2}}}=\frac{\pi}{6}+\left[\sqrt{4-x^{2}}\right]_{0}^{1}=\frac{\pi}{6}+\sqrt{3}-2\approx 0.256$ , the remaining integral being done with $w=4-x^{2}$ .

f)

Two rounds. $u=x^{2}$ , $dv=\cos x\,dx$ : $\left[x^{2}\sin x\right]_{0}^{\pi/2}-2\int_{0}^{\pi/2}x\sin x\,dx$ . Then $u=x$ , $dv=\sin x\,dx$ : $\int_{0}^{\pi/2}x\sin x\,dx=\left[-x\cos x\right]_{0}^{\pi/2}+\int_{0}^{\pi/2}\cos x\,dx=0+1=1$ . So the value is $\frac{\pi^{2}}{4}-2\approx 0.467$ . Pitfall: in the second round, $v=-\cos x$ , whose sign changes the sign of the remaining integral.

g)

Tabular method: differentiate $x^{3}$ down to zero ( $x^{3}$ , $3x^{2}$ , $6x$ , $6$ ), integrate $e^{x}$ as many times, and alternate the signs $+$ , $-$ , $+$ , $-$ . So $\int x^{3}e^{x}dx=e^{x}\left(x^{3}-3x^{2}+6x-6\right)+C$ , and $\left[e^{x}\left(x^{3}-3x^{2}+6x-6\right)\right]_{0}^{1}=e(1-3+6-6)-(-6)=6-2e\approx 0.563$ . Pitfall: the lower limit $0$ does not give $0$ , but $e^{0}\times(-6)=-6$ .

h)

$u=(\ln x)^{2}$ , $dv=x\,dx$ , $du=\frac{2\ln x}{x}dx$ , $v=\frac{x^{2}}{2}$ : $\left[\frac{x^{2}(\ln x)^{2}}{2}\right]_{1}^{e}-\int_{1}^{e}x\ln x\,dx$ . Second round: $\int_{1}^{e}x\ln x\,dx=\left[\frac{x^{2}\ln x}{2}\right]_{1}^{e}-\int_{1}^{e}\frac{x}{2}dx=\frac{e^{2}}{2}-\frac{e^{2}-1}{4}=\frac{e^{2}+1}{4}$ . So the value is $\frac{e^{2}}{2}-\frac{e^{2}+1}{4}=\frac{e^{2}-1}{4}\approx 1.597$ .

i)

$u=x$ , $dv=\sec^{2}x\,dx$ , $v=\tan x$ : $\left[x\tan x\right]_{0}^{\pi/4}-\int_{0}^{\pi/4}\tan x\,dx=\frac{\pi}{4}+\left[\ln|\cos x|\right]_{0}^{\pi/4}=\frac{\pi}{4}+\ln\frac{\sqrt{2}}{2}=\frac{\pi}{4}-\frac{\ln 2}{2}\approx 0.439$ . Pitfall: $\int\tan x\,dx=-\ln|\cos x|$ , and the minus sign of the formula turns it into a plus.

j)

Cyclic case. With $u=\sin(2x)$ , $dv=e^{-x}dx$ , $v=-e^{-x}$ : $I=\left[-e^{-x}\sin(2x)\right]_{0}^{\pi}+2\int_{0}^{\pi}e^{-x}\cos(2x)dx=2J$ . With $u=\cos(2x)$ , the SAME choice: $J=\left[-e^{-x}\cos(2x)\right]_{0}^{\pi}-2I=\left(1-e^{-\pi}\right)-2I$ . So $I=2\left(1-e^{-\pi}\right)-4I$ , that is $I=\frac{2\left(1-e^{-\pi}\right)}{5}\approx 0.383$ . Pitfall: switching the choice in the second round, which brings back the identity $I=I$ .

Check: every antiderivative can be checked by differentiating it with the product rule, and every value on a calculator with a fine Riemann sum; a negative integral of a positive function, as in d) without the boundary term, signals a sign error.

Tick the exercises you have done or want to review: a free account, no password, keeps your ticks and your right answers from one visit to the next, tells you which chapter to tackle next and lets you ask for the exercise you are missing. Create your space, an email is enough.

Exercise 0.2: Integrating trigonometric powers and products: ten direct calculations

For each definite integral, transform the integrand with the appropriate rule, integrate, then give the exact value and its value to three decimal places.

One method for all ten lines, with three questions asked in order. A product of sines and cosines of DIFFERENT angles is first turned into a sum. An ODD power of $\sin$ or $\cos$ (an even power of $\sec$ , or an odd power of $\tan$ together with $\sec$ ) gives up one factor, which becomes the $du$ , and the rest is written with the Pythagorean identity. If all the powers are even, reduce them with $\sin^{2}\theta=\frac{1-\cos 2\theta}{2}$ and $\cos^{2}\theta=\frac{1+\cos 2\theta}{2}$ .

a) $\int_{0}^{\pi/2}\sin^{2}\left(\frac{x}{2}\right)dx$ .
b) $\int_{0}^{\pi/2}\sin^{5}x\,dx$ .
c) $\int_{0}^{\pi/2}\sin^{3}x\sqrt{\cos x}\,dx$ .
d) $\int_{0}^{\pi/4}\cos(3x)\cos x\,dx$ .
e) $\int_{0}^{\pi/2}\sin(4x)\cos(2x)\,dx$ .
f) $\int_{0}^{\pi/3}\sin(3x)\sin x\,dx$ .
g) $\int_{0}^{\pi/4}\tan^{4}x\sec^{4}x\,dx$ .
h) $\int_{0}^{\pi/3}\tan x\sec^{3}x\,dx$ .
i) $\int_{0}^{\pi/2}\sin^{4}x\cos^{2}x\,dx$ .
j) $\int_{0}^{\pi/4}\tan^{6}x\,dx$ .

Type your answers, the page tells you right or wrong 0/10

a)

value

b)

value

c)

value

d)

value

e)

value

f)

value

g)

value

h)

value

i)

value

j)

value

Show the solution

Answers

a) $\left[\frac{x}{2}-\frac{\sin x}{2}\right]_{0}^{\pi/2}=\frac{\pi}{4}-\frac{1}{2}\approx 0.285$
b) $\int_{0}^{1}\left(1-u^{2}\right)^{2}du=\frac{8}{15}\approx 0.533$
c) $\int_{0}^{1}\left(u^{\frac{1}{2}}-u^{\frac{5}{2}}\right)du=\frac{8}{21}\approx 0.381$
d) $\frac{1}{2}\left[\frac{\sin(2x)}{2}+\frac{\sin(4x)}{4}\right]_{0}^{\pi/4}=\frac{1}{4}$
e) $\frac{1}{2}\left[-\frac{\cos(6x)}{6}-\frac{\cos(2x)}{2}\right]_{0}^{\pi/2}=\frac{2}{3}\approx 0.667$
f) $\frac{1}{2}\left[\frac{\sin(2x)}{2}-\frac{\sin(4x)}{4}\right]_{0}^{\pi/3}=\frac{3\sqrt{3}}{16}\approx 0.325$
g) $\int_{0}^{1}\left(u^{4}+u^{6}\right)du=\frac{12}{35}\approx 0.343$
h) $\int_{1}^{2}u^{2}du=\frac{7}{3}\approx 2.333$
i) $\frac{1}{8}\int_{0}^{\pi/2}\sin^{2}(2x)\,dx=\frac{\pi}{32}\approx 0.098$
j) $\frac{1}{5}-\frac{1}{3}+1-\frac{\pi}{4}=\frac{13}{15}-\frac{\pi}{4}\approx 0.081$

The move that decides everything: read the PARITIES before writing a single line. An odd power offers its split-off factor as $du$ ; two even powers offer none and force you to reduce the powers; two different angles force you to turn the product into a sum. Choosing the wrong rule does not make the computation wrong, it makes it endless.

a)

Everything is even: $\sin^{2}\left(\frac{x}{2}\right)=\frac{1-\cos x}{2}$ , since the DOUBLE of the angle $\frac{x}{2}$ is $x$ . So $\int_{0}^{\pi/2}\frac{1-\cos x}{2}dx=\left[\frac{x}{2}-\frac{\sin x}{2}\right]_{0}^{\pi/2}=\frac{\pi}{4}-\frac{1}{2}\approx 0.285$ . Pitfall: writing $\cos\left(\frac{x}{4}\right)$ or $\cos(2x)$ in the power reduction.

b)

Odd power: $\sin^{5}x=\left(1-\cos^{2}x\right)^{2}\sin x$ , and $u=\cos x$ , $du=-\sin x\,dx$ , with $u$ going from $1$ to $0$ . The minus sign flips the limits: $\int_{0}^{1}\left(1-u^{2}\right)^{2}du=\left[u-\frac{2u^{3}}{3}+\frac{u^{5}}{5}\right]_{0}^{1}=1-\frac{2}{3}+\frac{1}{5}=\frac{8}{15}\approx 0.533$ .

c)

The sine has the odd power, so split it off: $\sin^{3}x\sqrt{\cos x}=\left(1-\cos^{2}x\right)\cos^{\frac{1}{2}}x\,\sin x$ . With $u=\cos x$ : $\int_{0}^{1}\left(u^{\frac{1}{2}}-u^{\frac{5}{2}}\right)du=\frac{2}{3}-\frac{2}{7}=\frac{8}{21}\approx 0.381$ . The odd-power rule works whatever the other power is, even a fractional one.

d)

Different angles: $\cos A\cos B=\frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$ , so $\cos(3x)\cos x=\frac{1}{2}\left[\cos(2x)+\cos(4x)\right]$ . Then $\frac{1}{2}\left[\frac{\sin(2x)}{2}+\frac{\sin(4x)}{4}\right]_{0}^{\pi/4}=\frac{1}{2}\left(\frac{1}{2}+0\right)=\frac{1}{4}$ . Pitfall: integrating each factor separately, $\frac{\sin(3x)}{3}\sin x$ , a “product rule” for integrals that does not exist.

e)

$\sin A\cos B=\frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right]$ , so $\sin(4x)\cos(2x)=\frac{1}{2}\left[\sin(6x)+\sin(2x)\right]$ . Then $\frac{1}{2}\left[-\frac{\cos(6x)}{6}-\frac{\cos(2x)}{2}\right]_{0}^{\pi/2}=\frac{1}{2}\left[\left(\frac{1}{6}+\frac{1}{2}\right)-\left(-\frac{1}{6}-\frac{1}{2}\right)\right]=\frac{2}{3}\approx 0.667$ , since $\cos 3\pi=\cos\pi=-1$ .

f)

$\sin A\sin B=\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$ , so $\sin(3x)\sin x=\frac{1}{2}\left[\cos(2x)-\cos(4x)\right]$ . Then $\frac{1}{2}\left[\frac{\sin(2x)}{2}-\frac{\sin(4x)}{4}\right]_{0}^{\pi/3}=\frac{1}{2}\left(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{8}\right)=\frac{3\sqrt{3}}{16}\approx 0.325$ , since $\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}$ . Pitfall: the order of the difference, $\cos(A-B)-\cos(A+B)$ and not the other way around.

g)

The secant has an even power: split off $\sec^{2}x$ , and $\tan^{4}x\sec^{4}x=\tan^{4}x\left(1+\tan^{2}x\right)\sec^{2}x$ . With $u=\tan x$ , from $0$ to $1$ : $\int_{0}^{1}\left(u^{4}+u^{6}\right)du=\frac{1}{5}+\frac{1}{7}=\frac{12}{35}\approx 0.343$ .

h)

The secant power is odd, and so is the tangent power: split off $\sec x\tan x$ , and $\tan x\sec^{3}x=\sec^{2}x\cdot\sec x\tan x$ . With $u=\sec x$ , from $1$ to $\sec\frac{\pi}{3}=2$ : $\int_{1}^{2}u^{2}du=\frac{8-1}{3}=\frac{7}{3}\approx 2.333$ . Pitfall: $u=\tan x$ , which leaves a $\sec x$ without a partner.

i)

Two even powers: reduce them. $\sin^{2}x\cos^{2}x=\frac{\sin^{2}(2x)}{4}$ and $\sin^{2}x=\frac{1-\cos(2x)}{2}$ , so $\sin^{4}x\cos^{2}x=\frac{\sin^{2}(2x)}{8}-\frac{\sin^{2}(2x)\cos(2x)}{8}$ . On $\left[0, \frac{\pi}{2}\right]$ , $\int\sin^{2}(2x)dx=\frac{\pi}{4}$ (average value $\frac{1}{2}$ ), and $\int\sin^{2}(2x)\cos(2x)dx=\left[\frac{\sin^{3}(2x)}{6}\right]_{0}^{\pi/2}=0$ . So the value is $\frac{\pi}{32}\approx 0.098$ .

j)

No secant, no direct rule: lower the power with $\tan^{2}x=\sec^{2}x-1$ , twice. $\tan^{6}x=\tan^{4}x\sec^{2}x-\tan^{4}x$ , then $\tan^{4}x=\tan^{2}x\sec^{2}x-\sec^{2}x+1$ , so $\tan^{6}x=\tan^{4}x\sec^{2}x-\tan^{2}x\sec^{2}x+\sec^{2}x-1$ . With $u=\tan x$ on the first three terms: $\frac{1}{5}-\frac{1}{3}+1-\frac{\pi}{4}=\frac{13}{15}-\frac{\pi}{4}\approx 0.081$ .

Check: a calculator gives each integral through a fine Riemann sum, and the parity rule can be checked by differentiating the antiderivative you found. A negative result for a positive function, such as $\tan^{6}x$ or $\sin^{4}x\cos^{2}x$ , is a sure sign of a sign error in the identity.

Exercise 1: Antiderivatives and substitution

Substitution is the first reflex: look in the integral for a function AND its derivative (up to a constant factor).

a) Compute $\int\left(6x^{2}-\frac{4}{x}+e^{2x}\right)dx$ and check the result by differentiation.
b) Compute $\int x^{2}\sqrt{x^{3}+1}\,dx$ , then $\int\tan x\,dx$ , stating each time the substitution that unlocks everything.
c) Compute $\int\frac{dx}{x\ln x}$ . What structure did you have to recognize?
d) Compute $\int_{1}^{e}\frac{(\ln x)^{2}}{x}dx$ by transforming the limits of integration, and explain the classic mistake of the student who keeps the limits in $x$ while working in $u$ .

Show the solution

Answers

a) $2x^{3}-4\ln|x|+\frac{e^{2x}}{2}+C$
b) $\frac{2}{9}(x^{3}+1)^{3/2}+C$ ; $-\ln|\cos x|+C$
c) $\ln|\ln x|+C$ , the form $\frac{du}{u}$
d) $\int_{0}^{1}u^{2}du=\frac{1}{3}$

a)

Term by term: $\int 6x^{2}dx=2x^{3}$ , $\int\frac{4}{x}dx=4\ln|x|$ , $\int e^{2x}dx=\frac{e^{2x}}{2}$ (the inner derivative 2 is paid for with a factor $\frac{1}{2}$ ). Total: $2x^{3}-4\ln|x|+\frac{e^{2x}}{2}+C$ . Check: $\frac{d}{dx}\left(2x^{3}-4\ln|x|+\frac{e^{2x}}{2}\right)=6x^{2}-\frac{4}{x}+e^{2x}$ ✓. Checking by differentiation takes ten seconds and catches half of all constant-factor mistakes.

b)

First case: $u=x^{3}+1$ , $du=3x^{2}dx$ : the $x^{2}dx$ already present is $\frac{du}{3}$ : $\frac{1}{3}\int\sqrt{u}\,du=\frac{2}{9}u^{3/2}=\frac{2}{9}(x^{3}+1)^{3/2}+C$ . Second case: $\int\tan x\,dx=\int\frac{\sin x}{\cos x}dx$ with $u=\cos x$ , $du=-\sin x\,dx$ : $-\int\frac{du}{u}=-\ln|\cos x|+C$ . In both cases, the derivative of the inside was waiting as a factor: that is THE signature of a substitution.

c)

$\int\frac{dx}{x\ln x}=\int\frac{1}{\ln x}\cdot\frac{dx}{x}$ : the structure is $\frac{1}{u}\,du$ with $u=\ln x$ (whose derivative $\frac{1}{x}$ is there). Result: $\ln|\ln x|+C$ : a logarithm of a logarithm, perfectly legitimate. This integral is also the one that decides the divergence of $\int_{2}^{\infty}\frac{dx}{x\ln x}$ in the chapter on improper integrals: today's techniques are used everywhere afterwards.

d)

$u=\ln x$ , $du=\frac{dx}{x}$ ; limits: $x=1\Rightarrow u=0$ , $x=e\Rightarrow u=1$ . $\int_{0}^{1}u^{2}du=\frac{1}{3}$ . The classic mistake: writing $\int_{1}^{e}u^{2}du=\frac{e^{3}-1}{3}$ , keeping the limits for $x$ on an integral in $u$ : the limits belong to the VARIABLE, not to the integral sign. Two remedies: transform the limits immediately (the safest), or go back to $x$ before evaluating; but never a mix of the two.

Exercise 2: Integration by parts: simple, repeated, cyclic

The formula $\int u\,dv=uv-\int v\,du$ trades one integral for another, which you hope is simpler. Everything depends on the choice of $u$ : you differentiate what simplifies.

a) Compute $\int x\cos x\,dx$ , justifying the choice of $u$ and $dv$ .
b) Compute $\int x^{2}e^{x}dx$ (repeated parts). What happens to the polynomial at each round?
c) Compute $\int\ln x\,dx$ with the choice $dv=dx$ . Why is this trick unavoidable here?
d) Compute $I=\int e^{x}\cos x\,dx$ (cyclic case), then evaluate $\int_{0}^{\pi}e^{x}\cos x\,dx$ .

Show the solution

Answers

a) $x\sin x+\cos x+C$
b) $e^{x}(x^{2}-2x+2)+C$ , two rounds
c) $x\ln x-x+C$
d) $\frac{e^{x}(\sin x+\cos x)}{2}+C$ ; $-\frac{e^{\pi}+1}{2}\approx-12.07$

a)

$u=x$ (the polynomial, which dies out when differentiated), $dv=\cos x\,dx$ , $v=\sin x$ : $\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x+C$ . The opposite choice ( $u=\cos x$ ) would give $\int\frac{x^{2}}{2}\sin x\,dx$ : WORSE. The rule “differentiate what simplifies” is not decoration.

b)

First round ( $u=x^{2}$ ): $x^{2}e^{x}-2\int xe^{x}dx$ . Second round ( $u=x$ ): $\int xe^{x}dx=xe^{x}-e^{x}$ . Putting it together: $\int x^{2}e^{x}dx=e^{x}(x^{2}-2x+2)+C$ . The polynomial drops one degree at each round: degree 2, so exactly two rounds; the exponential takes it all without changing. Quick check: differentiating $e^{x}(x^{2}-2x+2)$ gives back $x^{2}e^{x}$ ✓.

c)

$\ln x$ has no obvious antiderivative but a very simple derivative: you force integration by parts with $u=\ln x$ and $dv=dx$ , $v=x$ : $\int\ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}dx=x\ln x-x+C$ . The trick is unavoidable because there is only ONE visible factor: integration by parts manufactures the second one ( $dv=dx$ ). Same recipe for $\arctan x$ and $\arcsin x$ : the functions “with a nice derivative” are all handled this way.

d)

Two rounds of integration by parts (always differentiating the trigonometric factor, or always the exponential, but WITHOUT switching sides): $I=e^{x}\sin x-\int e^{x}\sin x\,dx=e^{x}\sin x+e^{x}\cos x-I$ . The integral has come back: SOLVE the equation: $2I=e^{x}(\sin x+\cos x)$ , so $I=\frac{e^{x}(\sin x+\cos x)}{2}+C$ . Evaluation: $\int_{0}^{\pi}e^{x}\cos x\,dx=\left[\frac{e^{x}(\sin x+\cos x)}{2}\right]_{0}^{\pi}=\frac{e^{\pi}(0-1)}{2}-\frac{1(0+1)}{2}=-\frac{e^{\pi}+1}{2}\approx-12.07$ : negative, which is consistent with a cosine that is mostly negative on $[\frac{\pi}{2},\pi]$ , where $e^{x}$ weighs the most.

Exercise 3: Partial fractions: take it apart to integrate it

Every rational function can be integrated: you take it apart into elementary pieces whose antiderivatives are logarithms, powers and arctangents.

a) Compute $\int\frac{x+5}{x^{2}+x-2}dx$ (two simple factors).
b) Compute $\int\frac{x^{2}}{x-1}dx$ . What COMPULSORY step comes before the decomposition here, and why?
c) Compute $\int\frac{3x+5}{(x+1)^{2}}dx$ (repeated factor).
d) Compute $\int\frac{dx}{x^{2}+4}$ , then $\int_{0}^{2}\frac{dx}{x^{2}+4}$ . Why does this factor not decompose, and which function does it always lead to?

Show the solution

Answers

a) $-\ln|x+2|+2\ln|x-1|+C$
b) Divide first: $\frac{x^{2}}{2}+x+\ln|x-1|+C$
c) $3\ln|x+1|-\frac{2}{x+1}+C$
d) $\frac{1}{2}\arctan\frac{x}{2}+C$ ; $\frac{\pi}{8}\approx 0.393$

a)

$x^{2}+x-2=(x+2)(x-1)$ . Set $\frac{x+5}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}$ , that is $x+5=A(x-1)+B(x+2)$ . Method of convenient values: $x=1$ : $6=3B$ , $B=2$ ; $x=-2$ : $3=-3A$ , $A=-1$ . Hence $\int=-\ln|x+2|+2\ln|x-1|+C=\ln\frac{(x-1)^{2}}{|x+2|}+C$ .

b)

The degree of the numerator (2) reaches that of the denominator (1): polynomial LONG DIVISION first, since the decomposition is only defined for a proper rational function. $x^{2}=(x-1)(x+1)+1$ , so $\frac{x^{2}}{x-1}=x+1+\frac{1}{x-1}$ and $\int=\frac{x^{2}}{2}+x+\ln|x-1|+C$ . Forgetting the division is the number one mistake of the chapter: the decomposition set up on an improper fraction leads to a system with no solution, and you lose ten minutes finding that out.

c)

Repeated factor: $\frac{3x+5}{(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}$ , that is $3x+5=A(x+1)+B$ . With $x=-1$ : $B=2$ ; with the coefficients of $x$ : $A=3$ . Hence $\int=3\ln|x+1|-\frac{2}{x+1}+C$ : the simple factor gives the logarithm, the squared factor gives a power: each level of the repetition makes its own contribution.

d)

$x^{2}+4$ has NO real root: it is irreducible, it cannot be broken into first-degree factors. Its antiderivative is the arctangent: $\int\frac{dx}{x^{2}+4}=\frac{1}{2}\arctan\frac{x}{2}+C$ (the $a=2$ appears twice: in the argument AND as a factor). $\int_{0}^{2}\frac{dx}{x^{2}+4}=\frac{1}{2}\arctan 1-0=\frac{\pi}{8}\approx 0.393$ . Every irreducible quadratic leads to the arctangent (after completing the square if needed): it is the third and last family of antiderivatives of rational functions, along with logarithms and powers.

Exercise 4: Powers of tangent and secant

Alongside powers of sine and cosine (exercise 6), this is the other trigonometric family on the exam. It has its own rules, based on the identity $1+\tan^{2}=\sec^{2}$ .

a) Compute $\int\tan^{2}x\,dx$ using the identity $\tan^{2}x=\sec^{2}x-1$ .
b) Compute $\int\tan^{3}x\,dx$ by splitting off a factor $\tan x$ .
c) Compute $\int\sec^{3}x\,dx$ by integration by parts. Why is this case called cyclic?
d) State the two rules of the tangent-secant family, and say which of the two functions plays the role of $du$ .

Show the solution

Answers

a) $\tan x-x+C$
b) $\frac{\tan^{2}x}{2}+\ln|\cos x|+C$
c) $\frac{\sec x\tan x+\ln|\sec x+\tan x|}{2}+C$
d) Even secant: $du=\sec^{2}x\,dx$ ; odd tangent: $du=\sec x\tan x\,dx$

a)

The identity $1+\tan^{2}x=\sec^{2}x$ gives $\tan^{2}x=\sec^{2}x-1$ , so $\int\tan^{2}x\,dx=\int\sec^{2}x\,dx-\int dx=\tan x-x+C$ , since $\sec^{2}$ is the derivative of $\tan$ . An integral with no apparent antiderivative has become immediate through a simple rewriting: trigonometric identities are integration techniques in their own right.

b)

Split off a factor $\tan x$ and convert the rest: $\tan^{3}x=\tan x\cdot\tan^{2}x=\tan x\left(\sec^{2}x-1\right)=\tan x\sec^{2}x-\tan x$ . The first integral is done with $u=\tan x$ , $du=\sec^{2}x\,dx$ : $\int\tan x\sec^{2}x\,dx=\frac{\tan^{2}x}{2}$ . The second one is known: $\int\tan x\,dx=-\ln\left|\cos x\right|$ . Total: $\int\tan^{3}x\,dx=\frac{\tan^{2}x}{2}+\ln\left|\cos x\right|+C$ . Check by differentiation: $\tan x\sec^{2}x-\tan x=\tan x\left(\sec^{2}x-1\right)=\tan^{3}x$ ✓.

c)

Write $\sec^{3}x=\sec x\cdot\sec^{2}x$ and set $u=\sec x$ (so $du=\sec x\tan x\,dx$ ) and $dv=\sec^{2}x\,dx$ (so $v=\tan x$ ). Then $I=\int\sec^{3}x\,dx=\sec x\tan x-\int\sec x\tan^{2}x\,dx$ . Converting $\tan^{2}=\sec^{2}-1$ : $I=\sec x\tan x-\int\sec^{3}x\,dx+\int\sec x\,dx=\sec x\tan x-I+\ln\left|\sec x+\tan x\right|$ . The original integral has COME BACK: this is the cyclic case. You then SOLVE the equation in $I$ : $2I=\sec x\tan x+\ln\left|\sec x+\tan x\right|$ , hence $I=\frac{\sec x\tan x+\ln\left|\sec x+\tan x\right|}{2}+C$ . The mechanism is the same as for $\int e^{x}\cos x\,dx$ : when the integral reproduces itself, you treat it as an algebraic unknown.

d)

Two rules. EVEN POWER OF SECANT: split off $\sec^{2}x$ , which is the $du$ of $u=\tan x$ , and convert the rest into tangents with $\sec^{2}=1+\tan^{2}$ . ODD POWER OF TANGENT: split off $\sec x\tan x$ , which is the $du$ of $u=\sec x$ , and convert the rest into secants with $\tan^{2}=\sec^{2}-1$ . So it is sometimes $\sec^{2}x$ , sometimes $\sec x\tan x$ that plays the role of $du$ , depending on the parity you meet. When neither configuration shows up, as for $\int\sec^{3}x\,dx$ where the secant is odd and the tangent absent, you have to go through integration by parts, as in c). The logic is the same as for sine and cosine: first look for which factor can serve as $du$ , and parity decides.

Exercise 5: Products of sines and cosines of different angles

$\int\sin 3x\cos 5x\,dx$ calls for neither substitution nor integration by parts: you must first TURN the product into a sum. The formula is worth knowing, and its consequence underlies all of Fourier analysis.

a) Recall the three product-to-sum formulas for $\sin A\cos B$ , $\sin A\sin B$ and $\cos A\cos B$ .
b) Compute $\int\sin 3x\cos 5x\,dx$ .
c) Compute $\int_{0}^{2\pi}\sin(2x)\sin(3x)\,dx$ , then $\int_{0}^{2\pi}\sin^{2}(3x)\,dx$ . What do you notice?
d) This contrast has a name. State the general result and say what it is used for.

Show the solution

Answers

a) $\frac{1}{2}[\sin(A+B)+\sin(A-B)]$ , $\frac{1}{2}[\cos(A-B)\mp\cos(A+B)]$
b) $-\frac{\cos 8x}{16}+\frac{\cos 2x}{4}+C$
c) $0$ and $\pi$
d) Orthogonality: it isolates the Fourier coefficients

a)

The three formulas: $\sin A\cos B=\frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right]$ ; $\sin A\sin B=\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$ ; $\cos A\cos B=\frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$ . They come from adding or subtracting the sum formulas, and they turn a PRODUCT (which we cannot integrate) into a SUM (which we integrate term by term).

b)

With $A=3x$ and $B=5x$ : $\sin 3x\cos 5x=\frac{1}{2}\left[\sin(8x)+\sin(-2x)\right]=\frac{1}{2}\left[\sin 8x-\sin 2x\right]$ , using the fact that sine is odd. Hence $\int\sin 3x\cos 5x\,dx=\frac{1}{2}\left[-\frac{\cos 8x}{8}+\frac{\cos 2x}{2}\right]+C=-\frac{\cos 8x}{16}+\frac{\cos 2x}{4}+C$ . Check by differentiation: $\frac{8\sin 8x}{16}-\frac{2\sin 2x}{4}=\frac{\sin 8x-\sin 2x}{2}$ ✓, which is indeed the transformed integrand.

c)

First computation: $\sin 2x\sin 3x=\frac{1}{2}\left[\cos(-x)-\cos(5x)\right]=\frac{1}{2}\left[\cos x-\cos 5x\right]$ . On $\left[0,2\pi\right]$ , each cosine runs through a whole number of complete periods and its integral is ZERO, so $\int_{0}^{2\pi}\sin 2x\sin 3x\,dx=0$ . Second computation: $\sin^{2}(3x)=\frac{1-\cos 6x}{2}$ , so $\int_{0}^{2\pi}\sin^{2}(3x)dx=\frac{1}{2}\int_{0}^{2\pi}dx-\frac{1}{2}\int_{0}^{2\pi}\cos 6x\,dx=\frac{2\pi}{2}-0=\pi$ . What you notice: the product of two sines of DIFFERENT frequencies integrates to zero, whereas the square of a single sine gives $\pi$ .

d)

This contrast is called the ORTHOGONALITY of the trigonometric functions. General result: for positive integers $m$ and $n$ , $\int_{0}^{2\pi}\sin(mx)\sin(nx)\,dx=0$ if $m\neq n$ and $=\pi$ if $m=n$ ; the same relations hold for cosines, and $\int_{0}^{2\pi}\sin(mx)\cos(nx)\,dx=0$ always. What it is used for: it is the mechanism that EXTRACTS the coefficients of a Fourier series. If a function is written as an infinite sum $\sum a_{n}\sin(nx)$ , then multiplying by $\sin(mx)$ and integrating over one period cancels ALL the terms except the one with index $m$ , of which only $a_{m}\pi$ remains: this isolates each coefficient one at a time. It is exactly the idea of an orthogonal basis in linear algebra, where a coordinate is recovered by a dot product; here the integral plays the role of the dot product. This result, obtained with nothing but the formulas of a), underlies signal processing, audio compression and spectral analysis.

Exercise 6: Powers of sine and cosine

The figure shows $y=\sin^{2}x$ over one full period, with its average value $\frac{1}{2}$ as a dashed line: the power reduction $\sin^{2}x=\frac{1-\cos 2x}{2}$ can be READ on the graph.

a) Compute $\int\sin^{3}x\,dx$ with the odd-power method.
b) Compute $\int\cos^{2}x\,dx$ by power reduction, then $\int_{0}^{\pi}\sin^{2}x\,dx$ . Relate this last result to the dashed line in the figure.
c) Compute $\int\sin^{2}x\cos^{3}x\,dx$ : which power should you attack, and why?
d) Summarize the two rules (odd versus even) and explain WHY splitting off a factor from an odd power always works.

Show the solution

Answers

a) $-\cos x+\frac{\cos^{3}x}{3}+C$
b) $\frac{x}{2}+\frac{\sin 2x}{4}+C$ ; $\frac{\pi}{2}$
c) Break up $\cos^{3}x$ : $\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C$
d) Odd: split off and substitute; all even: reduce the powers

a)

Odd power: split off one sine and convert the rest: $\sin^{3}x=\sin x(1-\cos^{2}x)$ . With $u=\cos x$ , $du=-\sin x\,dx$ : $\int\sin^{3}x\,dx=-\int(1-u^{2})du=-u+\frac{u^{3}}{3}=-\cos x+\frac{\cos^{3}x}{3}+C$ .

b)

$\cos^{2}x=\frac{1+\cos 2x}{2}$ : $\int\cos^{2}x\,dx=\frac{x}{2}+\frac{\sin 2x}{4}+C$ . In the same way, $\int_{0}^{\pi}\sin^{2}x\,dx=\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_{0}^{\pi}=\frac{\pi}{2}$ . Reading the graph: $\sin^{2}$ oscillates SYMMETRICALLY around $\frac{1}{2}$ (the dashed line), the humps above making up for the dips below: its integral is that of the constant $\frac{1}{2}$ , namely $\frac{\pi}{2}$ on $[0,\pi]$ . This average $\frac{1}{2}$ is what produces the $\frac{A}{\sqrt{2}}$ of RMS voltages in electricity.

c)

The cosine carries the ODD power: that is the one you break up. $\cos^{3}x=\cos x(1-\sin^{2}x)$ , and with $u=\sin x$ : $\int u^{2}(1-u^{2})du=\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C$ . Attacking the sine (even power) would have led nowhere: the split-off factor must be the derivative of the other function.

d)

Rules: an ODD power somewhere: split off one factor of that function, convert the rest into the other function through $\sin^{2}+\cos^{2}=1$ , substitute. Everything EVEN: reduce with the double-angle formulas, repeating if necessary. Why odd always works: removing one factor from an odd power leaves an EVEN power, which can therefore be converted entirely into the other function; and the split-off factor is exactly the $du$ of the substitution. Parity is not a quirk of the recipe: it is what guarantees that the conversion comes out right.

Exercise 7: Trigonometric substitutions: the root becomes a triangle

The figure shows $y=\sqrt{9-x^{2}}$ on $[0,3]$ : the quarter circle of radius 3, whose area, known from geometry, will serve as a check on the computation.

a) Compute $\int\frac{dx}{\sqrt{9-x^{2}}}$ : with which substitution, and towards which function?
b) Compute $\int_{0}^{3}\sqrt{9-x^{2}}\,dx$ with the substitution $x=3\sin\theta$ , then check with the geometry of the figure.
c) Compute $\int_{0}^{3}\frac{dx}{\sqrt{x^{2}+9}}$ (you may use the antiderivative $\ln\left|x+\sqrt{x^{2}+9}\right|$ , to be checked by differentiation).
d) Draw up the table of the three substitutions: which root calls for which change of variable, and which right triangle hides behind each one?

Show the solution

Answers

a) $x=3\sin\theta$ : $\arcsin\frac{x}{3}+C$
b) $\frac{9\pi}{4}\approx 7.07$ , the quarter disk
c) $\ln(1+\sqrt{2})\approx 0.881$
d) $a\sin\theta$ , $a\tan\theta$ , $a\sec\theta$

a)

$x=3\sin\theta$ , $dx=3\cos\theta\,d\theta$ , and $\sqrt{9-9\sin^{2}\theta}=3\cos\theta$ : the integral becomes $\int\frac{3\cos\theta}{3\cos\theta}d\theta=\theta+C=\arcsin\frac{x}{3}+C$ . It is in fact an antiderivative worth knowing directly: $\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\arcsin\frac{x}{a}+C$ .

b)

Same ingredients: $\int_{0}^{3}\sqrt{9-x^{2}}\,dx=\int_{0}^{\pi/2}3\cos\theta\cdot 3\cos\theta\,d\theta=9\int_{0}^{\pi/2}\cos^{2}\theta\,d\theta=9\cdot\frac{\pi}{4}$ (power reduction from exercise 6: over a quarter period of the cosine, $\cos^{2}$ has an average value of $\frac{1}{2}$ ). Result: $\frac{9\pi}{4}\approx 7.07$ . Geometry: the curve is the quarter circle of radius 3, with area $\frac{\pi r^{2}}{4}=\frac{9\pi}{4}$ ✓: the computation and the figure confirm each other, and this is precisely how the formula for the area of a disk is proved.

c)

Checking the antiderivative: $\frac{d}{dx}\ln\left(x+\sqrt{x^{2}+9}\right)=\frac{1+\frac{x}{\sqrt{x^{2}+9}}}{x+\sqrt{x^{2}+9}}=\frac{\frac{\sqrt{x^{2}+9}+x}{\sqrt{x^{2}+9}}}{x+\sqrt{x^{2}+9}}=\frac{1}{\sqrt{x^{2}+9}}$ ✓ (it comes from the substitution $x=3\tan\theta$ ). Evaluation: $\ln(3+\sqrt{18})-\ln 3=\ln\frac{3+3\sqrt{2}}{3}=\ln(1+\sqrt{2})\approx 0.881$ .

d)

$\sqrt{a^{2}-x^{2}}$ : set $x=a\sin\theta$ (the triangle has hypotenuse $a$ and opposite side $x$ : the root is the adjacent side). $\sqrt{a^{2}+x^{2}}$ : set $x=a\tan\theta$ (legs $a$ and $x$ : the root is the hypotenuse). $\sqrt{x^{2}-a^{2}}$ : set $x=a\sec\theta$ (hypotenuse $x$ , leg $a$ : the root is the other leg). In all three cases, the identity $\sin^{2}+\cos^{2}=1$ (or one of its variants) makes the root VANISH, and the triangle is used for the return trip: read $\sin\theta$ , $\tan\theta$ , $\sec\theta$ off the sides to rewrite the answer in terms of $x$ .

Exercise 8: Strategy: choosing your technique in thirty seconds

The most discriminating format on the exam: integrals that look alike but call for different techniques. The diagnosis is worth half the marks.

a) Compute $\int xe^{x^{2}}dx$ , then $\int xe^{x}dx$ . Why two different techniques for two almost identical integrals?
b) Compute $\int\frac{dx}{x^{2}-1}$ , then $\int\frac{dx}{x^{2}+1}$ . Which detail changes the whole method?
c) Compute $\int\tan^{2}x\,dx$ : which identity turns the impossible integral into an immediate one?
d) Write the strategy checklist: in what order to question an unfamiliar integral, with a triggering example for each question.

Show the solution

Answers

a) $\frac{e^{x^{2}}}{2}+C$ ; $(x-1)e^{x}+C$
b) $\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C$ ; $\arctan x+C$
c) $\tan x-x+C$
d) Table, substitution, parts, partial fractions, parity, trigonometric substitution

a)

$\int xe^{x^{2}}dx$ : the $x$ is (up to a factor 2) the derivative of $x^{2}$ : SUBSTITUTION $u=x^{2}$ , result $\frac{e^{x^{2}}}{2}+C$ . $\int xe^{x}dx$ : no substitution works (the derivative of the exponent $x$ is $1$ , and the factor $x$ is left over): INTEGRATION BY PARTS, result $(x-1)e^{x}+C$ . The pair is the perfect diagnostic test: first ask “is the derivative of the inside there as a factor?”; integration by parts only comes after that check has failed. (And $\int e^{x^{2}}dx$ , without the $x$ , has no elementary antiderivative at all: the factor $x$ makes all the difference.)

b)

$x^{2}-1=(x-1)(x+1)$ factors: PARTIAL FRACTIONS, $\frac{1}{x^{2}-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$ , hence $\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C$ . $x^{2}+1$ does NOT factor: arctangent directly, $\arctan x+C$ . A single sign decides between logarithms and arctangent: always test whether the denominator factors before anything else.

c)

$\tan^{2}x=\sec^{2}x-1$ (the identity $1+\tan^{2}=\sec^{2}$ rearranged): $\int\tan^{2}x\,dx=\tan x-x+C$ , since $\sec^{2}$ is the derivative of $\tan$ . Trigonometric identities are integration techniques in their own right: they replace an integrand with no apparent antiderivative by another, equal one, that has one.

d)

The list, in order: 1) Is it a DIRECT form from the table (including $\arctan$ , $\arcsin$ )? 2) Do a function and its derivative appear together? SUBSTITUTION ( $xe^{x^{2}}$ , $\frac{\ln x}{x}$ , $\tan x$ ). 3) A product of different families (polynomial, exponential, trigonometric, logarithm)? INTEGRATION BY PARTS ( $xe^{x}$ , $x\cos x$ , $\ln x$ alone). 4) A rational function? Division if necessary, then PARTIAL FRACTIONS. 5) Trigonometric powers? Parity: split off or reduce. 6) A root $\sqrt{a^{2}\pm x^{2}}$ or $\sqrt{x^{2}-a^{2}}$ ? TRIGONOMETRIC SUBSTITUTION. Six questions, thirty seconds: diagnosis before treatment, as in medicine.

Exercise 9: Combining techniques: substitution, then parts

The toughest integrals on the exam give way to NO single technique: you have to chain two, and it is the choice of the first one that unlocks everything.

a) Compute $\int e^{\sqrt{x}}\,dx$ by substituting first, then integrating by parts.
b) Compute $\int x\arctan x\,dx$ : integration by parts first, then a division.
c) Compute $\int\frac{\ln x}{x^{2}}dx$ and check the result by differentiation.
d) Facing an integral that resists, in what order should you try the techniques? What does a root inside an exponential tell you?

Show the solution

Answers

a) $2(\sqrt{x}-1)e^{\sqrt{x}}+C$
b) $\frac{x^{2}}{2}\arctan x-\frac{x}{2}+\frac{\arctan x}{2}+C$
c) $-\frac{\ln x+1}{x}+C$
d) Substitute to clean up, integrate by parts to finish

a)

The root in the exponent rules out any direct antiderivative. Substitute $u=\sqrt{x}$ , so $x=u^{2}$ and $dx=2u\,du$ : $\int e^{\sqrt{x}}dx=\int e^{u}\cdot 2u\,du=2\int ue^{u}du$ . This last integral is a polynomial times an exponential, so INTEGRATION BY PARTS with $s=u$ and $dt=e^{u}du$ : $\int ue^{u}du=ue^{u}-e^{u}=(u-1)e^{u}$ . Hence $\int e^{\sqrt{x}}dx=2(u-1)e^{u}+C=2\left(\sqrt{x}-1\right)e^{\sqrt{x}}+C$ . Check by differentiation: $2\left[\frac{1}{2\sqrt{x}}e^{\sqrt{x}}+\left(\sqrt{x}-1\right)\frac{e^{\sqrt{x}}}{2\sqrt{x}}\right]=\frac{e^{\sqrt{x}}}{\sqrt{x}}\left[1+\sqrt{x}-1\right]=e^{\sqrt{x}}$ ✓.

b)

INTEGRATION BY PARTS first, with $u=\arctan x$ (the function with a nice derivative) and $dv=x\,dx$ , so $v=\frac{x^{2}}{2}$ : $\int x\arctan x\,dx=\frac{x^{2}}{2}\arctan x-\frac{1}{2}\int\frac{x^{2}}{1+x^{2}}dx$ . The remaining integral is an IMPROPER rational function (equal degrees), so you must DIVIDE: $\frac{x^{2}}{1+x^{2}}=1-\frac{1}{1+x^{2}}$ , whose integral is $x-\arctan x$ . Total: $\int x\arctan x\,dx=\frac{x^{2}}{2}\arctan x-\frac{x}{2}+\frac{\arctan x}{2}+C$ . Forgetting the division is the number one mistake here: it leads to a dead end when the computation was almost finished.

c)

Rewrite $\frac{\ln x}{x^{2}}=\ln x\cdot x^{-2}$ and integrate by parts with $u=\ln x$ (simple derivative) and $dv=x^{-2}dx$ , so $v=-\frac{1}{x}$ : $\int\frac{\ln x}{x^{2}}dx=-\frac{\ln x}{x}+\int\frac{1}{x}\cdot\frac{1}{x}dx=-\frac{\ln x}{x}+\int x^{-2}dx=-\frac{\ln x}{x}-\frac{1}{x}+C=-\frac{\ln x+1}{x}+C$ . Check: $\left(-\frac{\ln x+1}{x}\right)'=-\frac{\frac{1}{x}\cdot x-(\ln x+1)}{x^{2}}=-\frac{1-\ln x-1}{x^{2}}=\frac{\ln x}{x^{2}}$ ✓.

d)

The order: 1) simplify algebraically; 2) look for a SUBSTITUTION (a function and its derivative); 3) if the integrand is a product of different families, INTEGRATION BY PARTS; 4) if it is a rational function, division and then partial fractions; 5) if there are trigonometric powers, parity; 6) if there is a root of the form $\sqrt{a^{2}\pm x^{2}}$ , trigonometric substitution. Above all: after applying a technique, START THE LIST AGAIN on what is left, because that is how techniques are chained. A root INSIDE an exponential, a logarithm or a trigonometric function is a clear signal: no technique applies as long as the root is there, so you must start with the SUBSTITUTION $u=\sqrt{x}$ to get rid of it, even if the $dx=2u\,du$ brings in a factor that will then call for integration by parts. The general rule to remember: substitute to CLEAN UP, integrate by parts to FINISH.

Exercise 10: Completing the square and irreducible quadratics

A quadratic denominator with no real root does not factor. Completing the square brings it to the form $u^{2}+a^{2}$ , the only way in to the arctangent.

a) Compute $\int\frac{dx}{x^{2}+2x+5}$ by completing the square.
b) Compute $\int\frac{x+1}{x^{2}+4x+8}dx$ : the numerator forces a preliminary split. Which one, and why?
c) Compute $\int\frac{dx}{\sqrt{3-2x-x^{2}}}$ by completing the square under the root.
d) Facing a quadratic denominator, how do you recognize whether to factor or to complete the square? Which antiderivative does each case produce?

Show the solution

Answers

a) $\frac{1}{2}\arctan\frac{x+1}{2}+C$
b) $\frac{1}{2}\ln[(x+2)^{2}+4]-\frac{1}{2}\arctan\frac{x+2}{2}+C$
c) $\arcsin\frac{x+1}{2}+C$ on $(-3, 1)$
d) $\Delta>0$ logarithms, $\Delta<0$ arctangent, $\Delta=0$ power

a)

The discriminant is $4-20=-16<0$ : no real root, so no factorization is possible. Complete the square: $x^{2}+2x+5=\left(x^{2}+2x+1\right)+4=(x+1)^{2}+4$ . With the form $\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan\frac{u}{a}+C$ , where $u=x+1$ and $a=2$ : $\int\frac{dx}{x^{2}+2x+5}=\frac{1}{2}\arctan\frac{x+1}{2}+C$ . Numerical check at $x=1.1$ : the derivative of the answer equals $0.118906$ , and $\frac{1}{1.1^{2}+2(1.1)+5}=0.118906$ ✓.

b)

The denominator completes to $x^{2}+4x+8=(x+2)^{2}+4$ . The numerator $x+1$ is NOT proportional to the derivative of the denominator, which is $2x+4$ . So you SPLIT it into two pieces: $x+1=(x+2)-1$ . The first piece gives a form $\frac{u'}{u}$ up to a factor, hence a logarithm; the second gives an arctangent. Precisely: $\int\frac{x+2}{(x+2)^{2}+4}dx=\frac{1}{2}\ln\left[(x+2)^{2}+4\right]$ (substitution $u=(x+2)^{2}+4$ ), and $\int\frac{dx}{(x+2)^{2}+4}=\frac{1}{2}\arctan\frac{x+2}{2}$ . Total: $\int\frac{x+1}{x^{2}+4x+8}dx=\frac{1}{2}\ln\left[(x+2)^{2}+4\right]-\frac{1}{2}\arctan\frac{x+2}{2}+C$ . The split is needed because neither of the two usual antiderivatives can absorb a general linear numerator on its own: you must separate the “derivative of the denominator” part from the constant part.

c)

Under the root: $3-2x-x^{2}=-\left(x^{2}+2x-3\right)=-\left[(x+1)^{2}-4\right]=4-(x+1)^{2}$ . Watch the sign, this is the step where mistakes happen: you must factor out $-1$ BEFORE completing the square. You then recognize the form $\int\frac{du}{\sqrt{a^{2}-u^{2}}}=\arcsin\frac{u}{a}+C$ , with $u=x+1$ and $a=2$ : $\int\frac{dx}{\sqrt{3-2x-x^{2}}}=\arcsin\frac{x+1}{2}+C$ , valid on $-3<x<1$ , where the expression under the root is strictly positive.

d)

The test is the DISCRIMINANT of the quadratic. If $\Delta>0$ , there are two real roots, the quadratic FACTORS and you use partial fractions: the antiderivative is a sum of LOGARITHMS. If $\Delta<0$ , there is no real root, the quadratic is irreducible, you COMPLETE THE SQUARE and the antiderivative involves an ARCTANGENT (possibly together with a logarithm if the numerator is not constant, as in b). If $\Delta=0$ , the quadratic is a perfect square $(x-r)^{2}$ and the antiderivative is a POWER, in $-\frac{1}{x-r}$ . Three cases, three families of antiderivatives, and a single discriminant computation to decide: that is the reflex to have before anything else. Under a ROOT, the same completing work leads this time towards $\arcsin$ or towards a logarithm, depending on the sign of the square, which leads straight into trigonometric substitutions.

Exercise 11: The arches of x sin x: areas that grow in arithmetic progression

The curve of $f(x)=x\sin x$ oscillates between two lines, $y=x$ and $y=-x$ , in higher and higher arches. Each arch occupies an interval $[k\pi, (k+1)\pi]$ . The figure shows the first four.

a) Using integration by parts, find an antiderivative $F$ of $x\sin x$ .
b) Compute $\int_{\pi}^{2\pi}x\sin x\,dx$ . Why is the result negative?
c) Compute $\int_{2\pi}^{3\pi}x\sin x\,dx$ .
d) Show that $\int_{k\pi}^{(k+1)\pi}x\sin x\,dx=(-1)^{k}(2k+1)\pi$ for every integer $k\geq 0$ .
e) Deduce the total area between the curve and the axis on $[0, 4\pi]$ , then the value of $\int_{0}^{4\pi}x\sin x\,dx$ . Why are these two numbers so different?

Show the solution

Answers

a) $-x\cos x+\sin x+C$
b) $-3\pi$ : arch below the axis
c) $5\pi$
d) $(-1)^{k}(2k+1)\pi$
e) Area $16\pi$ , integral $-4\pi$

a)

$u=x$ , $dv=\sin x\,dx$ , so $du=dx$ and $v=-\cos x$ : $\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+C$ . Check: $\left(-x\cos x+\sin x\right)'=-\cos x+x\sin x+\cos x=x\sin x$ ✓.

b)

$F(2\pi)-F(\pi)=\left(-2\pi\times 1+0\right)-\left(-\pi\times(-1)+0\right)=-2\pi-\pi=-3\pi$ . On $(\pi, 2\pi)$ , $\sin x<0$ and $x>0$ : the arch lies BELOW the axis, and the integral counts its area negatively.

c)

$F(3\pi)-F(2\pi)=\left(-3\pi\times(-1)\right)-\left(-2\pi\times 1\right)=3\pi+2\pi=5\pi$ .

d)

Since $\sin(k\pi)=0$ and $\cos(k\pi)=(-1)^{k}$ : $F((k+1)\pi)-F(k\pi)=-(k+1)\pi(-1)^{k+1}+k\pi(-1)^{k}=(-1)^{k}\left[(k+1)\pi+k\pi\right]=(-1)^{k}(2k+1)\pi$ . The areas of the arches are $\pi$ , $3\pi$ , $5\pi$ , $7\pi$ ...: an arithmetic progression with common difference $2\pi$ .

e)

Total area: $\pi+3\pi+5\pi+7\pi=16\pi\approx 50.27$ . Integral: $\pi-3\pi+5\pi-7\pi=-4\pi$ , which $F(4\pi)-F(0)=-4\pi$ confirms. The integral adds up SIGNED areas that largely cancel out; the geometric area adds up their absolute values. For an area, split the interval at the zeros of the function before integrating.

Exercise 12: Two routes to the same integral: substitution or parts

We want to compute $I=\int x^{3}\sqrt{x^{2}+1}\,dx$ . Two methods work, and they do not give the answer in the same form.

a) Set $u=x^{2}+1$ and express the remaining $x^{2}$ in terms of $u$ to compute $I$ .
b) Compute $I$ by parts with $u=x^{2}$ and $dv=x\sqrt{x^{2}+1}\,dx$ .
c) Do the two answers differ by a constant? Show it by rewriting the second one in terms of $u=x^{2}+1$ .
d) Compute $\int_{0}^{\sqrt{3}}x^{3}\sqrt{x^{2}+1}\,dx$ .
e) Why is the choice $u=x^{3}$ , $dv=\sqrt{x^{2}+1}\,dx$ a dead end?

Show the solution

Answers

a) $\frac{1}{5}(x^{2}+1)^{5/2}-\frac{1}{3}(x^{2}+1)^{3/2}+C$
b) $\frac{x^{2}}{3}(x^{2}+1)^{3/2}-\frac{2}{15}(x^{2}+1)^{5/2}+C$
c) Identical once written in terms of $u$
d) $\frac{58}{15}\approx 3.867$
e) $dv$ must integrate without effort

a)

$du=2x\,dx$ , so $x^{3}dx=x^{2}\cdot x\,dx=(u-1)\frac{du}{2}$ : $I=\frac{1}{2}\int(u-1)u^{1/2}du=\frac{1}{2}\left(\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}\right)+C=\frac{1}{5}(x^{2}+1)^{5/2}-\frac{1}{3}(x^{2}+1)^{3/2}+C$ .

b)

$dv=x\sqrt{x^{2}+1}\,dx$ gives $v=\frac{1}{3}(x^{2}+1)^{3/2}$ (an immediate substitution), and $du=2x\,dx$ . Then $I=\frac{x^{2}}{3}(x^{2}+1)^{3/2}-\frac{2}{3}\int x(x^{2}+1)^{3/2}dx=\frac{x^{2}}{3}(x^{2}+1)^{3/2}-\frac{2}{15}(x^{2}+1)^{5/2}+C$ .

c)

With $x^{2}=u-1$ , the second one becomes $\frac{u-1}{3}u^{3/2}-\frac{2}{15}u^{5/2}=\frac{1}{3}u^{5/2}-\frac{1}{3}u^{3/2}-\frac{2}{15}u^{5/2}=\frac{1}{5}u^{5/2}-\frac{1}{3}u^{3/2}$ . They are IDENTICAL: the constant between them is zero. Two antiderivatives can look different without being different: before concluding there is a mistake, bring them back to the same variable.

d)

With the form from a) and the limits in $u$ , from $1$ to $4$ : $\left[\frac{1}{5}u^{5/2}-\frac{1}{3}u^{3/2}\right]_{1}^{4}=\left(\frac{32}{5}-\frac{8}{3}\right)-\left(\frac{1}{5}-\frac{1}{3}\right)=\frac{31}{5}-\frac{7}{3}=\frac{58}{15}\approx 3.867$ .

e)

With $dv=\sqrt{x^{2}+1}\,dx$ , you would first have to find $v=\int\sqrt{x^{2}+1}\,dx$ , which requires a trigonometric substitution and gives a logarithm; then $u=x^{3}$ only drops one degree per round. You trade one integral for a harder one. The rule: $dv$ must be the piece you can integrate WITHOUT effort, here $x\sqrt{x^{2}+1}$ , where the $x$ supplies the derivative of the inside.

Exercise 13: An integral with three poles: exact, improper, then forbidden

We study $f(x)=\frac{1}{x^{3}-x}$ , whose denominator vanishes at three points.

a) Factor $x^{3}-x$ and decompose $f(x)$ into partial fractions.
b) Give an antiderivative of $f$ and show that it can be written $\frac{1}{2}\ln\left|\frac{x^{2}-1}{x^{2}}\right|$ .
c) Compute $\int_{2}^{4}\frac{dx}{x^{3}-x}$ in exact form, then as a decimal.
d) Compute $\int_{2}^{+\infty}\frac{dx}{x^{3}-x}$ .
e) A student applies the same antiderivative on $\left[\frac{1}{2}, 2\right]$ and finds $-\ln 2$ . What is this integral really worth?

Show the solution

Answers

a) $-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}$
b) $\frac{1}{2}\ln\left|\frac{x^{2}-1}{x^{2}}\right|+C$
c) $\frac{1}{2}\ln\frac{5}{4}\approx 0.1116$
d) $\frac{1}{2}\ln\frac{4}{3}\approx 0.1438$
e) Pole at $1$ : the integral diverges

a)

$x^{3}-x=x(x-1)(x+1)$ , three simple poles. By the cover-up method: coefficient of $\frac{1}{x}$ , $\frac{1}{(0-1)(0+1)}=-1$ ; of $\frac{1}{x-1}$ , $\frac{1}{1\times 2}=\frac{1}{2}$ ; of $\frac{1}{x+1}$ , $\frac{1}{(-1)(-2)}=\frac{1}{2}$ . So $f(x)=-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}$ . Check at $x=2$ : $\frac{1}{6}$ and $-\frac{1}{2}+\frac{1}{2}+\frac{1}{6}=\frac{1}{6}$ ✓.

b)

$F(x)=-\ln|x|+\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|=\frac{1}{2}\ln|x^{2}-1|-\frac{1}{2}\ln x^{2}=\frac{1}{2}\ln\left|\frac{x^{2}-1}{x^{2}}\right|$ . Grouping the logarithms makes the limits much simpler to evaluate.

c)

$F(4)-F(2)=\frac{1}{2}\ln\frac{15}{16}-\frac{1}{2}\ln\frac{3}{4}=\frac{1}{2}\ln\left(\frac{15}{16}\times\frac{4}{3}\right)=\frac{1}{2}\ln\frac{5}{4}\approx 0.1116$ .

d)

$\int_{2}^{b}f=\frac{1}{2}\ln\frac{b^{2}-1}{b^{2}}-\frac{1}{2}\ln\frac{3}{4}$ , and $\frac{b^{2}-1}{b^{2}}\to 1$ as $b\to+\infty$ . The integral converges and equals $0-\frac{1}{2}\ln\frac{3}{4}=\frac{1}{2}\ln\frac{4}{3}\approx 0.1438$ . This is consistent: $f(x)$ behaves like $\frac{1}{x^{3}}$ at infinity.

e)

The student's computation, $\frac{1}{2}\ln\frac{3}{4}-\frac{1}{2}\ln 3=-\ln 2$ , makes no sense: the interval $\left[\frac{1}{2}, 2\right]$ contains the pole $x=1$ , where $f$ is not defined. Near $1$ , $f(x)\approx\frac{1}{2(x-1)}$ , whose integral diverges on both sides: the integral is improper and DIVERGES. The fundamental theorem requires a function that is continuous on the whole interval; check for poles BEFORE evaluating an antiderivative.

Exercise 14: Problem: the damped oscillator and its equilibrium position

An object attached to a spring is slowed down by a liquid. Its velocity is $v(t)=e^{-t/2}\cos t$ (in m/s, $t$ in seconds), and it starts from position $s(0)=0$ .

a) Compute $\int e^{-t/2}\cos t\,dt$ with two successive integrations by parts.
b) Deduce the position $s(t)$ .
c) Compute $s(\pi)$ .
d) Towards which position does the object settle?
e) At what instant does the object stop for the first time, and at what position? Is it the farthest position it will ever reach?

Show the solution

Answers

a) $\frac{2}{5}e^{-t/2}(2\sin t-\cos t)+C$
b) $s(t)=\frac{2}{5}e^{-t/2}(2\sin t-\cos t)+\frac{2}{5}$
c) $s(\pi)\approx 0.483$ m
d) $0.4$ m
e) $t=\frac{\pi}{2}$ , $s\approx 0.765$ m, the maximum

a)

First round, $u=\cos t$ , $dv=e^{-t/2}dt$ , $v=-2e^{-t/2}$ : $I=-2e^{-t/2}\cos t-2\int e^{-t/2}\sin t\,dt$ . Second round, again differentiating the trigonometric factor, $u=\sin t$ : $\int e^{-t/2}\sin t\,dt=-2e^{-t/2}\sin t+2I$ . Substituting back: $I=-2e^{-t/2}\cos t+4e^{-t/2}\sin t-4I$ , so $5I=e^{-t/2}(4\sin t-2\cos t)$ and $I=\frac{2}{5}e^{-t/2}\left(2\sin t-\cos t\right)+C$ .

b)

$s(0)=\frac{2}{5}(0-1)+C=0$ gives $C=\frac{2}{5}$ . So $s(t)=\frac{2}{5}e^{-t/2}\left(2\sin t-\cos t\right)+\frac{2}{5}$ .

c)

$s(\pi)=\frac{2}{5}e^{-\pi/2}(0+1)+\frac{2}{5}=0.4\left(e^{-\pi/2}+1\right)\approx 0.483$ m.

d)

$e^{-t/2}\to 0$ and $2\sin t-\cos t$ stays bounded by $\sqrt{5}$ : $s(t)\to\frac{2}{5}=0.4$ m. The equilibrium position is EXACTLY the constant of integration.

e)

$v(t)=0$ for the first time when $\cos t=0$ , that is $t=\frac{\pi}{2}\approx 1.57$ s. Then $s\left(\frac{\pi}{2}\right)=\frac{2}{5}e^{-\pi/4}\times 2+\frac{2}{5}\approx 0.765$ m. Before that, $v>0$ ; between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ , $v<0$ . This is the maximum: the object overshoots the equilibrium by $0.365$ m, then oscillates around it with an amplitude that shrinks like $e^{-t/2}$ . The figure in the solution shows the position and the velocity.

Exercise 15: Problem: the flood hydrograph of a river

After a storm, the flow of a river is modelled by $Q(t)=5+20\,t\,e^{-t/3}$ , in thousands of m $^{3}$ per hour, $t$ in hours since the start of the flood. The term $5$ is the base flow of the river.

a) Compute $\int t\,e^{-t/3}dt$ .
b) At what instant is the flow at its maximum? What is it?
c) Compute the volume of water that flows past during the first $12$ hours.
d) What is the average flow over these $12$ hours?
e) The volume DUE TO THE FLOOD is the part above the base flow. Compute it over $24$ hours, then over an unlimited time, and give the share that has flowed past within $24$ hours.

Show the solution

Answers

a) $-3(t+3)e^{-t/3}+C$
b) $t=3$ h, $Q\approx 27.07$
c) $240-900e^{-4}\approx 223.5$ thousand m $^{3}$
d) $\approx 18.63$ thousand m $^{3}$ per hour
e) $\approx 179.5$ over $24$ h, $180$ in total, $99.7$ %

a)

By parts, $u=t$ and $dv=e^{-t/3}dt$ , $v=-3e^{-t/3}$ : $\int te^{-t/3}dt=-3te^{-t/3}+3\int e^{-t/3}dt=-3te^{-t/3}-9e^{-t/3}+C=-3(t+3)e^{-t/3}+C$ . Check: $\left(-3(t+3)e^{-t/3}\right)'=-3e^{-t/3}+(t+3)e^{-t/3}=te^{-t/3}$ ✓.

b)

$Q'(t)=20e^{-t/3}\left(1-\frac{t}{3}\right)$ , positive before $t=3$ and negative after: the peak occurs at $t=3$ h, with $Q(3)=5+\frac{60}{e}\approx 27.07$ thousand m $^{3}$ per hour.

c)

$\int_{0}^{12}Q(t)dt=5\times 12+20\left[-3(t+3)e^{-t/3}\right]_{0}^{12}=60+20\left(-45e^{-4}+9\right)=240-900e^{-4}\approx 223.5$ thousand m $^{3}$ .

d)

Average flow $=\frac{1}{12}\int_{0}^{12}Q\approx\frac{223.52}{12}\approx 18.63$ thousand m $^{3}$ per hour, well above the base flow and well below the peak.

e)

Over $24$ hours: $20\left[-3(t+3)e^{-t/3}\right]_{0}^{24}=20\left(9-81e^{-8}\right)\approx 179.5$ thousand m $^{3}$ . Over an unlimited time, $(t+3)e^{-t/3}\to 0$ and the flood volume tends to $20\times 9=180$ thousand m $^{3}$ . Share that has flowed past within $24$ hours: $\frac{179.46}{180}\approx 99.7$ percent. The boundary term $-3(t+3)e^{-t/3}$ alone tells you how much water is still to come. The figure in the solution plots the hydrograph.

Corrected exercises: integration techniques (201-SN3-RE)

Course recap

Calculations: integrals by parts

Exercise 0.1: Integrating by parts: ten direct calculations

Calculations: trigonometric integrals

Exercise 0.2: Integrating trigonometric powers and products: ten direct calculations

Part A: Substitution, parts, partial fractions and trigonometry (/50)

Exercise 1: Antiderivatives and substitution

Exercise 2: Integration by parts: simple, repeated, cyclic

Exercise 3: Partial fractions: take it apart to integrate it

Exercise 4: Powers of tangent and secant

Exercise 5: Products of sines and cosines of different angles

Part B: Combined techniques, quadratics and strategy (/50)

Exercise 6: Powers of sine and cosine

Exercise 7: Trigonometric substitutions: the root becomes a triangle

Exercise 8: Strategy: choosing your technique in thirty seconds

Exercise 9: Combining techniques: substitution, then parts

Exercise 10: Completing the square and irreducible quadratics

Part C: the classics (/50)

Exercise 11: The arches of x sin x: areas that grow in arithmetic progression

Exercise 12: Two routes to the same integral: substitution or parts

Exercise 13: An integral with three poles: exact, improper, then forbidden

Exercise 14: Problem: the damped oscillator and its equilibrium position

Exercise 15: Problem: the flood hydrograph of a river

See also

Looking for an integral calculus tutor in Montreal?