Techniques of Integration: practice solutions, College – download the PDF

1 Parts with cosine ★★★

Take \(u=x\), \(dv=\cos x\,dx\), so \(du=dx\), \(v=\sin x\).

\( \int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x+C \).

2 Parts with sin 2x ★★★

Let \(u=x\), \(dv=\sin2x\,dx\); then \(du=dx\), \(v=-\tfrac12\cos2x\).

\( \int x\sin2x\,dx=-\tfrac x2\cos2x+\tfrac12\int\cos2x\,dx=-\tfrac x2\cos2x+\tfrac14\sin2x+C \).

3 Odd power of cosine ★★★

Write \( \cos^3x=(1-\sin^2x)\cos x \) and let \(u=\sin x\), \(du=\cos x\,dx\).

\( \int(1-u^2)du=u-\tfrac{u^3}3+C=\sin x-\tfrac{\sin^3x}{3}+C \).

4 Sine times a fourth power ★★★

Let \(u=\cos x\), so \(du=-\sin x\,dx\). Then \( -\int u^4du=-\dfrac{u^5}{5}+C=-\dfrac{\cos^5x}{5}+C \).

5 Simple partial fractions ★★★

\( \dfrac1{(x-2)(x+2)}=\dfrac A{x-2}+\dfrac B{x+2} \) with \(1=A(x+2)+B(x-2)\). At \(x=2\): \(A=\tfrac14\); at \(x=-2\): \(B=-\tfrac14\).

\( \int\dfrac{dx}{x^2-4}=\tfrac14\ln|x-2|-\tfrac14\ln|x+2|+C=\tfrac14\ln\Big|\dfrac{x-2}{x+2}\Big|+C \).

6 A first improper integral ★★★

\( \int_1^bx^{-3}dx=\Big[-\dfrac1{2x^2}\Big]_1^b=\tfrac12-\dfrac1{2b^2}\). As \(b\to\infty\) this tends to \( \tfrac12 \). The integral converges to \( \tfrac12 \) (p-test with \(p=3\)).

7 Pick the technique ★★★

1. \(u\)-substitution with \(u=x^2\), because \(2x\,dx=du\) is already present.
2. Integration by parts: a polynomial times an exponential.
3. Partial fractions: the denominator factors as \((x-3)(x+3)\).
4. The half-angle identity \( \sin^2x=\tfrac{1-\cos2x}2 \).

8 Double parts ★★★

First \(u=x^2\), \(dv=e^xdx\): \( \int x^2e^xdx=x^2e^x-2\int xe^xdx \). Second, \( \int xe^xdx=(x-1)e^x \). So an antiderivative is \( e^x(x^2-2x+2) \).

\( \int_0^1=e(1-2+2)-(1)(2)=e-2\approx0.718 \).

9 Root times logarithm ★★★

Take \(u=\ln x\), \(dv=x^{1/2}dx\), \(v=\tfrac23x^{3/2}\). Then \( \int\sqrt x\ln x\,dx=\tfrac23x^{3/2}\ln x-\tfrac23\int x^{1/2}dx=\tfrac23x^{3/2}\ln x-\tfrac49x^{3/2} \).

At 4: \( \tfrac{16}3\ln4-\tfrac{32}9 \). At 1: \( -\tfrac49 \). Difference: \( \tfrac{16}3\ln4-\tfrac{28}9\approx4.282 \).

10 Even powers ★★★

\( \sin x\cos x=\tfrac12\sin2x \), so \( \sin^2x\cos^2x=\tfrac14\sin^22x=\dfrac{1-\cos4x}{8} \).

\( \int=\dfrac x8-\dfrac{\sin4x}{32}+C \).

11 Area of a quarter disk ★★★

Let \(x=2\sin\theta\), \(dx=2\cos\theta\,d\theta\); \(x=0\to\theta=0\), \(x=2\to\theta=\tfrac\pi2\). The integrand is \(2\cos\theta\), so the integral is \( \int_0^{\pi/2}4\cos^2\theta\,d\theta=2\Big[\theta+\sin\theta\cos\theta\Big]_0^{\pi/2}=\pi \).

Geometrically this is the area of a quarter of a disk of radius 2: \( \tfrac14\pi\cdot2^2=\pi \).

12 Two linear factors ★★★

\(x^2+x-2=(x-1)(x+2)\). Write \(2x+3=A(x+2)+B(x-1)\). At \(x=1\): \(5=3A\), \(A=\tfrac53\). At \(x=-2\): \(-1=-3B\), \(B=\tfrac13\).

\( \int=\tfrac53\ln|x-1|+\tfrac13\ln|x+2|+C \).

13 An exponential tail ★★★

By parts, \(u=x\), \(dv=e^{-3x}dx\): \( \int xe^{-3x}dx=-\tfrac x3e^{-3x}-\tfrac19e^{-3x} \). At \(b\to\infty\) both terms tend to 0; at 0 the value is \( -\tfrac19 \).

So the integral equals \(0-(-\tfrac19)=\tfrac19\).

14 Integral of ln x near zero ★★★

Antiderivative: \(x\ln x-x\). So the integral is \( \lim_{t\to0^+}\big[x\ln x-x\big]_t^1=(0-1)-\lim_{t\to0^+}(t\ln t-t)\). Since \( t\ln t\to0 \), the limit is \(0\) and the integral equals \(-1\).

15 Drug exposure ★★★

By parts, with \(u=8t\), \(dv=e^{-0.5t}dt\), \(v=-2e^{-0.5t}\): \( \int8te^{-0.5t}dt=-16te^{-0.5t}+16\int e^{-0.5t}dt=-16te^{-0.5t}-32e^{-0.5t} \).

At \(\infty\) the value is 0 and at 0 it is \(-32\). The total exposure is \(32\) mg·h/L.

16 A cycle of exponential and cosine ★★★

Let \(I=\int e^{2x}\cos x\,dx\) and \(J=\int e^{2x}\sin x\,dx\). Parts gives \( I=\tfrac12e^{2x}\cos x+\tfrac12J \) and \( J=\tfrac12e^{2x}\sin x-\tfrac12I \).

Substitute: \( I=\tfrac12e^{2x}\cos x+\tfrac14e^{2x}\sin x-\tfrac14I \), so \( \tfrac54I=e^{2x}\big(\tfrac12\cos x+\tfrac14\sin x\big) \).

\( I=\dfrac{e^{2x}(2\cos x+\sin x)}{5}+C \).

17 Tangent substitution with a square ★★★

Let \(x=3\tan\theta\), \(dx=3\sec^2\theta\,d\theta\), \( \sqrt{x^2+9}=3\sec\theta \). Then \( \int\dfrac{3\sec^2\theta}{9\tan^2\theta\cdot3\sec\theta}d\theta=\dfrac19\int\dfrac{\cos\theta}{\sin^2\theta}d\theta=-\dfrac1{9\sin\theta}+C \).

From the triangle, \( \sin\theta=\dfrac{x}{\sqrt{x^2+9}} \), so \( \int=-\dfrac{\sqrt{x^2+9}}{9x}+C \).

18 Substitution then parts ★★★

Let \(u=x^2\), \(du=2x\,dx\): \( \int x^3e^{x^2}dx=\tfrac12\int ue^udu \). By parts, \( \int ue^udu=(u-1)e^u \).

So \( \int x^3e^{x^2}dx=\tfrac12(x^2-1)e^{x^2}+C \).

19 Partial fractions and a limit ★★★

Decompose: \( \dfrac1{x(x^2+1)}=\dfrac1x-\dfrac{x}{x^2+1} \). An antiderivative is \( \ln x-\tfrac12\ln(x^2+1)=\ln\dfrac{x}{\sqrt{x^2+1}} \).

As \(b\to\infty\), \( \dfrac b{\sqrt{b^2+1}}\to1 \), so the value at \(b\) tends to \(\ln1=0\). At 1 it is \( \ln\dfrac1{\sqrt2}=-\tfrac12\ln2 \).

The integral converges to \( \tfrac12\ln2\approx0.347 \).

20 Estimating pi ★★★

\(h=0.25\). Values: \(f(0)=4\), \(f(0.25)=3.7647\), \(f(0.5)=3.2\), \(f(0.75)=2.56\), \(f(1)=2\).

\( T_4=0.125\big[4+2(3.7647+3.2+2.56)+2\big]\approx3.1312 \) (error about \(0.0104\)).

\( S_4=\dfrac{0.25}{3}\big[4+4(3.7647)+2(3.2)+4(2.56)+2\big]\approx3.14157 \) (error about \(0.00002\)).

Simpson’s rule is more than 400 times more accurate here.

21 Telescoping improper integral ★★★

\( \dfrac1{x(x+1)}=\dfrac1x-\dfrac1{x+1} \), so an antiderivative is \( \ln\dfrac x{x+1} \). At \(\infty\) it tends to \(\ln1=0\); at 1 it is \(\ln\tfrac12\). The integral equals \( 0-\ln\tfrac12=\ln2\approx0.693 \).

For \(x\ge1\), \( x^2+x+1\gt x(x+1) \), so \( 0\lt\dfrac1{x^2+x+1}\lt\dfrac1{x(x+1)} \). By comparison, the second integral converges (and is less than \(\ln2\)).

22 A slow divergence ★★★

Let \(u=\ln x\), \(du=\dfrac{dx}x\): \( \int_2^b\dfrac{dx}{x\ln x}=\Big[\ln(\ln x)\Big]_2^b=\ln(\ln b)-\ln(\ln2) \).

As \(b\to\infty\), \( \ln b\to\infty \) and so does \( \ln(\ln b) \). The integral diverges, although very slowly: even for \(b=e^{1000}\) the value is only about 7.27.