Topic 8 Techniques of Integration

8.1 Integration by Parts

Exercise 8.1 Evaluate the integral.

∫xcos5xdx$$\int x\cos 5xdx$$

Solution.

∫xcos5xdx=15∫xd(sin5x)=15(xsin5x−∫sin5xdx)=15(xsin5x+15cos5x)+C=15xsin5x+125cos5x+C.$$\begin{array}{cc}& \int x\cos 5xdx\\ =& \frac{1}{5}\int xd\left(\sin 5x\right)\\ =& \frac{1}{5}(x\sin 5x-\int \sin 5xdx)\\ =& \frac{1}{5}(x\sin 5x+\frac{1}{5}\cos 5x)+C\\ =& \frac{1}{5}x\sin 5x+\frac{1}{25}\cos 5x+C.\end{array}$$

Exercise 8.2 Evaluate the integral.

∫te−3tdt$$\int t{e}^{-3t}dt$$

Solution.

∫te−3tdt=−13∫td(e−3t)=−13(te−3t−∫e−3tdt)=−13(te−3t+13e−3t)+C=−13te−3t−19e−3t+C$$\begin{array}{cc}& \int t{e}^{-3t}dt\\ =& -\frac{1}{3}\int td\left(e^{-3t}\right)\\ =& -\frac{1}{3}(t{e}^{-3t}-\int e^{-3t}dt)\\ =& -\frac{1}{3}(t{e}^{-3t}+\frac{1}{3}{e}^{-3t})+C\\ =& -\frac{1}{3}t{e}^{-3t}-\frac{1}{9}{e}^{-3t}+C\end{array}$$

Exercise 8.3 Evaluate the integral.

∫sin−1xdx$$\int {\sin }^{-1}xdx$$

Solution.

∫sin−1xdx=xsin−1x−∫xd(sin−1x)=xsin−1x−∫x√1−x2dx=xsin−1x+12∫u−1/2duu=1−x2=xsin−1x+√u+C=xsin−1x+√1−x2+C.$$\begin{array}{cc}& \int {\sin }^{-1}xdx\\ =& x{\sin }^{-1}x-\int xd\left({\sin }^{-1}x\right)\\ =& x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx\\ =& x{\sin }^{-1}x+\frac{1}{2}\int u^{-1/2}duu=1-x^2\\ =& x{\sin }^{-1}x+\sqrt{u}+C\\ =& x{\sin }^{-1}x+\sqrt{1-x^2}+C.\end{array}$$

Exercise 8.4 Evaluate the integral.

∫arctan4tdt$$\int \arctan 4tdt$$

Solution.

∫arctan4tdt=tarctan4t−∫td(arctan4t)=tarctan4t−∫4t1+(4t)2dt=tarctan4t−18∫1uduu=1+(4t)2=tarctan4t−18ln|u|+C=tarctan4t−18ln(1+16t2)+C$$\begin{array}{cc}& \int \arctan 4tdt\\ =& t\arctan 4t-\int td\left(\arctan 4t\right)\\ =& t\arctan 4t-\int \frac{4t}{1+(4t{)}^2}dt\\ =& t\arctan 4t-\frac{1}{8}\int \frac{1}{u}duu=1+(4t{)}^2\\ =& t\arctan 4t-\frac{1}{8}\ln |u|+C\\ =& t\arctan 4t-\frac{1}{8}\ln (1+16t^2)+C\end{array}$$

Exercise 8.5 Evaluate the integral.

∫e−θcos2θdθ$$\int e^{-\theta }\cos 2\theta d\theta$$

Solution.

Let I=∫e−θcos2θdθ$I=\int e^{-\theta }\cos 2\theta d\theta$.

I=−∫cos2θd(e−θ)=−(e−θcos2θ−∫e−θd(cos2θ))=−e−θcos2θ−2∫e−θsin2θdθ=−e−θcos2θ+2(∫sin2θd(e−θ))=−e−θcos2θ+2(e−θsin2θ−∫e−θd(sin2θ))=−e−θcos2θ+2(e−θsin2θ−2∫e−θcos2θdθ)=−e−θcos2θ+2e−θsin2θ−4I.$$\begin{array}{cc}I=& -\int \cos 2\theta d\left(e^{-\theta }\right)\\ =& -(e^{-\theta }\cos 2\theta -\int e^{-\theta }d(\cos 2\theta \left)\right)\\ =& -e^{-\theta }\cos 2\theta -2\int e^{-\theta }\sin 2\theta d\theta \\ =& -e^{-\theta }\cos 2\theta +2(\int \sin 2\theta d(e^{-\theta }\left)\right)\\ =& -e^{-\theta }\cos 2\theta +2(e^{-\theta }\sin 2\theta -\int e^{-\theta }d(\sin 2\theta \left)\right)\\ =& -e^{-\theta }\cos 2\theta +2(e^{-\theta }\sin 2\theta -2\int e^{-\theta }\cos 2\theta d\theta )\\ =& -e^{-\theta }\cos 2\theta +2e^{-\theta }\sin 2\theta -4I.\end{array}$$

Solve for I$I$, we get I=∫e−θcos2θdθ=15e−θ(2sin2θ−cos2θ)+C.$$I=\int e^{-\theta }\cos 2\theta d\theta =\frac{1}{5}{e}^{-\theta }(2\sin 2\theta -\cos 2\theta )+C.$$

Exercise 8.6 Evaluate the integral.

∫10(x2+1)e−xdx$${\int }_0^1(x^2+1)e^{-x}dx$$

Solution.

We first find an antiderivative. ∫(x2+1)e−xdx=−∫(x2+1)d(e−x)=−((x2+1)e−x−∫e−xd(x2+1))=−(x2+1)e−x+∫2xe−xdx=−(x2+1)e−x−∫2xd(e−x)=−(x2+1)e−x−(2xe−x−∫e−xd(2x))=−(x2+1)e−x−2xe−x+2∫e−xdx=−(x2+1)e−x−2xe−x−2e−x+C=−e−x(x2+2x+3)+C.$$\begin{array}{cc}& \int (x^2+1)e^{-x}dx\\ =& -\int (x^2+1)d\left(e^{-x}\right)\\ =& -\left(\right(x^2+1)e^{-x}-\int e^{-x}d(x^2+1\left)\right)\\ =& -(x^2+1)e^{-x}+\int 2x{e}^{-x}dx\\ =& -(x^2+1)e^{-x}-\int 2xd\left(e^{-x}\right)\\ =& -(x^2+1)e^{-x}-(2x{e}^{-x}-\int e^{-x}d(2x\left)\right)\\ =& -(x^2+1)e^{-x}-2x{e}^{-x}+2\int e^{-x}dx\\ =& -(x^2+1)e^{-x}-2x{e}^{-x}-2e^{-x}+C\\ =& -e^{-x}(x^2+2x+3)+C.\end{array}$$

By FTC, ∫10(x2+1)e−xdx=−(e−1(12+2⋅1+3)−3e0)=3−6e$$\begin{array}{cc}& {\int }_0^1(x^2+1)e^{-x}dx\\ =& -\left(e^{-1}\right(1^2+2\cdot 1+3)-3e^0)\\ =& 3-\frac{6}{e}\end{array}$$

Exercise 8.7 Evaluate the integral.

∫1/20cos−1xdx$${\int }_0^{1/2}{\cos }^{-1}xdx$$

Solution.

Find the indefinite integral first.

∫cos−1xdx=xcos−1x−∫xd(cos−1x)=xcos−1x+∫x√1−x2dx=xcos−1x−∫12√uduu=1−x2=xcos−1x−√u+C=xcos−1x−√1−x2+C$$\begin{array}{cc}& \int {\cos }^{-1}xdx\\ =& x{\cos }^{-1}x-\int xd\left({\cos }^{-1}x\right)\\ =& x{\cos }^{-1}x+\int \frac{x}{\sqrt{1-x^2}}dx\\ =& x{\cos }^{-1}x-\int \frac{1}{2\sqrt{u}}duu=1-x^2\\ =& x{\cos }^{-1}x-\sqrt{u}+C\\ =& x{\cos }^{-1}x-\sqrt{1-x^2}+C\end{array}$$

By FTC, ∫1/20cos−1xdx=((1/2)cos−1(1/2)−√1−(1/2)2)−(0⋅cos−10−√1−02)=π6−√32+1$$\begin{array}{cc}& {\int }_0^{1/2}{\cos }^{-1}xdx\\ =& \left(\right(1/2\left){\cos }^{-1}\right(1/2)-\sqrt{1-(1/2{)}^2})-(0\cdot {\cos }^{-1}0-\sqrt{1-0^2})\\ =& \frac{\pi }{6}-\frac{\sqrt{3}}{2}+1\end{array}$$

Exercise 8.8 Evaluate the integral.

∫cos√xdx$$\int \cos \sqrt{x}dx$$

Solution.

Apply the substitution method first and then the integration by parts.

∫cos√xdx=2∫ucosuduu=√x=2∫ud(sinu)=2(usinu−∫sinudu)=2(usinu+cosu)+C=2(√xsin√x+cos√x)+C$$\begin{array}{cc}& \int \cos \sqrt{x}dx\\ =& 2\int u\cos uduu=\sqrt{x}\\ =& 2\int ud\left(\sin u\right)\\ =& 2(u\sin u-\int \sin udu)\\ =& 2(u\sin u+\cos u)+C\\ =& 2(\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x})+C\end{array}$$

Exercise 8.9 Evaluate the integral.

∫xln(1+x)dx$$\int x\ln (1+x)dx$$

Solution.

Apply the substitution method first and then the integration by parts.

∫xln(1+x)dx=∫(u−1)lnuduu=x+1=∫ulnudu−∫lnudu=12∫lnudu2−(ulnu−∫ud(lnu))=12(u2lnu−∫u2d(lnu))−(ulnu−∫du)=12(u2lnu−∫udu)−(ulnu−u)+C=12(u2lnu−12u2)−(ulnu−u)+C=12(u2−2u)lnu+u−14u2+C=12(x2−1)ln(x+1)−14(x2−2x−3)+C.$$\begin{array}{cc}& \int x\ln (1+x)dx\\ =& \int (u-1)\ln uduu=x+1\\ =& \int u\ln udu-\int \ln udu\\ =& \frac{1}{2}\int \ln ud{u}^2-(u\ln u-\int ud(\ln u\left)\right)\\ =& \frac{1}{2}(u^2\ln u-\int u^{2}d(\ln u\left)\right)-(u\ln u-\int du)\\ =& \frac{1}{2}(u^2\ln u-\int udu)-(u\ln u-u)+C\\ =& \frac{1}{2}(u^2\ln u-\frac{1}{2}{u}^2)-(u\ln u-u)+C\\ =& \frac{1}{2}(u^2-2u)\ln u+u-\frac{1}{4}{u}^2+C\\ =& \frac{1}{2}(x^2-1)\ln (x+1)-\frac{1}{4}(x^2-2x-3)+C.\end{array}$$

Exercise 8.10 Prove the reduction formula

∫cosnxdx=1ncosn−1xsinx+n−1n∫cosn−2xdx$$\int {\cos }^{n}xdx=\frac{1}{n}{\cos }^{n-1}x\sin x+\frac{n-1}{n}\int {\cos }^{n-2}xdx$$

Solution.

Let f(x)=cosn−1x$f\left(x\right)={\cos }^{n-1}x$ and g(x)=sinx$g\left(x\right)=\sin x$. Then cosnx=f(x)g′(x)dx${\cos }^{n}x=f\left(x\right)g^{\prime }\left(x\right)dx$. We apply the integration by parts to the left-hand side. I=∫cosnxdx=∫cosn−1xd(sinx)=cosn−1xsinx−∫sinxd(cosn−1x)=cosn−1xsinx+(n−1)∫sin2xcosn−2xdx=cosn−1xsinx+(n−1)∫(1−cos2x)cosn−2xdx=cosn−1xsinx+(n−1)∫cosn−2xdx−(n−1)∫cosnxdx=cosn−1xsinx+(n−1)∫cosn−2xdx−(n−1)I$$\begin{array}{cc}I=& \int {\cos }^{n}xdx\\ =& \int {\cos }^{n-1}xd\left(\sin x\right)\\ =& {\cos }^{n-1}x\sin x-\int \sin xd\left({\cos }^{n-1}x\right)\\ =& {\cos }^{n-1}x\sin x+(n-1)\int {\sin }^{2}x{\cos }^{n-2}xdx\\ =& {\cos }^{n-1}x\sin x+(n-1)\int (1-{\cos }^{2}x){\cos }^{n-2}xdx\\ =& {\cos }^{n-1}x\sin x+(n-1)\int {\cos }^{n-2}xdx-(n-1)\int {\cos }^{n}xdx\\ =& {\cos }^{n-1}x\sin x+(n-1)\int {\cos }^{n-2}xdx-(n-1)I\end{array}$$

Solve for I$I$, we get I=1ncosn−1xsinx+n−1n∫cosn−2xdx.$$I=\frac{1}{n}{\cos }^{n-1}x\sin x+\frac{n-1}{n}\int {\cos }^{n-2}xdx.$$

Therefor, ∫cosnxdx=1ncosn−1xsinx+n−1n∫cosn−2xdx.$$\int {\cos }^{n}xdx=\frac{1}{n}{\cos }^{n-1}x\sin x+\frac{n-1}{n}\int {\cos }^{n-2}xdx.$$

Exercise 8.11 Evaluate the integral.

∫sin2xcos3xdx$$\int {\sin }^{2}x{\cos }^{3}xdx$$

Solution.

Since not both cosx$\cos x$ has an odd power, we may use the substitution t=sinx$t=\sin x$ and the Pythagorean identity sin2x+cos2=1${\sin }^{2}x+{\cos }^2=1$ to change the integrand into a polynomial of t$t$. ∫sin2xcos3xdx=∫sin2xcos2xd(sinx)=∫t2(1−t2)dtt=sinx=∫t2−t4dt=t33−t55+C=sin3x3−sin5x5+C$$\begin{array}{cc}& \int {\sin }^{2}x{\cos }^{3}xdx\\ =& \int {\sin }^{2}x{\cos }^{2}xd\left(\sin x\right)\\ =& \int t^2(1-t^2)dtt=\sin x\\ =& \int t^2-t^{4}dt\\ =& \frac{t^3}{3}-\frac{t^5}{5}+C\\ =& \frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+C\end{array}$$

Exercise 8.12 Evaluate the integral.

∫2π0sin2(13x)dx$${\int }_0^{2\pi }{\sin }^2\left(\frac{1}{3}x\right)dx$$

Solution.

Again, we first find the indefinite integral.

We use the double angle formula to reduce the power of the trigonometric function in the integrand.

∫sin2(13x)dx=12∫(1−cos(23x))dx=x2−12∫cos(23x)dx=x2−34∫cos(t)dtt=23x=x2−34sint+C=x2−34sin(23x)+C$$\begin{array}{cc}& \int {\sin }^2\left(\frac{1}{3}x\right)dx\\ =& \frac{1}{2}\int (1-\cos \left(\frac{2}{3}x\right))dx\\ =& \frac{x}{2}-\frac{1}{2}\int \cos \left(\frac{2}{3}x\right)dx\\ =& \frac{x}{2}-\frac{3}{4}\int \cos \left(t\right)dtt=\frac{2}{3}x\\ =& \frac{x}{2}-\frac{3}{4}\sin t+C\\ =& \frac{x}{2}-\frac{3}{4}\sin \left(\frac{2}{3}x\right)+C\end{array}$$

By FTC, ∫2π0sin2(13x)dx=π−34sin(4π3)=π+3√38$${\int }_0^{2\pi }{\sin }^2\left(\frac{1}{3}x\right)dx=\pi -\frac{3}{4}\sin \left(\frac{4\pi }{3}\right)=\pi +\frac{3\sqrt{3}}{8}$$

`

Exercise 8.13 Evaluate the integral.

∫sin2xcos4xdx$$\int {\sin }^{2}x{\cos }^{4}xdx$$

Solution.

We use trigonometric identities to reduce the power in the integrand. sin2xcos4x=(sinxcosx)2cos2x=14(sin(2x)cosx)2=116(sin(3x)+sinx)2=116(sin2(3x)+2sin3xsinx+sin2x)=116(12(1−cos6x)+cos(2x)−cos(4x)+12(1−cos2x))=116−132cos6x+132cos(2x)−116cos(4x)$$\begin{array}{cc}& {\sin }^{2}x{\cos }^{4}x\\ =& (\sin x\cos x{)}^2{\cos }^{2}x\\ =& \frac{1}{4}\left(\sin \right(2x)\cos x{)}^2\\ =& \frac{1}{16}\left(\sin \right(3x)+\sin x{)}^2\\ =& \frac{1}{16}\left({\sin }^2\right(3x)+2\sin 3x\sin x+{\sin }^{2}x)\\ =& \frac{1}{16}\left(\frac{1}{2}\right(1-\cos 6x)+\cos (2x)-\cos (4x)+\frac{1}{2}(1-\cos 2x\left)\right)\\ =& \frac{1}{16}-\frac{1}{32}\cos 6x+\frac{1}{32}\cos \left(2x\right)-\frac{1}{16}\cos \left(4x\right)\end{array}$$

Then the integral is ∫sin2xcos4xdx=∫(116−132cos6x+132cos(2x)−116cos(4x))dx=x16+164sin2x−164sin4x−1192sin6x+C$$\begin{array}{cc}& \int {\sin }^{2}x{\cos }^{4}xdx\\ =& \int (\frac{1}{16}-\frac{1}{32}\cos 6x+\frac{1}{32}\cos (2x)-\frac{1}{16}\cos (4x\left)\right)dx\\ =& \frac{x}{16}+\frac{1}{64}\sin 2x-\frac{1}{64}\sin 4x-\frac{1}{192}\sin 6x+C\end{array}$$

Remark: Note that sin2xcos4x=(1−cos2x)cos4x=cos4x−cos6x${\sin }^{2}x{\cos }^{4}x=(1-{\cos }^{2}x){\cos }^{4}x={\cos }^{4}x-{\cos }^{6}x$. So we may evaluate the integral using the formula (n>0$n>0$) ∫cosnxdx=1nsinxcosn−1x−n−1n∫cosn−2xdx$$\int {\cos }^{n}xdx=\frac{1}{n}\sin x{\cos }^{n-1}x-\frac{n-1}{n}\int {\cos }^{n-2}xdx$$ which is obtain via integration by parts.

Exercise 8.14 Evaluate the integral.

∫tanxsec3xdx$$\int \tan x{\sec }^{3}xdx$$

Solution.

Note that d(secx)=(tanxsecx)dx$d\left(\sec x\right)=\left(\tan x\sec x\right)dx$. Let t=secx$t=\sec x$. Then ∫tanxsec3xdx=∫t2dtt=secx=t33+C=sec3x3+C$$\begin{array}{cc}\int \tan x{\sec }^{3}xdx=& \int t^{2}dtt=\sec x\\ =& \frac{t^3}{3}+C\\ =& \frac{{\sec }^{3}x}{3}+C\end{array}$$

Exercise 8.15 Evaluate the integral.

∫π/40sec4xtan4xdx$${\int }_0^{\pi /4}{\sec }^{4}x{\tan }^{4}xdx$$

Solution.

Let t=tanx$t=\tan x$. Then dt=sec2xdx$dt={\sec }^{2}xdx$. Using the Pythagorean identity 1+tan2x=sec2x$1+{\tan }^{2}x={\sec }^{2}x$, we get ∫sec4xtan4xdx=∫(1+t2)t2dt=∫(t2+t4)dt=t33+t44+C=tan3x3+tan4x4+C$$\begin{array}{cc}& \int {\sec }^{4}x{\tan }^{4}xdx\\ =& \int (1+t^2)t^{2}dt\\ =& \int (t^2+t^4)dt\\ =& \frac{t^3}{3}+\frac{t^4}{4}+C\\ =& \frac{{\tan }^{3}x}{3}+\frac{{\tan }^{4}x}{4}+C\end{array}$$

By FTC, ∫π/40sec4xtan4xdx=tan3(π/4)3+tan4(π/4)4=13+14=712.$${\int }_0^{\pi /4}{\sec }^{4}x{\tan }^{4}xdx=\frac{{\tan }^3\left(\pi /4\right)}{3}+\frac{{\tan }^4\left(\pi /4\right)}{4}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}.$$

Exercise 8.16 Evaluate the integral.

∫tan5xdx$$\int {\tan }^{5}xdx$$

Solution.

One way to evaluate the integral is to rewrite tan5x=sin5xcos5x${\tan }^{5}x=\frac{{\sin }^{5}x}{{\cos }^{5}x}$ and then substitute t=cosx$t=\cos x$.

Another way is to use the identity tan2x=1−sec2x${\tan }^{2}x=1-{\sec }^{2}x$ and ddxtanx=sec2$\frac{d}{dx}\tan x={\sec }^2$ to deduce a reduction formula. ∫tan5xdx=∫tan3x(sec2x−1)dx=∫(tan3xsec2x−tan3x)dx=∫tan3xsec2xdx−∫tan3xdx=∫t3dt−∫tan3xdxt=tanx=t44−∫tanx(sec2x−1)dx=tan4x4−∫tanxsec2xdx+∫tanxdx=tan4x4−tan2x2−ln|cosx|+C.$$\begin{array}{cc}& \int {\tan }^{5}xdx\\ =& \int {\tan }^{3}x({\sec }^{2}x-1)dx\\ =& \int ({\tan }^{3}x{\sec }^{2}x-{\tan }^{3}x)dx\\ =& \int {\tan }^{3}x{\sec }^{2}xdx-\int {\tan }^{3}xdx\\ =& \int t^{3}dt-\int {\tan }^{3}xdxt=\tan x\\ =& \frac{t^4}{4}-\int \tan x({\sec }^{2}x-1)dx\\ =& \frac{{\tan }^{4}x}{4}-\int \tan x{\sec }^{2}xdx+\int \tan xdx\\ =& \frac{{\tan }^{4}x}{4}-\frac{{\tan }^{2}x}{2}-\ln |\cos x|+C.\end{array}$$

Remark: In general (n≠1$n\ne 1$), we have ∫tannxdx=1n−1tann−1x−∫tann−2xdx$$\int {\tan }^{n}xdx=\frac{1}{n-1}{\tan }^{n-1}x-\int {\tan }^{n-2}xdx$$

Exercise 8.17 Evaluate the integral.

∫xsecxtanxdx$$\int x\sec x\tan xdx$$

Solution.

Let f(x)=x$f\left(x\right)=x$ and g′(x)=secxtanx$g^{\prime }\left(x\right)=\sec x\tan x$. Then g(x)=secx$g\left(x\right)=\sec x$. By integration by parts, we have ∫xsecxtanxdx=xsecx−∫secxdx=xsecx−ln|secx+tanx|+C.$$\begin{array}{cc}& \int x\sec x\tan xdx\\ =& x\sec x-\int \sec xdx\\ =& x\sec x-\ln |\sec x+\tan x|+C.\end{array}$$

Exercise 8.18 Evaluate the integral.

∫sin5xsinxdx$$\int \sin 5x\sin xdx$$

Solution.

Using the sum of angles formula for cosine, we have ∫sin5xsinxdx=12∫(cos4x−cos6x)dx=18sin4x−112sin6x+C.$$\begin{array}{cc}& \int \sin 5x\sin xdx\\ =& \frac{1}{2}\int (\cos 4x-\cos 6x)dx\\ =& \frac{1}{8}\sin 4x-\frac{1}{12}\sin 6x+C.\end{array}$$

8.2 Trigonometric Substitution

When having √a2−b2x2$\sqrt{a^2-b^2{x}^2}$, √a2+b2x2$\sqrt{a^2+b^2{x}^2}$ or √b2x2−a2$\sqrt{b^2{x}^2-a^2}$, we consider trigonometric substitution.

Assume that a$a$ and b$b$ are both positive. Then we have the following table of substitutions and transformations.

Expression	Substitution	Transformation
√a2−b2x2$\sqrt{a^2-b^2{x}^2}$	x=absinθ,−π2≤θ≤π2$x=\frac{a}{b}\sin \theta ,-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$	√a2−b2x2=acos(θ)$\sqrt{a^2-b^2{x}^2}=a\cos \left(\theta \right)$
√a2+b2x2$\sqrt{a^2+b^2{x}^2}$	x=abtanθ,−π2<θ<π2$x=\frac{a}{b}\tan \theta ,-\frac{\pi }{2}<\theta <\frac{\pi }{2}$	√a2+b2x2=asec(θ)$\sqrt{a^2+b^2{x}^2}=a\sec \left(\theta \right)$
√b2x2−a2$\sqrt{b^2{x}^2-a^2}$	x=absecθ,0≤θ<π2 or π≤θ<3π2$x=\frac{a}{b}\sec \theta ,0\le \theta <\frac{\pi }{2}\text{or}\pi \le \theta <\frac{3\pi }{2}$	√b2x2−a2=atan(θ)$\sqrt{b^2{x}^2-a^2}=a\tan \left(\theta \right)$

Exercise 8.19 Evaluate the integral.

∫x2√9−x2dx$$\int \frac{x^2}{\sqrt{9-x^2}}dx$$

Solution.

Let x=3sint$x=3\sin t$ with −π/2<t<π/2$-\pi /2<t<\pi /2$.

Then ∫x2√9−x2dx=∫9sin2t3cost3costdt=∫9sin2tdt=92∫(1−cos2t)dt=92t−94sin2t+C=92arcsin(x3)−12x√9−x2+C$$\begin{array}{cc}& \int \frac{x^2}{\sqrt{9-x^2}}dx\\ =& \int \frac{9{\sin }^{2}t}{3\cos t}3\cos tdt\\ =& \int 9{\sin }^{2}tdt\\ =& \frac{9}{2}\int (1-\cos 2t)dt\\ =& \frac{9}{2}t-\frac{9}{4}\sin 2t+C\\ =& \frac{9}{2}\arcsin \left(\frac{x}{3}\right)-\frac{1}{2}x\sqrt{9-x^2}+C\end{array}$$

Exercise 8.20 Evaluate the integral.

∫a0dx(a2+x2)3/2,a>0$${\int }_0^a\frac{dx}{{(a^2+x^2)}^{3/2}},a>0$$

Solution.

Let x=atant$x=a\tan t$ with −π/2<t<π/2$-\pi /2<t<\pi /2$. Then dx=asec2tdt$dx=a{\sec }^{2}tdt$ and (a2+x2)3/2=(asect)3$(a^2+x^2{)}^{3/2}=(a\sec t{)}^3$.

Then ∫dx(a2+x2)3/2=∫asec2tdta3sec3t=∫costdta2=sinta2+C=1a2√x2a2+x2+C$$\begin{array}{cc}& \int \frac{dx}{{(a^2+x^2)}^{3/2}}\\ =& \int \frac{a{\sec }^{2}tdt}{a^3{\sec }^{3}t}\\ =& \int \frac{\cos tdt}{a^2}\\ =& \frac{\sin t}{a^2}+C\\ =& \frac{1}{a^2}\sqrt{\frac{x^2}{a^2+x^2}}+C\end{array}$$

Remark: Since tant=xa$\tan t=\frac{x}{a}$, then sec2t=1+x2a2${\sec }^{2}t=1+\frac{x^2}{a^2}$ and cos2t=a2a2+x2${\cos }^{2}t=\frac{a^2}{a^2+x^2}$. So sint=√1−cos2t=√x2a2+x2$\sin t=\sqrt{1-{\cos }^{2}t}=\sqrt{\frac{x^2}{a^2+x^2}}$.

By FTC, ∫a0dx(a2+x2)3/2=1a2√a2a2+a2=√22a2.$${\int }_0^a\frac{dx}{{(a^2+x^2)}^{3/2}}=\frac{1}{a^2}\sqrt{\frac{a^2}{a^2+a^2}}=\frac{\sqrt{2}}{2a^2}.$$

Exercise 8.21 Evaluate the integral.

∫32dx(x2−1)3/2$${\int }_2^3\frac{dx}{{(x^2-1)}^{3/2}}$$

Solution.

Let x=sect$x=\sec t$ with 0<t<π/2$0<t<\pi /2$ or π<t<3π/2$\pi <t<3\pi /2$. Then dx=secttantdt$dx=\sec t\tan tdt$ and (x2−1)3/2=tan3t$(x^2-1{)}^{3/2}={\tan }^{3}t$.

Then ∫dx(x2−1)3/2=∫secttantdttan3t=∫costdtsin2t=−1sint+C=x√x2−1+C$$\begin{array}{cc}& \int \frac{dx}{{(x^2-1)}^{3/2}}\\ =& \int \frac{\sec t\tan tdt}{{\tan }^{3}t}\\ =& \int \frac{\cos tdt}{{\sin }^{2}t}\\ =& -\frac{1}{\sin t}+C\\ =& \frac{x}{\sqrt{x^2-1}}+C\end{array}$$

Remark: Here we used the identities sint=√1−cos2t=√1−1sec2t=√1−1x2$\sin t=\sqrt{1-{\cos }^{2}t}=\sqrt{1-\frac{1}{{\sec }^{2}t}}=\sqrt{1-\frac{1}{x^2}}$.

By FTC, ∫32dx(x2−1)3/2=3√92−1−2√22−1=8√3−9√212.$${\int }_2^3\frac{dx}{{(x^2-1)}^{3/2}}=\frac{3}{\sqrt{9^2-1}}-\frac{2}{\sqrt{2^2-1}}=\frac{8\sqrt{3}-9\sqrt{2}}{12}.$$

Exercise 8.22 Evaluate the integral.

∫dx√x2+2x+5$$\int \frac{dx}{\sqrt{x^2+2x+5}}$$

Solution.

By completing the square, we may rewrite x2+2x+5=1+(x+1)2$x^2+2x+5=1+(x+1{)}^2$. Let x+1=sect$x+1=\sec t$. Then dx=secttantdt$dx=\sec t\tan tdt$ and √x2+2x+5=tant$\sqrt{x^2+2x+5}=\tan t$. Therefore, ∫dx√x2+2x+5=∫secttantdttant=∫sectdt=ln|sect+tant|+C=ln|x+1+√1−(x+1)2|+C$$\begin{array}{cc}& \int \frac{dx}{\sqrt{x^2+2x+5}}\\ =& \int \frac{\sec t\tan tdt}{\tan t}\\ =& \int \sec tdt\\ =& \ln |\sec t+\tan t|+C\\ =& \ln |x+1+\sqrt{1-(x+1{)}^2}|+C\end{array}$$

Exercise 8.23 Evaluate the integral.

∫x√1−x4dx$$\int x\sqrt{1-x^4}dx$$

Solution.

Let x2=sint$x^2=\sin t$, then 2xdx=costdt$2xdx=\cos tdt$ and √1−x4=cost$\sqrt{1-x^4}=\cos t$. Then ∫x√1−x4dx=12∫cos2tdt=14∫(1+cos2t)dt=14t+18sin2t+C=14arcsin(x2)+12x2√1−x4+C$$\begin{array}{cc}& \int x\sqrt{1-x^4}dx\\ =& \frac{1}{2}\int {\cos }^{2}tdt\\ =& \frac{1}{4}\int (1+\cos 2t)dt\\ =& \frac{1}{4}t+\frac{1}{8}\sin 2t+C\\ =& \frac{1}{4}\arcsin \left(x^2\right)+\frac{1}{2}{x}^2\sqrt{1-x^4}+C\end{array}$$

Remark: Instead of substitute x2$x^2$ by sint$\sin t$, you may use substitution twice, u=x2$u=x^2$, u=sint$u=\sin t$.

Exercise 8.24 Evaluate the integral.

∫π/20cosx√1+sin2xdx$${\int }_0^{\pi /2}\frac{\cos x}{\sqrt{1+{\sin }^{2}x}}dx$$

Solution.

Let sinx=tant$\sin x=\tan t$. Then cosxdx=sec2tdt$\cos xdx={\sec }^{2}tdt$ and √1+sin2x=√1+tan2t=sect$\sqrt{1+{\sin }^{2}x}=\sqrt{1+{\tan }^{2}t}=\sec t$. Therefore, ∫cosx√1+sin2xdx=∫sec2tdtsect=∫sectdt=ln|sect+tant|+C=ln|sinx+√1+sin2x|+C$$\begin{array}{ccc}& \int \frac{\cos x}{\sqrt{1+{\sin }^{2}x}}dx& \\ =& \int \frac{{\sec }^{2}tdt}{\sec t}& \\ =& \int \sec tdt& \\ =& \ln |\sec t+\tan t|+C=& \ln |\sin x+\sqrt{1+{\sin }^{2}x}|+C\end{array}$$

By FTC, ∫π/20cosx√1+sin2xdx=ln|1+√1+1|−0=ln(1+√2)$${\int }_0^{\pi /2}\frac{\cos x}{\sqrt{1+{\sin }^{2}x}}dx=\ln |1+\sqrt{1+1}|-0=\ln (1+\sqrt{2})$$

Remark: Instead of the substitution sinx=tant$\sin x=\tan t$, you may first set u=sinx$u=\sin x$ and then set u=tant$u=\tan t$.

8.3 Integrations of Rational Functions by Partial Fractions

In Algebra, there is a theorem called the Fundamental Theorem of Algebra which implies that for rational function P(x)/Q(x)$P\left(x\right)/Q\left(x\right)$ can be written as a sum of partial fractions of the form A(ax+b)i or Ax+B(ax2+bx+c)j,$$\frac{A}{(ax+b{)}^i}\text{or}\frac{Ax+B}{{(a{x}^2+bx+c)}^j},$$ where ax2+bx+c$a{x}^2+bx+c$ is a irreducible quadratic form, that is the equation ax2+bx+c=0$a{x}^2+bx+c=0$ has no real roots.

After rewriting the rational function into a sum of partial fractions, we may use substitution methods to find the integral.

Exercise 8.25 Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.

x2−2x3+2x2+x$$\frac{x^2-2}{x^3+2x^2+x}$$

Solution.

The denominator can be factor into x(x+1)2$x(x+1{)}^2$. Then the sum of partial fractions for the rational function can be written in the following form x2−2x3+2x2+x=Ax+Bx+1+C(x+1)2.$$\frac{x^2-2}{x^3+2x^2+x}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}.$$

Remark: To determine the values of coefficients, we clear the denominator first. From here, there are two ways you may try.

Method one: substitute x$x$ by the roots of the GCD and/or special numbers and then solve for the undetermined coefficients.

Method two: simplify both sides and get system of equations of the undetermined coefficients by comparing coefficients of polynomials in both sides.

Exercise 8.26 Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.

x4(x2−x+1)(x2+2)2$$\frac{x^4}{(x^2-x+1){(x^2+2)}^2}$$

Solution.

Since the denominators has irreducible quadratic functions, the partial fraction form for the rational function can be written as

x4(x2−x+1)(x2+2)2=A1x+B1x2−x+1+A2x+B2x2+2+A3x+C3(x2+2)2.$$\frac{x^4}{(x^2-x+1){(x^2+2)}^2}=\frac{A_{1}x+B_1}{x^2-x+1}+\frac{A_{2}x+B_2}{x^2+2}+\frac{A_{3}x+C_3}{(x^2+2{)}^2}.$$

Remark: Although an irreducible quadratic form has no real roots, it has complex roots. One may also substitute x$x$ by a complex root to solve of the undetermined coefficients.

Exercise 8.27 Evaluate the integral ∫5x+1(2x+1)(x−1)dx$$\int \frac{5x+1}{(2x+1)(x-1)}dx$$

Solution.

We first write the rational function in the partial fraction form 5x+1(2x+1)(x−1)=A2x+1+Bx−1.$$\frac{5x+1}{(2x+1)(x-1)}=\frac{A}{2x+1}+\frac{B}{x-1}.$$

Clear the denominators, we have 5x+1=A(x−1)+B(2x+1).$$5x+1=A(x-1)+B(2x+1).$$

Let x=1$x=1$, we get 6=3B$6=3B$. So B=2$B=2$.

Let x=−12$x=-\frac{1}{2}$, we get −52+1=A(−12−1)$-\frac{5}{2}+1=A(-\frac{1}{2}-1)$. So A=1$A=1$.

Then the integral is ∫5x+1(2x+1)(x−1)dx=∫12x+1dx+∫2x−1dx=12ln|2x+1|+2ln|x−1|+C.$$\begin{array}{cc}& \int \frac{5x+1}{(2x+1)(x-1)}dx\\ =& \int \frac{1}{2x+1}dx+\int \frac{2}{x-1}dx\\ =& \frac{1}{2}\ln |2x+1|+2\ln |x-1|+C.\end{array}$$

Exercise 8.28 Evaluate the integral

∫1022x2+3x+1dx$${\int }_0^1\frac{2}{2x^2+3x+1}dx$$

Solution.

The denominator can be factored into (2x+1)(x+1)$(2x+1)(x+1)$.

Write the rational function in the partial fraction form 22x2+3x+1=A2x+1+Bx+1.$$\frac{2}{2x^2+3x+1}=\frac{A}{2x+1}+\frac{B}{x+1}.$$

Clear the denominators, we have 2=A(x+1)+B(2x+1).$$2=A(x+1)+B(2x+1).$$

Let x=−1$x=-1$, we get 2=−B$2=-B$. So B=−2$B=-2$.

Let x=−12$x=-\frac{1}{2}$, we get 2=A(−12+1)$2=A(-\frac{1}{2}+1)$. So A=4$A=4$.

Then the indefinite integral is ∫2(2x+1)(x+1)dx=∫42x+1dx+∫−2x−1dx=2ln|2x+1|−2ln|x+1|+C.$$\begin{array}{cc}& \int \frac{2}{(2x+1)(x+1)}dx\\ =& \int \frac{4}{2x+1}dx+\int \frac{-2}{x-1}dx\\ =& 2\ln |2x+1|-2\ln |x+1|+C.\end{array}$$

By FTC, ∫1022x2+3x+1dx=2ln3−2ln2=ln(94).$${\int }_0^1\frac{2}{2x^2+3x+1}dx=2\ln 3-2\ln 2=\ln \left(\frac{9}{4}\right).$$

Exercise 8.29 Evaluate the integral

∫10x2+x+1(x+1)2(x+2)dx$${\int }_0^1\frac{x^2+x+1}{(x+1{)}^2(x+2)}dx$$

Solution.

Write the rational function in the partial fraction form x2+x+1(x+1)2(x+2)=Ax+2+Bx+1+C(x+1)2.$$\frac{x^2+x+1}{(x+1{)}^2(x+2)}=\frac{A}{x+2}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}.$$

Clear the denominators, we have x2+x+1=A(x+1)2+B(x+1)(x+2)+C(x+2).$$x^2+x+1=A(x+1{)}^2+B(x+1\left)\right(x+2)+C(x+2).$$

Let x=−1$x=-1$, we get 1=C$1=C$.

Let x=−2$x=-2$, we get 3=A$3=A$.

Comparing coefficients of degree two terms in both sides, we get 1=A+B$1=A+B$. So B=−2$B=-2$.

Then the indefinite integral is ∫x2+x+1(x+1)2(x+2)dx=∫3x+2dx+∫−2x+1dx+∫1(x+1)2dx=3ln|x+2|−2ln|x+1|−1x+1+C.$$\begin{array}{cc}& \int \frac{x^2+x+1}{(x+1{)}^2(x+2)}dx\\ =& \int \frac{3}{x+2}dx+\int \frac{-2}{x+1}dx+\int \frac{1}{(x+1{)}^2}dx\\ =& 3\ln |x+2|-2\ln |x+1|-\frac{1}{x+1}+C.\end{array}$$

By FTC, ∫1022x2+3x+1dx=(3ln3−2ln2−12)−(3ln2−2ln1−1)=12+ln(2732).$$\begin{array}{cc}{\int }_0^1\frac{2}{2x^2+3x+1}dx=& (3\ln 3-2\ln 2-\frac{1}{2})-(3\ln 2-2\ln 1-1)\\ =& \frac{1}{2}+\ln \left(\frac{27}{32}\right).\end{array}$$

Exercise 8.30 Evaluate the integral

∫x2−x+6x3+3xdx$$\int \frac{x^2-x+6}{x^3+3x}dx$$

Solution.

The denominator can be factored into x(x2+3)$x(x^2+3)$.

We may write the rational function in the partial fraction form x2−x+6x3+3x=Ax+Bx+Cx2+3.$$\frac{x^2-x+6}{x^3+3x}=\frac{A}{x}+\frac{Bx+C}{x^2+3}.$$

Clear the denominators, we have x2−x+6=A(x2+3)+Bx2+Cx.$$x^2-x+6=A(x^2+3)+B{x}^2+Cx.$$

Let x=0$x=0$. We find A=2$A=2$. Let x=i√3$x=i\sqrt{3}$, we get x2=−3$x^2=-3$ and −3−i√3+6=−3B+i√3C.$$-3-i\sqrt{3}+6=-3B+i\sqrt{3}C.$$ Since B$B$ and C$C$ are real numbers, 3=−3B$3=-3B$ and −i√3=i√3C$-i\sqrt{3}=i\sqrt{3}C$. So B=−1$B=-1$ and C=−1$C=-1$.

Then the indefinite integral is ∫x2−x+6x3+3xdx=∫2xdx−∫xx2+3dx−∫1x2+3dx=2ln|x|−12ln(x2+3)−√33arctan(x√3)+C=ln(x2√x3+3)−√33arctan(x√3)+C.$$\begin{array}{cc}& \int \frac{x^2-x+6}{x^3+3x}dx\\ =& \int \frac{2}{x}dx-\int \frac{x}{x^2+3}dx-\int \frac{1}{x^2+3}dx\\ =& 2\ln |x|-\frac{1}{2}\ln (x^2+3)-\frac{\sqrt{3}}{3}\arctan \left(\frac{x}{\sqrt{3}}\right)+C\\ =& \ln \left(\frac{x^2}{\sqrt{x^3+3}}\right)-\frac{\sqrt{3}}{3}\arctan \left(\frac{x}{\sqrt{3}}\right)+C.\end{array}$$

Exercise 8.31 Evaluate the integral

∫x2+x+1(x2+1)2dx$$\int \frac{x^2+x+1}{{(x^2+1)}^2}dx$$

Solution.

The rational function can be written in the partial fraction form x2+x+1(x2+1)2=x2+1(x2+1)2+x(x2+1)2=1x2+1+x(x2+1)2.$$\begin{array}{cc}\frac{x^2+x+1}{{(x^2+1)}^2}& =\frac{x^2+1}{{(x^2+1)}^2}+\frac{x}{{(x^2+1)}^2}\\ & =\frac{1}{x^2+1}+\frac{x}{(x^2+1{)}^2}.\end{array}$$

Clear the denominators, we have x2+x+1=(Ax+B)(x2+1)+Cx+D.$$x^2+x+1=(Ax+B)(x^2+1)+Cx+D.$$

By comparing the top degree terms, we find that A=0$A=0$. So the equality become x2+x+1=B(x2+1)+Cx+D.$$x^2+x+1=B(x^2+1)+Cx+D.$$ Comparing the same degree terms, we find B=1$B=1$, C=1$C=1$ and 1=B+D$1=B+D$ which implies D=0$D=0$. –>

Therefore, ∫x2+x+1(x2+1)2dx=∫1x2+1dx+∫x(x2+1)2dx=arctanx−12(x2+1)+C.$$\begin{array}{cc}& \int \frac{x^2+x+1}{{(x^2+1)}^2}dx\\ =& \int \frac{1}{x^2+1}dx+\int \frac{x}{(x^2+1{)}^2}dx\\ =& \arctan x-\frac{1}{2(x^2+1)}+C.\end{array}$$

Exercise 8.32 Evaluate the integral

∫x3+6x−2x4+6x2dx$$\int \frac{x^3+6x-2}{x^4+6x^2}dx$$

Solution.

The rational function can be written in the partial fraction form x3+6x−2x4+6x2=Ax+Bx2+Cx+Dx2+6.$$\begin{array}{cc}\frac{x^3+6x-2}{x^4+6x^2}& =\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+6}.\end{array}$$

Clear the denominators, we have x3+6x−2=Ax(x2+6)+B(x2+6)+Cx3+Dx2.$$x^3+6x-2=Ax(x^2+6)+B(x^2+6)+C{x}^3+D{x}^2.$$

Let x=0$x=0$. We find B=−13$B=-\frac{1}{3}$.

Let x=i√6$x=i\sqrt{6}$. Then x2=−6$x^2=-6$, x3=−6√6 i$x^3=-6\sqrt{6}i$ and −2=−6√6C i−6D.$$-2=-6\sqrt{6}Ci-6D.$$ As C$C$ and D$D$ are real numbers, we must have C=0$C=0$ and D=13$D=\frac{1}{3}$.

Comparing degree 3 terms, we find A=1$A=1$.

Therefore, ∫x3+6x−2x4+6x2dx=∫1xdx+∫−13x2dx+∫13x2+6dx=ln|x|+13x+√618arctan(x√6)+C.$$\begin{array}{cc}& \int \frac{x^3+6x-2}{x^4+6x^2}dx\\ =& \int \frac{1}{x}dx+\int \frac{-\frac{1}{3}}{x^2}dx+\int \frac{\frac{1}{3}}{x^2+6}dx\\ =& \ln |x|+\frac{1}{3x}+\frac{\sqrt{6}}{18}\arctan \left(\frac{x}{\sqrt{6}}\right)+C.\end{array}$$

Exercise 8.33 Evaluate the integral

∫dxx√x−1$$\int \frac{dx}{x\sqrt{x-1}}$$

Solution.

Although the integrand is note a rational function, we may apply a linear substitution u=√x−1$u=\sqrt{x-1}$ to rewrite the integral as ∫dxx√x−1=∫2uduu(u2+1)$$\int \frac{dx}{x\sqrt{x-1}}=\int \frac{2udu}{u(u^2+1)}$$

Simplify and evaluate the resulting integral, we get ∫dxx√x−1=2arctan(√x−1)+C.$$\int \frac{dx}{x\sqrt{x-1}}=2\arctan \left(\sqrt{x-1}\right)+C.$$

Exercise 8.34 Evaluate the integral

∫sinxcos2x−3cosxdx$$\int \frac{\sin x}{{\cos }^{2}x-3\cos x}dx$$

Solution.

Let u=cosx$u=\cos x$. Then ∫sinxcos2x−3cosxdx=∫−dxx2−3x=13∫(1x−1x−3)dx=13(ln|x|−ln|x−3|)+C=13(ln|cosx|−ln|cosx−3|)+C.$$\begin{array}{cc}& \int \frac{\sin x}{{\cos }^{2}x-3\cos x}dx\\ =& \int \frac{-dx}{x^2-3x}\\ =& \frac{1}{3}\int (\frac{1}{x}-\frac{1}{x-3})dx\\ =& \frac{1}{3}(\ln |x|-\ln |x-3|)+C\\ =& \frac{1}{3}(\ln |\cos x|-\ln |\cos x-3|)+C.\end{array}$$

8.4 Improper Integrals

8.4.1 Infinite Intervals

Definition 8.1 Suppose the integral ∫taf(x)dx${\int }_a^{t}f\left(x\right)dx$ exists for any t$t$.

Then ∫∞af(x)dx=limt→∞∫taf(x)dx,$${\int }_a^{\infty }f\left(x\right)dx=\underset{t\to \infty }{lim}{\int }_a^{t}f\left(x\right)dx,$$ and ∫a−∞f(x)dx=limt→−∞∫atf(x)dx.$${\int }_{-\infty }^{a}f\left(x\right)dx=\underset{t\to -\infty }{lim}{\int }_t^{a}f\left(x\right)dx.$$

An improper integral is called convergent if the corresponding limit exists and divergent if the limit does not exist.

If both ∫∞af(x)dx${\int }_a^{\infty }f\left(x\right)dx$ and ∫a−∞f(x)dx${\int }_{-\infty }^{a}f\left(x\right)dx$ are convergent, then ∫∞−∞f(x)dx=∫∞af(x)dx+∫a−∞f(x)dx.$${\int }_{-\infty }^{\infty }f\left(x\right)dx={\int }_a^{\infty }f\left(x\right)dx+{\int }_{-\infty }^{a}f\left(x\right)dx.$$

Exercise 8.35 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent. ∫∞014√1+xdx$${\int }_0^{\infty }\frac{1}{\sqrt[4]{41+x}}dx$$

Solution.

We first find the indefinite integral. ∫14√1+xdx=∫u−14duu=x+1=43u34+C=43(1+x)34+C.$$\begin{array}{cc}& \int \frac{1}{\sqrt[4]{41+x}}dx\\ =& \int u^{-\frac{1}{4}}duu=x+1\\ =& \frac{4}{3}{u}^{\frac{3}{4}}+C\\ =& \frac{4}{3}(1+x{)}^{\frac{3}{4}}+C.\end{array}$$

Then limt→∞∫t014√1+xdx=limt→∞(43(1+t)34−43)=∞.$$\begin{array}{cc}& \underset{t\to \infty }{lim}{\int }_0^t\frac{1}{\sqrt[4]{41+x}}dx\\ =& \underset{t\to \infty }{lim}\left(\frac{4}{3}\right(1+t{)}^{\frac{3}{4}}-\frac{4}{3})\\ =& \infty .\end{array}$$

By definition, ∫∞014√1+xdx=limt→∞∫t014√1+xdx=∞$${\int }_0^{\infty }\frac{1}{\sqrt[4]{41+x}}dx=\underset{t\to \infty }{lim}{\int }_0^t\frac{1}{\sqrt[4]{41+x}}dx=\infty$$ is divergent.

Exercise 8.36 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent. ∫0−∞2xdx$${\int }_{-\infty }^0{2}^{x}dx$$

Solution.

We first find the indefinite integral. ∫2xdx=2xln2+C$$\begin{array}{cc}& \int 2^{x}dx\\ =& \frac{2^x}{\ln 2}+C\end{array}$$

Then limt→−∞∫0t2xdx=limt→∞(1ln2−2xln2)=1ln2.$$\begin{array}{cc}& \underset{t\to -\infty }{lim}{\int }_t^0{2}^{x}dx\\ =& \underset{t\to \infty }{lim}(\frac{1}{\ln 2}-\frac{2^x}{\ln 2})\\ =& \frac{1}{\ln 2}.\end{array}$$

By definition, ∫∞014√1+xdx=limt→∞∫t014√1+xdx=1ln2.$${\int }_0^{\infty }\frac{1}{\sqrt[4]{41+x}}dx=\underset{t\to \infty }{lim}{\int }_0^t\frac{1}{\sqrt[4]{41+x}}dx=\frac{1}{\ln 2}.$$

Exercise 8.37 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent. ∫∞−∞x29+x6dx$${\int }_{-\infty }^{\infty }\frac{x^2}{9+x^6}dx$$

Solution.

We first find the indefinite integral. ∫x29+x6dx=19∫11+u2du3u=x3=19arctanu+C=19arctan(x33)+C$$\begin{array}{cc}& \int \frac{x^2}{9+x^6}dx\\ =& \frac{1}{9}\int \frac{1}{1+u^2}du3u=x^3\\ =& \frac{1}{9}\arctan u+C\\ =& \frac{1}{9}\arctan \left(\frac{x^3}{3}\right)+C\end{array}$$

Then limt→−∞∫0t19arctan(x33)=limt→−∞−19arctan(t33)=π18$$\begin{array}{cc}& \underset{t\to -\infty }{lim}{\int }_t^0\frac{1}{9}\arctan \left(\frac{x^3}{3}\right)\\ =& \underset{t\to -\infty }{lim}-\frac{1}{9}\arctan \left(\frac{t^3}{3}\right)\\ =& \frac{\pi }{18}\end{array}$$ and limt→∞∫t019arctan(x33)=limt→∞19arctan(t33)=π18.$$\begin{array}{cc}& \underset{t\to \infty }{lim}{\int }_0^t\frac{1}{9}\arctan \left(\frac{x^3}{3}\right)\\ =& \underset{t\to \infty }{lim}\frac{1}{9}\arctan \left(\frac{t^3}{3}\right)\\ =& \frac{\pi }{18}.\end{array}$$

By definition, ∫∞−∞x29+x6dx=π18+π18=π9.$${\int }_{-\infty }^{\infty }\frac{x^2}{9+x^6}dx=\frac{\pi }{18}+\frac{\pi }{18}=\frac{\pi }{9}.$$

Exercise 8.38 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent. ∫∞2xe−3xdx$${\int }_2^{\infty }x{e}^{-3x}dx$$

Solution.

We first find the indefinite integral. ∫xe−3xdx=∫x(e−3x−3)′dx=−13(xe−3x−∫e−3xdx)=−13(xe−3x+13e−3x)+C=−13xe−3x−19e−3x+C$$\begin{array}{cc}& \int x{e}^{-3x}dx\\ =& \int x{\left(\frac{e^{-3x}}{-3}\right)}^{\prime }dx\\ =& -\frac{1}{3}(x{e}^{-3x}-\int e^{-3x}dx)\\ =& -\frac{1}{3}(x{e}^{-3x}+\frac{1}{3}{e}^{-3x})+C\\ =& -\frac{1}{3}x{e}^{-3x}-\frac{1}{9}{e}^{-3x}+C\end{array}$$

Then ∫∞2xe−3xdx=limt→−∞[(−13te−3t−19e−3t)−(−132e−6−19e−6)]=79e−6−13limt→−∞te−3t=79e−6by L'Hospital's rule$$\begin{array}{cc}& {\int }_2^{\infty }x{e}^{-3x}dx\\ =& \underset{t\to -\infty }{lim}[(-\frac{1}{3}t{e}^{-3t}-\frac{1}{9}{e}^{-3t})-(-\frac{1}{3}2e^{-6}-\frac{1}{9}{e}^{-6})]\\ =& \frac{7}{9}{e}^{-6}-\frac{1}{3}\underset{t\to -\infty }{lim}t{e}^{-3t}\\ =& \frac{7}{9}{e}^{-6}\text{by L'Hospital's rule}\end{array}$$

8.4.2 Discountinuous Integrands

Definition 8.2 Suppose the function f$f$ is continuous on [a,b)$[a,b)$ but discontinuous at b$b$. Then ∫baf(x)dx=limt→b−∫taf(x)dx.$${\int }_a^{b}f\left(x\right)dx=\underset{t\to b^{-}}{lim}{\int }_a^{t}f\left(x\right)dx.$$

Suppose the function f$f$ is continuous on (a,b]$(a,b]$ but discontinuous at a$a$. Then ∫baf(x)dx=limt→a+∫btf(x)dx.$${\int }_a^{b}f\left(x\right)dx=\underset{t\to a^{+}}{lim}{\int }_t^{b}f\left(x\right)dx.$$

An improper integral is called convergent if the corresponding limit exists and divergent if the limit does not exist.

If f$f$ has a discontinuity c$c$ within the interval (a,b)$(a,b)$ and both ∫caf(x)dx${\int }_a^{c}f\left(x\right)dx$ and ∫bcf(x)dx${\int }_c^{b}f\left(x\right)dx$ are convergent, then ∫baf(x)dx=∫caf(x)dx+∫bcf(x)dx$${\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx$$

Exercise 8.39 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent.

∫321√3−xdx$${\int }_2^3\frac{1}{\sqrt{3-x}}dx$$

Solution.

We first find the indefinite integral. ∫1√3−xdx=−∫u−12duu=3−x=−2√u+C=−2√3−x+C$$\begin{array}{cc}& \int \frac{1}{\sqrt{3-x}}dx\\ =& -\int u^{-\frac{1}{2}}duu=3-x\\ =& -2\sqrt{u}+C\\ =& -2\sqrt{3-x}+C\end{array}$$

Then ∫321√3−xdx=limt→3−∫t21√3−x=limt→3−(−2√3−t+2)=2$$\begin{array}{cc}& {\int }_2^3\frac{1}{\sqrt{3-x}}dx\\ =& \underset{t\to 3^{-}}{lim}{\int }_2^t\frac{1}{\sqrt{3-x}}\\ =& \underset{t\to 3^{-}}{lim}(-2\sqrt{3-t}+2)\\ =& 2\end{array}$$

Exercise 8.40 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent.

∫10lnx√xdx$${\int }_0^1\frac{\ln x}{\sqrt{x}}dx$$

Solution.

We first find the indefinite integral. ∫lnx√xdx=2∫ln(u2)duu=√x=4∫lnudu=4(ulnu−u)+C=2√xlnx−4√x+C.$$\begin{array}{cc}& \int \frac{\ln x}{\sqrt{x}}dx\\ =& 2\int \ln \left(u^2\right)duu=\sqrt{x}\\ =& 4\int \ln udu\\ =& 4(u\ln u-u)+C\\ =& 2\sqrt{x}\ln x-4\sqrt{x}+C.\end{array}$$

Then ∫10lnx√xdx=limt→0+∫1tlnx√xdx=−4−limt→0+(2√tlnt−4√t)=−4−limt→0+(2√tlnt)=−4−2limt→0+(lnt)′(1√t)′L'Hospitcal's rule=−4−2⋅0=−4.$$\begin{array}{cc}& {\int }_0^1\frac{\ln x}{\sqrt{x}}dx\\ =& \underset{t\to 0^{+}}{lim}{\int }_t^1\frac{\ln x}{\sqrt{x}}dx\\ =& -4-\underset{t\to 0^{+}}{lim}(2\sqrt{t}\ln t-4\sqrt{t})\\ =& -4-\underset{t\to 0^{+}}{lim}\left(2\sqrt{t}\ln t\right)\\ =& -4-2\underset{t\to 0^{+}}{lim}\frac{(\ln t{)}^{\prime }}{{\left(\frac{1}{\sqrt{t}}\right)}^{\prime }}\text{L'Hospitcal's rule}\\ =& -4-2\cdot 0\\ =& -4.\end{array}$$

Exercise 8.41 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent.

∫50xx−2dx$${\int }_0^5\frac{x}{x-2}dx$$

Solution.

We first find the indefinite integral. ∫xx−2dx=∫(1+2x−2)dx=x+2ln|x−2|.$$\begin{array}{cc}& \int \frac{x}{x-2}dx\\ =& \int (1+\frac{2}{x-2})dx\\ =& x+2\ln |x-2|.\end{array}$$

Since limt→2+ln(t−2)=−∞$\underset{t\to 2^{+}}{lim}\ln (t-2)=-\infty$, limt→2+∫5txx−2dx=5+2ln3−limt→2+(t+2ln(t−2))=−∞.$$\begin{array}{cc}& \underset{t\to 2^{+}}{lim}{\int }_t^5\frac{x}{x-2}dx\\ =& 5+2\ln 3-\underset{t\to 2^{+}}{lim}(t+2\ln (t-2\left)\right)\\ =& -\infty .\end{array}$$

Therefore, the given improper integral is divergent.

Exercise 8.42 Determine whether the integral is convergent or divergent. Evaluate the integral if it is convergent.

∫30dxx2−6x+5$${\int }_0^3\frac{dx}{x^2-6x+5}$$

Solution.

We first find the indefinite integral by partial fractions. ∫dxx2−6x+5=∫−14x−1+14x−5dx=−14∫1x−1−1x−5dx=−14(ln|x−1|−ln|x−5|)+C.$$\begin{array}{cc}& \int \frac{dx}{x^2-6x+5}\\ =& \int \frac{-\frac{1}{4}}{x-1}+\frac{\frac{1}{4}}{x-5}dx\\ =& -\frac{1}{4}\int \frac{1}{x-1}-\frac{1}{x-5}dx\\ =& -\frac{1}{4}(\ln |x-1|-\ln |x-5|)+C.\end{array}$$

Since limt→1+ln(t−1)=−∞$\underset{t\to 1^{+}}{lim}\ln (t-1)=-\infty$, similar to the previous question, the given improper integral is divergent.

8.4.3 Comparison Test

Theorem 8.1 Suppose f$f$ and g$g$ are continuos functions and f(x)≥g(x)≥0$f\left(x\right)\ge g\left(x\right)\ge 0$ for all x≥a$x\ge a$.

1. If ∫∞af(x)dx${\int }_a^{\infty }f\left(x\right)dx$ is convergent, then ∫∞ag(x)dx${\int }_a^{\infty }g\left(x\right)dx$ is convergent.

2. If ∫∞ag(x)dx${\int }_a^{\infty }g\left(x\right)dx$ is divergent, then ∫∞af(x)dx${\int }_a^{\infty }f\left(x\right)dx$ is divergent.

The same conclusion holds true for improper integrals of functions with discontinuities.

We have the following results. ∫∞11xpdx is convergent if p>1 and divergent if p≤1$${\int }_1^{\infty }\frac{1}{x^p}dx\text{is convergent if}p>1\text{and divergent if}p\le 1$$

∫∞11ekxdx is convergent if k>0 and divergent if k<0.$${\int }_1^{\infty }\frac{1}{e^{kx}}dx\text{is convergent if}k>0\text{and divergent if}k<0.$$

Exercise 8.43 Determine whether the integral is convergent or divergent.

∫∞0sin2x1+x2dx$${\int }_0^{\infty }\frac{{\sin }^{2}x}{1+x^2}dx$$

Solution.

We know that ∫∞011+x2dx=limt→∞∫t011+x2dx=limt→∞arctant=π2.$${\int }_0^{\infty }\frac{1}{1+x^2}dx=\underset{t\to \infty }{lim}{\int }_0^t\frac{1}{1+x^2}dx=\underset{t\to \infty }{lim}\arctan t=\frac{\pi }{2}.$$

Since 0≤sin2x≤1$0\le {\sin }^{2}x\le 1$, by comparison theorem, the integral is convergent.

Exercise 8.44 Determine whether the integral is convergent or divergent.

∫∞0arctanx2+exdx$${\int }_0^{\infty }\frac{\arctan x}{2+e^x}dx$$

Solution.

We know that ∫∞01exdx=limt→∞∫t0e−xdx=limt→∞−et+1=1.$${\int }_0^{\infty }\frac{1}{e^x}dx=\underset{t\to \infty }{lim}{\int }_0^t{e}^{-x}dx=\underset{t\to \infty }{lim}-e^t+1=1.$$

Since 0≤arctanx2+ex≤π21ex,$$0\le \frac{\arctan x}{2+e^x}\le \frac{\pi }{2}\frac{1}{e^x},$$ by comparison theorem, the integral is convergent.