009B Sample Final 2, Problem 6

Evaluate the following integrals:

(a) $\int {\frac {dx}{x^{2}{\sqrt {x^{2}-16}}}}$

(b) $\int _{-\pi }^{\pi }\sin ^{3}x\cos ^{3}x~dx$

(c) $\int _{0}^{1}{\frac {x-3}{x^{2}+6x+5}}~dx$

Foundations:
1. For Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {dx}{x^{2}{\sqrt {x^{2}-16}}}},} what would be the correct trig substitution?
The correct substitution is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=4\sec ^{2}\theta .}
2. We have the Pythagorean identity
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \cos ^{2}(x)=1-\sin ^{2}(x).}
3. Through partial fraction decomposition, we can write the fraction
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}}
for some constants Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A,B.}

Solution:

(a)

Step 1:
We start by using trig substitution.
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=4\sec \theta.}
Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dx=4\sec \theta \tan \theta ~d\theta.}
So, the integral becomes
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta \sqrt{16\sec^2 \theta -16}}~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta (4\tan \theta)} ~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{1}{16\sec \theta} ~d\theta.} \end{array}}

Step 2:

Now, we integrate to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\frac{1}{16}\cos\theta~d\theta}\\ &&\\ & = & \displaystyle{\frac{1}{16}\sin \theta +C}\\ &&\\ & = & \displaystyle{\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C.} \end{array}}

(b)

Step 1:

First, we write

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x \cos^2x \cos x~dx}\\ &&\\ & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x (1-\sin^2x)\cos x~dx.} \end{array}}

Step 2:
Now, we use $u$ -substitution.
Let $u=\sin x.$ Then, $du=\cos x~dx.$
Since this is a definite integral, we need to change the bounds of integration.
Then, we have
$u_{1}=\sin(-\pi )=0$ and $u_{2}=\sin(\pi )=0.$
So, we have
${\begin{array}{rcl}\displaystyle {\int _{-\pi }^{\pi }\sin ^{3}x\cos ^{3}x~dx}&=&\displaystyle {\int _{0}^{0}u^{3}(1-u^{2})~du}\\&&\\&=&\displaystyle {0.}\end{array}}$

(c)

Step 1:
First, we write
$\int _{0}^{1}{\frac {x-3}{x^{2}+6x+5}}~dx=\int _{0}^{1}{\frac {x-3}{(x+1)(x+5)}}~dx.$
Now, we use partial fraction decomposition. Wet set
${\frac {x-3}{(x+1)(x+5)}}={\frac {A}{x+1}}+{\frac {B}{x+5}}.$
If we multiply both sides of this equation by $(x+1)(x+5),$ we get
$x-3=A(x+5)+B(x+1).$
If we let $x=-1,$ we get $A=-1.$
If we let $x=-5,$ we get $B=2.$
So, we have
${\frac {x-3}{(x+1)(x+5)}}={\frac {-1}{x+1}}+{\frac {2}{x+5}}.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {x-3}{(x+1)(x+5)}}~dx}&=&\displaystyle {\int _{0}^{1}{\frac {-1}{x+1}}+{\frac {2}{x+5}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{1}{\frac {-1}{x+1}}~dx+\int _{0}^{1}{\frac {2}{x+5}}~dx.}\end{array}}$
Now, we use $u$ -substitution for both of these integrals.
Let $u=x+1.$ Then, $du=dx.$
Let $t=x+5.$ Then, $dt=dx.$
Since these are definite integrals, we need to change the bounds of integration.
We have $u_{1}=0+1=1$ and $u_{2}=1+1=2.$
Also, $t_{1}=0+5=5$ and $u_{2}=1+5=6.$
Therefore, we get
${\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {x-3}{(x+1)(x+5)}}~dx}&=&\displaystyle {\int _{1}^{2}{\frac {-1}{u}}~du+\int _{5}^{6}{\frac {2}{t}}~dt}\\&&\\&=&\displaystyle {-\ln \|u\|{\bigg \|}_{1}^{2}+2\ln \|t\|{\bigg \|}_{5}^{6}}\\&&\\&=&\displaystyle {-\ln(2)+2\ln(6)-2\ln(5).}\end{array}}$