# 009B Sample Final 2, Problem 6

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Evaluate the following integrals:

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

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

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

Foundations:
1. For  ${\displaystyle \int {\frac {dx}{x^{2}{\sqrt {x^{2}-16}}}},}$  what would be the correct trig substitution?
The correct substitution is  ${\displaystyle x=4\sec ^{2}\theta .}$
2. Recall the Pythagorean identity
${\displaystyle \cos ^{2}(x)=1-\sin ^{2}(x).}$
3. Through partial fraction decomposition, we can write the fraction
${\displaystyle {\frac {1}{(x+1)(x+2)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}}$
for some constants ${\displaystyle A,B.}$

Solution:

(a)

Step 1:
We start by using trig substitution.
Let  ${\displaystyle x=4\sec \theta .}$
Then,  ${\displaystyle dx=4\sec \theta \tan \theta ~d\theta .}$
So, the integral becomes
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {1}{x^{2}{\sqrt {x^{2}-16}}}}~dx}&=&\displaystyle {\int {\frac {4\sec \theta \tan \theta }{16\sec ^{2}\theta {\sqrt {16\sec ^{2}\theta -16}}}}~d\theta }\\&&\\&=&\displaystyle {\int {\frac {4\sec \theta \tan \theta }{16\sec ^{2}\theta (4\tan \theta )}}~d\theta }\\&&\\&=&\displaystyle {\int {\frac {1}{16\sec \theta }}~d\theta .}\end{array}}}$
Step 2:
Now, we integrate to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {1}{x^{2}{\sqrt {x^{2}-16}}}}~dx}&=&\displaystyle {\int {\frac {1}{16}}\cos \theta ~d\theta }\\&&\\&=&\displaystyle {{\frac {1}{16}}\sin \theta +C}\\&&\\&=&\displaystyle {{\frac {1}{16}}{\bigg (}{\frac {\sqrt {x^{2}-16}}{x}}{\bigg )}+C.}\end{array}}}$

(b)

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

(c)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {x-3}{(x+1)(x+5)}}~dx}&=&\displaystyle {\int _{0}^{1}{\frac {-1}{x+1}}+{\frac {2}{x+5}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{1}{\frac {-1}{x+1}}~dx+\int _{0}^{1}{\frac {2}{x+5}}~dx.}\end{array}}}$

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

(a)    ${\displaystyle {\frac {1}{16}}{\bigg (}{\frac {\sqrt {x^{2}-16}}{x}}{\bigg )}+C}$
(b)    ${\displaystyle 0}$
(c)    ${\displaystyle -\ln(2)+2\ln(6)-2\ln(5)}$