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

Solution:

(a)

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

Step 2:
Now, we integrate to get
${\begin{array}{rcl}\displaystyle {\int {\frac {1}{x^{2}{\sqrt {x^{2}-16}}}}~dx}&=&\displaystyle {\int {\frac {1}{16}}\cos \theta ~d\theta }\\&&\\&=&\displaystyle {{\frac {1}{16}}\sin \theta +C}\\&&\\&=&\displaystyle {{\frac {1}{16}}{\bigg (}{\frac {\sqrt {x^{2}-16}}{x}}{\bigg )}+C.}\end{array}}$

(b)

Step 1:
First, we write
${\begin{array}{rcl}\displaystyle {\int _{-\pi }^{\pi }\sin ^{3}x\cos ^{3}x~dx}&=&\displaystyle {\int _{-\pi }^{\pi }\sin ^{3}x\cos ^{2}x\cos x~dx}\\&&\\&=&\displaystyle {\int _{-\pi }^{\pi }\sin ^{3}x(1-\sin ^{2}x)\cos x~dx.}\end{array}}$

Step 2:
Now, we use  $u$-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  $t_{2}=1+5=6.$
Therefore, we get
${\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {x-3}{(x+1)(x+5)}}~dx}&=&\displaystyle {\int _{1}^{2}{\frac {-1}{u}}~du+\int _{5}^{6}{\frac {2}{t}}~dt}\\&&\\&=&\displaystyle {-\ln |u|{\bigg |}_{1}^{2}+2\ln |t|{\bigg |}_{5}^{6}}\\&&\\&=&\displaystyle {-\ln(2)+2\ln(6)-2\ln(5).}\end{array}}$

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

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