009B Sample Final 2, Problem 6 Detailed Solution

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

---

Background Information:
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
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_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  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 u} -substitution for both of these integrals.
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 u=x+1.}   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 du=dx.}
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 t=x+5.}   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 dt=dx.}
Since these are definite integrals, we need to change the bounds of integration.
We have
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 u_1=0+1=1}   and  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 u_2=1+1=2.}
Also,
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 t_1=0+5=5}   and  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 t_2=1+5=6.}
Therefore, we 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_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)    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}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C}
(b)    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 0}
(c)    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 -\ln(2)+2\ln(6)-2\ln(5)}

Return to Sample Exam