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:

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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
${\begin{array}{rcl}\displaystyle {\int _{-\pi }^{\pi }\sin ^{3}x\cos ^{3}x~dx}&=&\displaystyle {\int _{-\pi }^{\pi }\sin ^{3}x\cos ^{2}x\cos x~dx}\\&&\\&=&\displaystyle {\int _{-\pi }^{\pi }\sin ^{3}x(1-\sin ^{2}x)\cos x~dx.}\end{array}}$

Step 2:
Now, we use  $u$-substitution.
Let  $u=\sin x.$  Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=\cos xdx.}
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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  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 u_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)}

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