007B Sample Midterm 3, Problem 5 Detailed Solution

Evaluate the following integrals.

(a)   $\int x\sin x~dx$

(b)   $\int {\frac {1}{(x-3)(x-2)}}~dx$

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Background Information:
1. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$
2. 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 proceed using integration by parts.
Let  $u=x$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=\sin x~dx.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=dx}   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 v=-\cos x.}
Therefore, 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 \int x \sin x~dx=-x\cos x+\int \cos x~dx.}

Step 2:
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 x\sin x~dx} & = & \displaystyle{-x\cos x+\int \cos x~dx}\\ &&\\ & = & \displaystyle{-x\cos x+\sin x +C.} \end{array}}

(b)

Step 1:
We need to use partial fraction decomposition for this integral.
We start by letting
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-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}.}
Multiplying both sides of the last equation by  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x-3)(x+2),}
we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1=A(x+2)+B(x-3).}
If we let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=3,}   the last equation becomes  $1=5A.$  Thus,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A={\frac {1}{5}}.}
If we let  $x=-2,$  then we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1=(-5)B.}   Thus,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle B=-{\frac {1}{5}}.}
So, in summation, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{(x-3)(x+2)}}={\frac {\frac {1}{5}}{x-3}}+{\frac {-{\frac {1}{5}}}{x+2}}.}

Step 2:
Now, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {1}{(x-3)(x+2)}}~dx=\int {\frac {\frac {1}{5}}{x-3}}~dx+\int {\frac {-{\frac {1}{5}}}{x+2}}~dx.}
Now, we use  $u$-substitution to evaluate these integrals.
For the first integral, we substitute  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x-3.}
For the second integral, the substitution is  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle t=x+2.}
Then, we integrate to get
${\begin{array}{rcl}\displaystyle {\int {\frac {1}{(x-3)(x+2)}}~dx}&=&\displaystyle {{\frac {1}{5}}\int {\frac {1}{u}}~du-{\frac {1}{5}}\int {\frac {1}{t}}~dt}\\&&\\&=&\displaystyle {{\frac {1}{5}}\ln |u|-{\frac {1}{5}}\ln |t|+C}\\&&\\&=&\displaystyle {{\frac {1}{5}}\ln |x-3|-{\frac {1}{5}}\ln |x+2|+C.}\end{array}}$

Final Answer:
(a)     Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -x\cos x+\sin x+C}
(b)     Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{5}}\ln |x-3|-{\frac {1}{5}}\ln |x+2|+C}

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