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  $dv=\sin x~dx.$
Then,  $du=dx$  and  $v=-\cos x.$
Therefore, we have
$\int x\sin x~dx=-x\cos x+\int \cos x~dx.$

Step 2:
We integrate to get
${\begin{array}{rcl}\displaystyle {\int x\sin x~dx}&=&\displaystyle {-x\cos x+\int \cos x~dx}\\&&\\&=&\displaystyle {-x\cos x+\sin x+C.}\end{array}}$

(b)

Step 1:
We need to use partial fraction decomposition for this integral.
We start by letting
${\frac {1}{(x-3)(x+2)}}={\frac {A}{x-3}}+{\frac {B}{x+2}}.$
Multiplying both sides of the last equation by  $(x-3)(x+2),$
we get
$1=A(x+2)+B(x-3).$
If we let  $x=3,$  the last equation becomes  $1=5A.$  Thus,  $A={\frac {1}{5}}.$
If we let  $x=-2,$  then we get  $1=(-5)B.$  Thus,  $B=-{\frac {1}{5}}.$
So, in summation, we have
${\frac {1}{(x-3)(x+2)}}={\frac {\frac {1}{5}}{x-3}}+{\frac {-{\frac {1}{5}}}{x+2}}.$

Step 2:
Now, we have
$\int {\frac {1}{(x-3)(x+2)}}~dx=\int {\frac {\frac {1}{5}}{x-3}}~dx+\int {\frac {-{\frac {1}{5}}}{x+2}}~dx.$
Now, we use  $u$-substitution to evaluate these integrals.
For the first integral, we substitute  $u=x-3.$
For the second integral, the substitution is  $t=x+2.$
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)     $-x\cos x+\sin x+C$
(b)     ${\frac {1}{5}}\ln |x-3|-{\frac {1}{5}}\ln |x+2|+C$

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