# 009B Sample Midterm 3, Problem 5 Detailed Solution

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Evaluate the indefinite and definite integrals.

(a)   ${\displaystyle \int x\ln x~dx}$

(b)   ${\displaystyle \int _{0}^{\pi }\sin ^{2}x~dx}$

Background Information:
1. Integration by parts tells us that
${\displaystyle \int u~dv=uv-\int v~du.}$
2. Recall the trig identity
${\displaystyle \sin ^{2}(x)={\frac {1-\cos(2x)}{2}}.}$

Solution:

(a)

Step 1:
To evaluate this integral, we use integration by parts.
Let  ${\displaystyle u=\ln x}$  and  ${\displaystyle dv=x~dx.}$
Then,  ${\displaystyle du={\frac {1}{x}}dx}$  and  ${\displaystyle v={\frac {x^{2}}{2}}.}$
Step 2:
Using integration by parts, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {\int x\ln x~dx}&=&\displaystyle {{\frac {x^{2}\ln x}{2}}-\int {\frac {x}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}\ln x}{2}}-{\frac {x^{2}}{4}}+C.}\end{array}}}$

(b)

Step 1:
One of the double angle formulas is
${\displaystyle \cos(2x)=1-2\sin ^{2}(x).}$
Solving for  ${\displaystyle \sin ^{2}(x),}$  we get
${\displaystyle \sin ^{2}(x)={\frac {1-\cos(2x)}{2}}.}$
Plugging this identity into our integral, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {\int _{0}^{\pi }{\frac {1-\cos(2x)}{2}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi }{\frac {1}{2}}~dx-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx.}\\\end{array}}}$

Step 2:
If we integrate the first integral, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {\left.{\frac {x}{2}}\right|_{0}^{\pi }-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx.}\\\end{array}}}$

Step 3:
For the remaining integral, we need to use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=2x.}$
Then,  ${\displaystyle du=2~dx}$  and  ${\displaystyle {\frac {du}{2}}=dx.}$
Also, since this is a definite integral and we are using ${\displaystyle u}$-substitution,
we need to change the bounds of integration.
We have  ${\displaystyle u_{1}=2(0)=0}$  and  ${\displaystyle u_{2}=2(\pi )=2\pi .}$
So, the integral becomes

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {{\frac {\pi }{2}}-\int _{0}^{2\pi }{\frac {\cos(u)}{4}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-\left.{\frac {\sin(u)}{4}}\right|_{0}^{2\pi }}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-{\bigg (}{\frac {\sin(2\pi )}{4}}-{\frac {\sin(0)}{4}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}.}\\\end{array}}}$

Final Answer:
(a)     ${\displaystyle {\frac {x^{2}\ln x}{2}}-{\frac {x^{2}}{4}}+C}$
(b)     ${\displaystyle {\frac {\pi }{2}}}$