# 009B Sample Midterm 3, Problem 5 Detailed Solution

Evaluate the indefinite and definite integrals.

(a)   $\int x\ln x~dx$ (b)   $\int _{0}^{\pi }\sin ^{2}x~dx$ Background Information:
1. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$ 2. Recall the trig identity
$\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}.$ Solution:

(a)

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

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

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

${\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {\int _{0}^{\pi }{\frac {1-\cos(2x)}{2}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi }{\frac {1}{2}}~dx-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx.}\\\end{array}}$ Step 2:
If we integrate the first integral, we get

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

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