009B Sample Midterm 3, Problem 5 Detailed Solution

Evaluate the indefinite and definite integrals.

(a)   $\int x\ln x~dx$

(b)   $\int _{0}^{\pi }\sin ^{2}x~dx$

---

ExpandBackground 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)

ExpandStep 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  $v={\frac {x^{2}}{2}}.$

ExpandStep 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)

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

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

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

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

Return to Sample Exam