# 009B Sample Midterm 1, Problem 2 Detailed Solution

Evaluate the indefinite and definite integrals.

(a)   ${\displaystyle \int x^{2}{\sqrt {1+x^{3}}}~dx}$

(b)   ${\displaystyle \int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\cos(x)}{\sin ^{2}(x)}}~dx}$

Background Information:
How would you integrate   ${\displaystyle \int {\frac {\ln x}{x}}~dx?}$

You can use  ${\displaystyle u}$-substitution.

Let  ${\displaystyle u=\ln(x).}$
Then,  ${\displaystyle du={\frac {1}{x}}dx.}$

Thus,

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

Solution:

(a)

Step 1:
We use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=1+x^{3}.}$
Then,  ${\displaystyle du=3x^{2}dx}$  and  ${\displaystyle {\frac {du}{3}}=x^{2}dx.}$
Therefore, the integral becomes
${\displaystyle {\frac {1}{3}}\int {\sqrt {u}}~du.}$
Step 2:
We now have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int x^{2}{\sqrt {1+x^{3}}}~dx}&=&\displaystyle {{\frac {1}{3}}\int {\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {2}{9}}u^{\frac {3}{2}}+C}\\&&\\&=&\displaystyle {{\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C.}\end{array}}}$

(b)

Step 1:
We use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=\sin(x).}$
Then,  ${\displaystyle du=\cos(x)dx.}$
Also, we need to change the bounds of integration.
Plugging in our values into the equation  ${\displaystyle u=\sin(x),}$  we get
${\displaystyle u_{1}=\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}={\frac {\sqrt {2}}{2}}}$  and  ${\displaystyle u_{2}=\sin {\bigg (}{\frac {\pi }{2}}{\bigg )}=1.}$
Therefore, the integral becomes
${\displaystyle \int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du.}$
Step 2:
We now have

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\cos(x)}{\sin ^{2}(x)}}~dx}&=&\displaystyle {\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du}\\&&\\&=&\displaystyle {\left.{\frac {-1}{u}}\right|_{\frac {\sqrt {2}}{2}}^{1}}\\&&\\&=&\displaystyle {-{\frac {1}{1}}-{\frac {-1}{\frac {\sqrt {2}}{2}}}}\\&&\\&=&\displaystyle {-1+{\sqrt {2}}.}\end{array}}}$

(a)     ${\displaystyle {\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C}$
(b)     ${\displaystyle -1+{\sqrt {2}}}$