# 009B Sample Final 3, Problem 2

Evaluate the following integrals.

(a)  ${\displaystyle \int _{0}^{\frac {\sqrt {3}}{4}}{\frac {1}{1+16x^{2}}}~dx}$

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

(c)  ${\displaystyle \int _{1}^{e}{\frac {\cos(\ln(x))}{x}}~dx}$

Foundations:
1. Recall
${\displaystyle \int {\frac {1}{1+x^{2}}}~dx=\arctan(x)+C}$
2. How would you integrate   ${\displaystyle \int {\frac {\ln x}{x}}~dx?}$

You could 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:
First, we notice
${\displaystyle \int _{0}^{\frac {\sqrt {3}}{4}}{\frac {1}{1+16x^{2}}}~dx=\int _{0}^{\frac {\sqrt {3}}{4}}{\frac {1}{1+(4x)^{2}}}~dx.}$
Now, we use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=4x.}$
Then,  ${\displaystyle du=4dx}$  and  ${\displaystyle {\frac {du}{4}}=dx.}$
Also, we need to change the bounds of integration.
Plugging in our values into the equation  ${\displaystyle u=4x,}$  we get
${\displaystyle u_{1}=4(0)=0}$  and  ${\displaystyle u_{2}=4{\bigg (}{\frac {\sqrt {3}}{4}}{\bigg )}={\sqrt {3}}.}$
Therefore, the integral becomes
${\displaystyle {\frac {1}{4}}\int _{0}^{\sqrt {3}}{\frac {1}{1+u^{2}}}~du.}$
Step 2:
We now have

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\frac {\sqrt {3}}{4}}{\frac {1}{1+16x^{2}}}~dx}&=&\displaystyle {{\frac {1}{4}}\int _{0}^{\sqrt {3}}{\frac {1}{1+u^{2}}}~du}\\&&\\&=&\displaystyle {{\frac {1}{4}}\arctan(u){\bigg |}_{0}^{\sqrt {3}}}\\&&\\&=&\displaystyle {{\frac {1}{4}}\arctan({\sqrt {3}})-{\frac {1}{4}}\arctan(0)}\\&&\\&=&\displaystyle {{\frac {1}{4}}{\bigg (}{\frac {\pi }{3}}{\bigg )}-0}\\&&\\&=&\displaystyle {{\frac {\pi }{12}}.}\end{array}}}$

(b)

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
${\displaystyle {\frac {1}{3}}\int {\frac {1}{u^{2}}}~du.}$
Step 2:
We now have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {x^{2}}{(1+x^{3})^{2}}}~dx}&=&\displaystyle {{\frac {1}{3}}\int {\frac {1}{u^{2}}}~du}\\&&\\&=&\displaystyle {-{\frac {1}{3u}}+C}\\&&\\&=&\displaystyle {-{\frac {1}{3(1+x^{3})}}+C.}\end{array}}}$

(c)

Step 1:
We use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=\ln(x).}$
Then,  ${\displaystyle du={\frac {1}{x}}dx.}$
Also, we need to change the bounds of integration.
Plugging in our values into the equation  ${\displaystyle u=\ln(x),}$
we get
${\displaystyle u_{1}=\ln(1)=0}$  and  ${\displaystyle u_{2}=\ln(e)=1.}$
Therefore, the integral becomes
${\displaystyle \int _{0}^{1}\cos(u)~du.}$
Step 2:
We now have

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{e}{\frac {\cos(\ln(x))}{x}}~dx}&=&\displaystyle {\int _{0}^{1}\cos(u)~du}\\&&\\&=&\displaystyle {\sin(u){\bigg |}_{0}^{1}}\\&&\\&=&\displaystyle {\sin(1)-\sin(0)}\\&&\\&=&\displaystyle {\sin(1).}\end{array}}}$

(a)    ${\displaystyle {\frac {\pi }{12}}}$
(b)    ${\displaystyle -{\frac {1}{3(1+x^{3})}}+C}$
(c)    ${\displaystyle \sin(1)}$