# 009B Sample Final 3, Problem 2

Evaluate the following integrals.

(a)  $\int _{0}^{\frac {\sqrt {3}}{4}}{\frac {1}{1+16x^{2}}}~dx$ (b)  $\int {\frac {x^{2}}{(1+x^{3})^{2}}}~dx$ (c)  $\int _{1}^{e}{\frac {\cos(\ln(x))}{x}}~dx$ Foundations:
1. Recall
$\int {\frac {1}{1+x^{2}}}~dx=\arctan(x)+C$ 2. How would you integrate   $\int {\frac {\ln x}{x}}~dx?$ You could use  $u$ -substitution.

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

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

(a)

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

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

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

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

${\begin{array}{rcl}\displaystyle {\int _{1}^{e}{\frac {\cos(\ln(x))}{x}}~dx}&=&\displaystyle {\int _{0}^{1}\cos(u)~du}\\&&\\&=&\displaystyle {\sin(u){\bigg |}_{0}^{1}}\\&&\\&=&\displaystyle {\sin(1)-\sin(0)}\\&&\\&=&\displaystyle {\sin(1).}\end{array}}$ (a)    ${\frac {\pi }{12}}$ (b)    $-{\frac {1}{3(1+x^{3})}}+C$ (c)    $\sin(1)$ 