009B Sample Final 3, Problem 2

Evaluate the following integrals.

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

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

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

Foundations:
1.
$\int {\frac {1}{1+x^{2}}}~dx=\arctan(x)+C$
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:

Step 2:

(c)

Step 1:

Step 2:

Final Answer:
(a)    ${\frac {\pi }{12}}$
(b)
(c)

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