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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Final Answer:
(a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\pi}{12}}
(b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{3(1+x^3)}+C}
(c)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(1)}

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