009B Sample Midterm 3, Problem 5

Evaluate the indefinite and definite integrals.

a)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{3}x~dx}

b)   $\int _{0}^{\pi }\sin ^{2}x~dx$

Foundations:
1. Recall the trig identity
$\tan ^{2}x+1=\sec ^{2}x$
2. Recall the trig identity
$\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}$
3. How would you integrate $\tan x~dx?$
You could use $u$-substitution. First, write $\int \tan x~dx=\int {\frac {\sin x}{\cos x}}~dx.$
Now, let $u=\cos(x).$ Then, $du=-\sin(x)dx.$ Thus,
${\begin{array}{rcl}\displaystyle {\int \tan x~dx}&=&\displaystyle {\int {\frac {-1}{u}}~du}\\&&\\&=&\displaystyle {-\ln(u)+C}\\&&\\&=&\displaystyle {-\ln |\cos x|+C.}\\\end{array}}$

Solution:

(a)

Step 1:
We start by writing
$\int \tan ^{3}x~dx=\int \tan ^{2}x\tan x~dx.$
Since $\tan ^{2}x=\sec ^{2}x-1,$ we have
${\begin{array}{rcl}\displaystyle {\int \tan ^{3}x~dx}&=&\displaystyle {\int (\sec ^{2}x-1)\tan x~dx}\\&&\\&=&\displaystyle {\int \sec ^{2}x\tan x~dx-\int \tan x~dx.}\\\end{array}}$

Step 2:
Now, we need to use $u$-substitution for the first integral.
Let $u=\tan(x).$ Then, $du=\sec ^{2}x~dx.$ So, we have
${\begin{array}{rcl}\displaystyle {\int \tan ^{3}x~dx}&=&\displaystyle {\int u~du-\int \tan x~dx}\\&&\\&=&\displaystyle {{\frac {u^{2}}{2}}-\int \tan x~dx}\\&&\\&=&\displaystyle {{\frac {\tan ^{2}x}{2}}-\int \tan x~dx.}\\\end{array}}$

Step 3:
For the remaining integral, we also need to use $u$-substitution.
First, we write
$\int \tan ^{3}x~dx={\frac {\tan ^{2}x}{2}}-\int {\frac {\sin x}{\cos x}}~dx.$
Now, we let $u=\cos x.$ Then, $du=-\sin x~dx.$
Therefore, we get
${\begin{array}{rcl}\displaystyle {\int \tan ^{3}x~dx}&=&\displaystyle {{\frac {\tan ^{2}x}{2}}+\int {\frac {1}{u}}~dx}\\&&\\&=&\displaystyle {{\frac {\tan ^{2}x}{2}}+\ln |u|+C}\\&&\\&=&\displaystyle {{\frac {\tan ^{2}x}{2}}+\ln |\cos x|+C.}\end{array}}$

(b)

Step 1:
One of the double angle formulas is $\cos(2x)=1-2\sin ^{2}(x).$
Solving for $\sin ^{2}(x),$ we get
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^2(x)=\frac{1-\cos(2x)}{2}.}
Plugging this identity into our integral, we get
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_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\ &&\\ & = & \displaystyle{\int_0^\pi \frac{1}{2}~dx-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\ \end{array}}

Step 2:
If we integrate the first integral, we get
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_0^\pi \sin^2x~dx} & = & \displaystyle{\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\ \end{array}}

Step 3:
For the remaining integral, we need to use 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 u} -substitution.
Let 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 u=2x.} Then, 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 du=2~dx} and 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{du}{2}=dx.}
Also, since this is a definite integral and we are using 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 u} -substitution,
we need to change the bounds of integration.
We 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 u_1=2(0)=0} and 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 u_2=2(\pi)=2\pi.}
So, the integral becomes
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_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}-\left.\frac{\sin(u)}{4}\right|_0^{2\pi}}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}-\bigg(\frac{\sin(2\pi)}{4}-\frac{\sin(0)}{4}\bigg)}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}.}\\ \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{\tan^2x}{2}+\ln |\cos x|+C}
(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{\pi}{2}}

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