009B Sample Midterm 3, Problem 5

Evaluate the indefinite and definite integrals.

a) $\int \tan ^{3}x~dx$

b) $\int _{0}^{\pi }\sin ^{2}x~dx$

Foundations:
Recall the trig identities:
1. $\tan ^{2}x+1=\sec ^{2}x$
2. $\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}$
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, $du=-\sin x~dx.$ So, 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 $\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}.$
Plugging this identity into our integral, we get
${\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {\int _{0}^{\pi }{\frac {1-\cos(2x)}{2}}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi }{\frac {1}{2}}~dx-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx.}\\\end{array}}$

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

Step 3:
For the remaining integral, we need to use $u$-substitution. Let $u=2x.$ Then, $du=2~dx$ and ${\frac {du}{2}}=dx.$ Also, since this is a definite integral
and we are using $u$-substitution, we need to change the bounds of integration. We have $u_{1}=2(0)=0$ and $u_{2}=2(\pi )=2\pi .$
So, the integral becomes
${\begin{array}{rcl}\displaystyle {\int _{0}^{\pi }\sin ^{2}x~dx}&=&\displaystyle {{\frac {\pi }{2}}-\int _{0}^{2\pi }{\frac {\cos(u)}{4}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-\left.{\frac {\sin(u)}{4}}\right|_{0}^{2\pi }}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}-{\bigg (}{\frac {\sin(2\pi )}{4}}-{\frac {\sin(0)}{4}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {\pi }{2}}.}\\\end{array}}$

Final Answer:
(a) ${\frac {\tan ^{2}x}{2}}+\ln |\cos x|+C$
(b) ${\frac {\pi }{2}}$

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