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, $\int \tan x~dx=\int {\frac {-1}{u}}~du=-\ln(u)+C=-\ln |\cos x|+C$.

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 $\int \tan ^{3}x~dx=\int (\sec ^{2}x-1)\tan x~dx=\int \sec ^{2}x\tan x~dx-\int \tan x~dx$.

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

Step 3:
For the remaining integral, we also need to use $u$-substitution. First, we write Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{3}x~dx={\frac {\tan ^{2}x}{2}}-\int {\frac {\sin x}{\cos x}}~dx} .
Now, we let $u=\cos x$. Then, $du=-\sin xdx$. So, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{3}x~dx={\frac {\tan ^{2}x}{2}}+\int {\frac {1}{u}}~dx={\frac {\tan ^{2}x}{2}}+\ln |u|+C={\frac {\tan ^{2}x}{2}}+\ln |\cos x|+C} .

(b)

Step 1:
One of the double angle formulas is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \cos(2x)=1-2\sin ^{2}(x)} . Solving for Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin ^{2}(x)} , we get $\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}$.
Plugging this identity into our integral, we get $\int _{0}^{\pi }\sin ^{2}x~dx=\int _{0}^{\pi }{\frac {1-\cos(2x)}{2}}~dx=\int _{0}^{\pi }{\frac {1}{2}}~dx-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx$.

Step 2:
If we integrate the first integral, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\pi }\sin ^{2}x~dx=\left.{\frac {x}{2}}\right|_{0}^{\pi }-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx={\frac {\pi }{2}}-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx} .

Step 3:
For the remaining integral, we need to use $u$-substitution. Let $u=2x$. Then, $du=2dx$ 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{2}=2(\pi )=2\pi } .
So, the integral becomes
$\int _{0}^{\pi }\sin ^{2}x~dx={\frac {\pi }{2}}-\int _{0}^{2\pi }{\frac {\cos(u)}{4}}~du={\frac {\pi }{2}}-\left.{\frac {\sin(u)}{4}}\right|_{0}^{2\pi }={\frac {\pi }{2}}-{\bigg (}{\frac {\sin(2\pi )}{4}}-{\frac {\sin(0)}{4}}{\bigg )}={\frac {\pi }{2}}$

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

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