009B Sample Midterm 2, Problem 5

Evaluate the integral:

$\int \tan ^{4}x~dx$

Foundations:
Recall:
1. $\sec ^{2}x=tan^{2}x+1$
2. $\int \sec ^{2}x~dx=\tan x+C$
How would you integrate $\int \sec ^{2}(x)\tan(x)~dx$?
You could use $u$-substitution. Let $u=\tan x$. Then, $du=\sec ^{2}(x)dx$.
Thus, $\int \sec ^{2}(x)\tan(x)~dx=\int u~du={\frac {u^{2}}{2}}+C={\frac {\tan ^{2}x}{2}}+C$.

Solution:

Step 1:
First, we write $\int \tan ^{4}(x)~dx=\int \tan ^{2}(x)\tan ^{2}(x)~dx$.
Using the trig identity $\sec ^{2}(x)=\tan ^{2}(x)+1$, we have $\tan ^{2}(x)=\sec ^{2}(x)-1$.
Plugging in the last identity into one of the $\tan ^{2}(x)$, we get
$\int \tan ^{4}(x)~dx=\int \tan ^{2}(x)(\sec ^{2}(x)-1)~dx=\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int \tan ^{2}(x)~dx=\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int (\sec ^{2}x-1)~dx$,
using the identity again on the last equality.

Step 2:
So, we have $\int \tan ^{4}(x)~dx=\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int (\sec ^{2}x-1)~dx$.
For the first integral, we need to use $u$-substitution. Let $u=\tan(x)$. Then, $du=\sec ^{2}(x)dx$.
So, we have
$\int \tan ^{4}(x)~dx=\int u^{2}~du-\int (\sec ^{2}(x)-1)~dx$.

Step 3:
We integrate to get
$\int \tan ^{4}(x)~dx={\frac {u^{3}}{3}}-(\tan(x)-x)+C={\frac {\tan ^{3}(x)}{3}}-\tan(x)+x+C$.

Final Answer:
${\frac {\tan ^{3}(x)}{3}}-\tan(x)+x+C$

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