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
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 \int \tan^4(x)~dx= \frac{u^3}{3}-(\tan(x)-x)+C=\frac{\tan^3(x)}{3}-\tan(x)+x+C} .

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

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