009B Sample Midterm 2, Problem 5

Evaluate the integral:

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

Foundations:
1. Recall the trig identity
$\sec ^{2}x=\tan ^{2}x+1$
2. Also,
$\int \sec ^{2}x~dx=\tan x+C$
3. How would you integrate  $\int \sec ^{2}(x)\tan(x)~dx?$
You can use  $u$-substitution.
Let  $u=\tan x.$
Then,  $du=\sec ^{2}(x)dx.$
Thus,
${\begin{array}{rcl}\displaystyle {\int \sec ^{2}(x)\tan(x)~dx}&=&\displaystyle {\int u~du}\\&&\\&=&\displaystyle {{\frac {u^{2}}{2}}+C}\\&&\\&=&\displaystyle {{\frac {\tan ^{2}x}{2}}+C.}\end{array}}$

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
${\begin{array}{rcl}\displaystyle {\int \tan ^{4}(x)~dx}&=&\displaystyle {\int \tan ^{2}(x)(\sec ^{2}(x)-1)~dx}\\&&\\&=&\displaystyle {\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int \tan ^{2}(x)~dx}\\&&\\&=&\displaystyle {\int \tan ^{2}(x)\sec ^{2}(x)~dx-\int (\sec ^{2}x-1)~dx}\end{array}}$
by 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
${\begin{array}{rcl}\displaystyle {\int \tan ^{4}(x)~dx}&=&\displaystyle {{\frac {u^{3}}{3}}-(\tan(x)-x)+C}\\&&\\&=&\displaystyle {{\frac {\tan ^{3}(x)}{3}}-\tan(x)+x+C.}\end{array}}$

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

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