009B Sample Midterm 2, Problem 5 Detailed Solution

Evaluate the integral:

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

Background Information:
1. Recall the trig identity
$\sec ^{2}x=\tan ^{2}x+1$
2. Recall
$\int \sec ^{2}x~dx=\tan x+C$
3. How would you integrate  $\int \sec ^{2}(x)\tan(x)~dx?$
You can use  $u$-substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\tan x.}
Then,  $du=\sec ^{2}(x)dx.$
Thus,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{4}(x)~dx=\int \tan ^{2}(x)\tan ^{2}(x)~dx.}
Using the trig identity  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sec ^{2}(x)=\tan ^{2}(x)+1,}
we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \tan ^{2}(x)=\sec ^{2}(x)-1.}
Plugging in the last identity into one of the  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \tan ^{2}(x),}   we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\tan(x).}
Then,  $du=\sec ^{2}(x)dx.$
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \tan ^{4}(x)~dx=\int u^{2}~du-\int (\sec ^{2}(x)-1)~dx.}

Step 3:
We integrate to get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {\tan ^{3}(x)}{3}}-\tan(x)+x+C}

Return to Sample Exam