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 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 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 \tan^2(x),} we 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 \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^2x-1)~dx} \end{array}}
by using the identity again on the last equality.

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

Return to Sample Exam