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,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \sec ^{2}x~dx=\tan x+C}
3. How would you integrate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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,
${\begin{array}{rcl}\displaystyle {\int \sec ^{2}(x)\tan(x)~dx}&=&\displaystyle {\int u~du}\\&&\\&=&\displaystyle {{\frac {u^{2}}{2}}+C}\\&&\\&=&\displaystyle {{\frac {\tan ^{2}x}{2}}+C.}\end{array}}$

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  $\tan ^{2}(x),$  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
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)\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
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}

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