Question
Find the integral of the function tan4 x

Answer

tanx = tanx.tanx
= (secx - 1) tanx
= secx tanx - tanx
= secx tanx - (secx - 1)
= secx tanx - secx + 1
Now, $\int \tan ^{4} x d x=\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x+\int 1 d x$ 
$= \int \sec ^{2} x \tan ^{2} x d x-\tan x+x+C$ 
Now, let tanx = t
$\Rightarrow$ sec2x dx = dt
$\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}$ 
$\Rightarrow \int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+C$

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