Find the integral of the function ^4 x - Vidyadip

Question

Find the integral of the function $\tan^4 x$

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Answer

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

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