Question
Find the integral of the function $\tan^4 x$
✓
Answer
$\tan^{4 }x = \tan^{2 }x.\tan^{2 }x$
$= (\sec^{2 }x - 1) \tan^{2 }x$
$= \sec^{2 }x \tan^{2 }x - \tan^{2 }x$
$= \sec^{2 }x \tan^{2 }x - (\sec^{2 }x - 1)$
$= \sec^{2 }x \tan^{2 }x - \sec^{2 }x + 1$
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 $\tan x = t$
$\Rightarrow \sec^2x dx = dt$
$\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}$
$\Rightarrow \int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+C$
$= (\sec^{2 }x - 1) \tan^{2 }x$
$= \sec^{2 }x \tan^{2 }x - \tan^{2 }x$
$= \sec^{2 }x \tan^{2 }x - (\sec^{2 }x - 1)$
$= \sec^{2 }x \tan^{2 }x - \sec^{2 }x + 1$
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 $\tan x = t$
$\Rightarrow \sec^2x dx = dt$
$\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}$
$\Rightarrow \int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+C$
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