Write a value


Write a value of $\int \tan ^{3} x \sec ^{2} x d x$.


let $\tan x=t$

Differentiating on both sides we get,

$\sec ^{2} x d x=d t$

Substituting above equation in $\int \tan ^{3} x \sec ^{2} x d x$ we get,

$=\int \mathrm{t}^{3} \mathrm{dt}$


$=\frac{\tan ^{4} x}{4}+c$

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