Question:
Write a value of $\int \tan x \sec ^{3} x d x$.
Solution:
given $\int \tan x \sec ^{3} x d x$
$=\int(\tan x \sec x) \sec ^{2} x d x$
Let $\sec x=t$
Differentiating on both sides we get,
$\tan x \sec x d x=d t$
Substituting above equation in $\int \tan x \sec ^{3} x d x$ we get,
$=\int t^{2} d t$
$=\frac{t^{3}}{3}+c$
$=\frac{\sec ^{3} x}{3}+c$
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