Question

Evaluate: $\int {\tan }^{3}xdx$

Answer 1

Here $\int {\tan }^{3}xdx$

$=\int \tan x.{\tan }^{3}xdx=\int \tan x({\sec }^{2}x-1)dx$

$=\int \tan x.{\sec }^{2}xdx-\int \tan xdx$

In the first integral, put $u=\tan x$ so that

$\frac{du}{dx}={\sec }^{2}x$

$=\int u.du-\log \left|\sec x\right|+c=\frac{u^2}{2}-\log \left|\sec x\right|+c$

$\therefore \int {\tan }^{3}xdx=\frac{{\tan }^{2}x}{2}-\log \left|\sec x\right|+c$