Question

Evaluate $\int {\tan }^{4}xdx$

Answer 1

$\int {\tan }^{4}xdx$

$=\int {\tan }^2{\tan }^{2}xdx$

$=\int \left({\sec }^{2}x-1\right){\tan }^{2}xdx$

$=\int \left({\sec }^{2}x{\tan }^{2}x-{\tan }^{2}x\right)dx$

$=\int \left[{\sec }^{2}x{\tan }^{2}x-\left({\sec }^{2}x-1\right)\right]dx$

$=\int \left({\sec }^{2}x{\tan }^{2}x-{\sec }^{2}x+1\right)dx$

$=\int {\sec }^{2}x{\tan }^{2}xdx-\int {\sec }^{2}xdx+\int 1dx$

Putting $tanx=t$

$se{c}^{2}xdx=dt$

$=\int t^{2}dt-\tan x+x$

$=\frac{t^3}{3}-\tan x+x+c$

$=\frac{1}{3}{\tan }^{3}x-\tan x+x+c$