Question

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

Answer 1

Consider the given integral.

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

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

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

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

$I=I_1-I_2$ ....... $\left(1\right)$

Let $I_1=\int {\sec }^{2}x{\tan }^{2}xdx$

Put $t=\tan x$
$dt={\sec }^{2}xdx$

Therefore,

$I_1=\int t^{2}dt$

$I_1=\frac{t^3}{3}+C$

$I_1=\frac{{\tan }^{3}x}{3}+C$

Now,

$I_2=\int {\tan }^{2}xdx$

$I_2=\int ({\sec }^{2}x-1)dx$

$I_2=\tan x-x+C$

From equation $\left(1\right)$,

$I=\frac{{\tan }^{3}x}{3}-\tan x+x+C$

Hence, this is the answer.