Question

Find the integral of the function
${\tan }^{4}x$

Answer 1

Consider the given integral.

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

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

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

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

$I=I_1-I_2$ ……… (1)

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

Put $t=\tan x$

$dt={\sec }^{2}xdx$

Therefore,

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

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

On putting the value of $t$, we get

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

Now,

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

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

$I_2=\tan x-x+C$

From equation (1),

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

Hence, this is the answer.