Question

Find the integrals of the functions ${\tan }^{4}x$

Answer 1

${\tan }^{4}x={\tan }^{2}x\cdot {\tan }^{2}x$
$=({\sec }^{2}x-1){\tan }^{2}x={\sec }^{2}x{\tan }^{2}x-{\tan }^{2}x$
$={\sec }^{2}x{\tan }^{2}x-({\sec }^{2}x-1)$
$={\sec }^{2}x{\tan }^{2}x-{\sec }^{2}x+1$
$\therefore \int {\tan }^{4}xdx=\int {\sec }^{2}x{\tan }^{2}xdx-\int {\sec }^{2}xdx+\int 1\cdot dx$
$=\int {\sec }^{2}x{\tan }^{2}xdx-\tan x+x+C$ .........(i)
Consider $\int {\sec }^{2}x{\tan }^{2}xdx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow \int {\sec }^{2}x{\tan }^{2}xdx=\int t^{2}dt=\frac{t^3}{3}=\frac{{\tan }^{3}x}{3}$
From equation (1), we obtain
$\int {\tan }^{4}xdx=\frac{1}{3}{\tan }^{3}x-\tan x+x+C$