Question

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

Answer 1

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

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

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

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

Let $I_1=\int \tan x{\sec }^{2}xdx$ and $I_2=\int \tan xdx$

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

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

$\Rightarrow \int tdt=\frac{t^2}{2}+c=\frac{{\tan }^{2}x}{2}+c$

$\therefore I_1=\frac{{\tan }^{2}x}{2}+c$

Consider $I_2=\int \tan xdx$

$=\int \frac{\sin x}{\cos x}dx$

Let $t=\cos x\Rightarrow dt=-\sin xdx$

$\Rightarrow I_2=-\int \frac{dt}{t}=-\log t=\log \frac{1}{t}=\log \frac{1}{\cos x}=\log \sec x+c$

$\therefore \int {\tan }^{3}xdx=I_1+I_2$

$=\frac{{\tan }^{2}x}{2}+\log \sec x+c$