Question

Find the integral of the function $\frac{1}{\sin x{\cos }^{3}x}$

Answer 1

$\frac{1}{\sin x{\cos }^{3}x}=\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x{\cos }^{3}x}$
$=\frac{\sin x}{{\cos }^{3}x}+\frac{1}{\sin x\cos x}$

$=\tan x{\sec }^{2}x+\frac{\frac{1}{{\cos }^{2}x}}{\frac{\sin x\cos x}{{\cos }^{2}x}}$

$=\tan x{\sec }^{2}x+\frac{{\sec }^{2}x}{\tan x}$
$\therefore \int \frac{1}{\sin x{\cos }^{3}x}dx=\int \tan x{\sec }^{2}xdx+\int \frac{{\sec }^{2}x}{\tan x}dx$

Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow \int \frac{1}{\sin x{\cos }^{3}x}dx=\int tdt+\int \frac{1}{t}dt$

$=\frac{t^2}{2}+\log |t|+C=\frac{1}{2}{\tan }^{2}x+\log |\tan x|+C$