Question

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

Answer 1

$\int \frac{1}{\sin x{\cos }^{3}x}.dx$

$=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x{\cos }^{3}x}.dx$

$[\because {\sin }^{2}x+{\cos }^{2}x=1]$

$=\int (\frac{{\sin }^{2}x}{\sin x.{\cos }^{3}x}+\frac{{\cos }^{2}x}{\sin x{\cos }^{3}x}).dx$

$=\int \frac{1}{{\cos }^{2}x}(\frac{{\sin }^{2}x}{\sin x.\cos x}+\frac{{\cos }^{2}x}{\sin x.\cos x}).dx$

$=\int {\sec }^{2}x(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}).dx$

$=\int {\sec }^{2}x(\tan x+\cot x).dx$

$=\int {\sec }^{2}x(\tan x+\frac{1}{\tan x}).dx$

$=\int (\tan x+\frac{1}{\tan x}).{\sec }^{2}x.dx$

Putting $\tan x=t$

Differentiating w.r.t.$x$

$\Rightarrow {\sec }^{2}x=\frac{dt}{dx}\Rightarrow se{c}^{2}xdx=dt$

Thus,

$=\int (\tan x+\frac{1}{\tan x}).{\sec }^{2}x.dx$

$=\int (t+\frac{1}{t}).dt$

$=\int t.dt+\int \frac{1}{t}.dt$

$=\frac{t^2}{2}+\text{log}|t|+C$

Putting value of $t=\tan x$

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

Where $C$ is constant of integration.