# Find the integrals of the functions$\frac{1}{\sin x\cos^3x}$

$\begin{array}{1 1} \frac{\tan^2x}{2}+\log|\tan x|+c \\ \frac{\tan^2x}{2}-\log|\tan x|+c \\ \frac{\cot^2x}{2}+\log|\tan x|+c \\ \frac{\tan^2x}{2}+\log|\cot x|+c \end{array}$

Toolbox:
• $(i)\;\sin^2x+\cos^2x=1.$
• $(ii)\;Method\;of\;substitution.$
• Let f(x)=t.
• Therefore f'(x)dx=dt.Thus $\int f(x)dx=\int t.dt.$
Given:$I=\int\frac{1}{\sin x\cos x}dx.$

This can be written as,

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

Separating the terms we get,

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

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

$\;\;\;=\int\tan x.\sec xdx+\int\frac{\sec^2}{\tan x}dx.$

Let tan x=t.

On differentiating we get,

$\sec^2xdx=dt.$

By substituting t we get,

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

On integrating we get,

$\;\;\;=\frac{t^2}{2}+log |t|+c.$

Substituting for t we get,

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