# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{5 \cos ^{3} x+6 \sin ^{3} x}{2 \sin ^{2} x \cos ^{2} x} d x$

Solution:

Given:

$\int \frac{5 \cos ^{3} x+6 \sin ^{3} x}{2 \sin ^{2} x \cos ^{2} x} d x$

By Splitting we get,

$\Rightarrow \int \frac{5 \cos ^{3} x}{2 \sin ^{2} x \cos ^{2} x} d x+\int \frac{6 \sin ^{3} x}{2 \sin ^{2} x \cos ^{2} x} d x$

$\Rightarrow \frac{5}{2} \int \frac{\cos x \cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+3 \int \frac{\sin ^{2} x \sin ^{1} x}{\sin ^{2} x \cos ^{2} x} d x$

$\Rightarrow \frac{5}{2} \int \frac{\cos x}{\sin ^{2} x} d x+3 \int \frac{\sin ^{1} x}{1 \cos ^{2} x} d x$

We know that,

$\int 1 \frac{\cos x}{\sin x} d x=\cot x$

$\int \frac{\sin x}{\cos x} d x=\tan x$

$\int 1 \frac{1}{\sin x} d x=\sec x$

$\int 1 \frac{1}{\sin x} d x=\operatorname{cosecx}$

$\Rightarrow \frac{5}{2} \int \cot x \operatorname{cosec} x d x+3 \int \sec x \tan x d x$

We know that,

$\int \cot x \operatorname{cosec} x d x=-\operatorname{cosec} x$

$\int \sec x \tan x d x=\sec x$

$\Rightarrow \frac{5}{2}(-\operatorname{cosec} x)+3 \sec x+c$

$I=-\frac{5}{2} \operatorname{cosec} x+3 \sec x+c$