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# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\sin x \cos ^{3} x} d x$We know, $\sin ^{2} x+\cos ^{2} x=1$

Solution:

We know, $\sin ^{2} x+\cos ^{2} x=1$

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

Divide each term of numerator separately by $\sin x \cos ^{3} x$

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

$=\frac{\sin x}{\cos x} *\left(\frac{1}{\cos ^{2} x}\right)+\frac{\frac{1}{\cos ^{2} x}}{\frac{\sin x \cos x}{\cos ^{2} x}}$ (divide second term each by $\cos ^{2} x$ )

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

Therefore,

$\int \frac{1}{\sin x \cos ^{3} x} d x=\int\left(\tan x \sec ^{2} x+\frac{\sec ^{2} x}{\tan x}\right) d x$

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

Put $\tan x=t, d t=\sec ^{2} x d x$

$=\int \tan x \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x=\int t d t+\int \frac{1}{t} d t$

$=\frac{\mathrm{t}^{2}}{2}+\operatorname{lot} \mathrm{t}+\mathrm{c}=\frac{1}{2} \tan ^{2} \mathrm{x}+\log (\tan \mathrm{x})+\mathrm{c}$