# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{1-\sin x} d x$

Solution:

Let $\mathrm{I}=\int \frac{1}{1-\sin \mathrm{x}} \mathrm{dx}$

On multiplying and dividing $(1+\sin x)$, we can write the integral as

$I=\int \frac{1}{1-\sin x}\left(\frac{1+\sin x}{1+\sin x}\right) d x$

$\Rightarrow I=\int \frac{1+\sin x}{(1-\sin x)(1+\sin x)} d x$

$\Rightarrow I=\int \frac{1+\sin x}{1-\sin ^{2} x} d x$

$\Rightarrow I=\int \frac{1+\sin x}{\cos ^{2} x} d x\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\Rightarrow I=\int\left(\frac{1}{\cos ^{2} x}+\frac{\sin x}{\cos ^{2} x}\right) d x$

$\Rightarrow I=\int\left(\frac{1}{\cos ^{2} x}+\frac{1}{\cos x} x \frac{\sin x}{\cos x}\right) d x$

$\Rightarrow I=\int\left(\sec ^{2} x+\sec x \tan x\right) d x$

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

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

We also have $\int \sec x \tan x d x=\sec x+c$

$\therefore I=\tan x+\sec x+c$

Thus, $\int \frac{1}{1-\sin x} d x=\tan x+\sec x+c$