Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\sin x \cos ^{2} x} d x$

Solution:

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

$\Rightarrow \int \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos ^{2} x}$

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

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

$\Rightarrow \int \tan x \sec x d x+\int \csc x d x$

$d(\sec x)=\tan x \cdot \sec x$

$\therefore \int \tan x \sec x d x=\sec x+c$

$\therefore \int \tan x \sec x d x+\int \csc x d x$

$\because \int \csc x d x=\log \left|\tan \frac{x}{2}\right|+c$

$\Rightarrow \sec x+\log \left|\tan \frac{x}{2}\right|+c$

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