Prove
Question:

$\frac{1}{\sin x \cos ^{3} x}$

Solution:

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

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

$\therefore \int \frac{1}{\sin x \cos ^{3} x} d x=\int \tan x \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$\Rightarrow \int \frac{1}{\sin x \cos ^{3} x} d x=\int t d t+\int_{t}^{1} d t$

$=\frac{t^{2}}{2}+\log |t|+\mathrm{C}$

$=\frac{1}{2} \tan ^{2} x+\log |\tan x|+\mathrm{C}$

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