Question

Find : $\int \frac{{\sin }^{2}x-{\cos }^{2}x}{\sin x\cos x}dx$

Answer 1

$\int \frac{{\sin }^{2}x-{\cos }^{2}x}{\sin x\cos x}dx$

= $-\int \frac{{\cos }^{2}x-{\sin }^{2}x}{\sin x\cos x}dx$

= $-2\int \frac{\cos 2x}{2\sin x\cos x}dx$

$=-2\int \frac{\cos 2x}{\sin 2x}dx$

$=\text{-2}\int \cot 2xdx$

$=-2\frac{\log |\sin 2x|}{2}+C=\text{-}\log |\sin 2x|+C=\log |\text{cosec}2x|+C$