Q.Evaluate : $\int \dfrac{1}{\sin^{2}x\,\cos^{2}x}\,dx$

Replace $1$ by $\sin^2x+\cos^2x$ in the numerator; the integral splits into $\int\sec^2x\,dx+\int\operatorname{cosec}^2x\,dx=\tan x-\cot x+c$.

Rewrite the numerator. Since $\sin^2x+\cos^2x=1$,

$\int\frac{dx}{\sin^2x\cos^2x}=\int\frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}\,dx.$

Split the fraction. …