Question

Find the value of $\int\frac{dx}{\,sin^2x\,cos^2x}$ = ?

Answer 1

Let I = $\int\frac{dx}{\,sin^2x\,cos^2x}$
=$\int\frac{1}{\,sin^2x\,cos^2x}dx$
=$\int\frac{sin^2x+cos^2x}{\,sin^2x\,cos^2x}$
=$\int\frac{sin^2x}{\,sin^2x\,cos^2x}dx+\int\frac{cos^2x}{\,sin^2x\,cos^2x}dx$
=$\int sec^2x\,dx+ \int cosec^2x\,dx$
= $tan\,x-cot\,x +c$