Question

evaluate the following integral
integral of cos^2x/sin^4x dx

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{{\cos }^{2}x}{{\sin }^{4}x}dx \\ =\int \frac{{\cos }^{2}x}{{\sin }^{2}x}\times \frac{1}{{\sin }^{2}x}dx \\ =\int {\cot }^{2}x{\mathrm{cosec}}^{2}xdx \\ \mathrm{Putting}\cot x=t \\ \Rightarrow -{\mathrm{cosec}}^{2}xdx=dt \\ \therefore I=\int -t^{2}dt \\ =\frac{-t^3}{3}+C \\ =-\frac{1}{3}{\cot }^{3}x+C \end{gathered}$$