Question

Find the following integral: $\int \frac{\sec^2 x}{\cosec^2 x}dx$

Answer 1

$\int \frac{\sec^2 x}{\cosec^2 x}dx$

= $\int \frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}}dx$

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

=$\int \tan^2 x dx$

= $\int (\sec^2 x-1)dx$

= $\int \sec^2 xdx - \int 1dx$

= $\tan x -x+C$