Question

What is the integral of `int secx/tan^2x dx`?

Answer 1

`-cscx+C`

Explanation:

Let's rewrite the integrand:

`intsecx/tan^2xdx=int(1/cosx)/(sin^2x/cos^2x)dx=intcosx/sin^2x`

Here, recognize that `intcosx/sin^2xdx=intcosx/sinx1/sinxdx=intcotxcscxdx=-cscx+C`.

Another way we can solve `intcosx/sin^2xdx` is by letting `u=sinx` so `du=cosxdx`. Then,

`intcosx/sin^2xdx=intu^-2du=-u^-1+C=-1/sinx+C=-cscx+C`.