Question

How to find integral of `sinx/cos^2x`?

Answer 1

`intsinx/(cos^2x)dx=secx`

Explanation:

To find `intsinx/(cos^2x)dx`, let `t=cosx`, then as `dt=-sinxdx`

`intsinx/(cos^2x)dx=int(-dt)/t^2`

= `-(t^(-1))/(-1)`

= `1/t`

= `1/cosx`

= `secx`

Observe that `d/(dx)secx=secxtanx=sinx/cos^2x`