Question

How do I evaluate `int(sec^2(x))/(tan^2(x)) d x`?

Answer 1

The answer is: `-1/tanx+c`.

`intsec^2x/tan^2xdx=int1/cos^2x*tan^(-2)xdx=`

`=tan^(-2+1)x/(-2+1)+c=-1/tanx+c`.

This is because:

`int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c` and the derivative of the function `tanx` is `1/cos^2x.`