Question

What is the integral of `Sin(2x)/(1+cos^2(x))`?

Answer 1

`int(sin2x)/(1+cos^2x)dx=-ln(1+cos^2x)+C`

Explanation:

At first glance, this one seems like a toughie - but it in fact can be solved with a clever application of trig identities.

Note that `d/dxcos^2x=d/dx(cosx)^2=2*-sinx*cosx=-2sinxcosx`
and `sin2x=2sinxcosx`.

These two results are almost exactly the same, differing only by a negative sign. But what does it mean in the course of our problem?

Well, look what happens when we let `u=cos^2x`. We also need to replace `dx` in this `u`-substitution, as follows:
`u=cos^2x`
`(du)/dx=-2sinxcosx->du=-2sinxcosxdx`

Before we apply this substitution, look at the modified integral (which has `sin2x` replaced with its equivalent `2sinxcosx`):
`int(2sinxcosx)/(1+cos^2x)dx`

Hm...that numerator looks familiar. It's almost the expression for `du`!
`du=color(blue)(-2sinxcosxdx)`
`int(color(blue)(2sinxcosx))/(1+cos^2x)color(blue)(dx)`

All we have to do is apply a negative sign inside the integral, and one outside (to balance it), and...
`du=color(blue)(-2sinxcosxdx)`
`-int(color(blue)(-2sinxcosx))/(1+cos^2x)color(blue)(dx)`

We can replace `-2sinxcosxdx` with `du` (also remember that `u=cos^2x`):
`=-int(du)/(1+u)`

This evaluates to:
`-ln(1+u)+C`

Because `u=cos^2x`:
`int(sin2x)/(1+cos^2x)dx=-ln(1+cos^2x)+C`