Question

How do you find the integral of `(Sin2x) / (1 + cos^2x) dx`?

Answer 1

`=> -ln|cos2x + 3 | + c`

Explanation:

Use trig identities...

`cos2x = cos^2 x - sin^2 x = 2cos^2 x -1`

`=> ( cos2x +1 )/2 +1 = cos^2 x + 1`

`=> ( cos2x + 3 )/2 = cos^2 x + 1`

`=> int (sin2x) / (( cos2x +3 )/2) dx`

`=> int (2sin2x )/ (cos2x +3 ) dx`

Let `u = cos2x +3`

`du = -2sin2x dx`

`=> -du = 2sin2x dx`

`=> -int (du)/u`

`=> -ln|u| + c`

`=> -ln|cos2x + 3 | + c`

`c - "constant"`

Answer 2

Second method

Explanation:

`u = 1 + cos^2 x`

`=> du = 2cosx * -sinx dx`

`=> -du = 2sinxcosx dx`

`=> -du = sin2x dx`

`=> -int (du)/u`

`=> -ln|u| + c`

`=> -ln|1+cos^2 x | + c`