Numerator: sin(3x) = sin(2x) cos(x) + sin(x) cos(2x).
Denominator: sin(2x) = 2 sin(x) cos(x).
Substitute the above, break up the integral into two parts and simplify. The following will be useful: integrate Cos[2x]/Cos[x] - Wolfram|Alpha
Clcik on Show steps.
Thanks for the hints guys, just a Q, archie, How did you evaluate Sin3x into Sin2xCosx+Cos2xSinx?
Edit: This problem is indeed eating my mind, how do we go from here to get and to evaluate to get integral value of: 2 sinx + 1/4 ln[1-sinx] / [1 + sinx ] + c (The Right answer)
Thanks for your answers