Question

Find: `int (sin2x)/sqrt(9 - cos^4x) dx`

Answer 1

Put `cos^2x` = t ⇒ `−2cosxsinxdx` = dt ⇒ `sin2xdx = -dt`

The given integral = `- int (dt)/sqrt(3^2 - t^2) = - sin^(-1)  t/3 + c = - sin^(-1)  (cos^2x)/3 + c`