indefinite integral math help?

∫ (sec2x + tan2x dx
= ∫ 1/(cos2x) + sin2x/cos2x dx
= ∫ (1+sin2x) / cos2x dx
= ∫(1+sin2x) / cos2x ∙ (1 - sin2x) / (1 - sin2x) dx
= ∫ 1 - sin²2x / (cos2x(1 - sin2x)) dx
= ∫ cos²2x / (cos2x(1 - sin2x)) dx
= ∫ cos2x / (1 - sin2x) dx

Let u = 1 - sin2x
du = -2cos2x dx → -½du = cos2x dx
In terms of u, the integral is now:
-½ ∫ du/u
= -½ ln|u| + C
= -½ ln |1 - sin2x| + C

You can check it by taking the derivative of -½ ln |1 - sin2x| + C

hint : sec 2x + tan 2x = cos 2x / [ 1 - sin 2x]...let w = bottom