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Integral of [sin 2x/(1 + cos2 x)] dx (substitute 2sinx cos x for sin 2x)
= Integral of [2sin x cos x/(1 + cos2 x)] dx
Let u = 1 + cos2 x, so that du = -2cos x sinx dx, and dx = du/-2sin x cos x.
(Substitute u for 1 + cos x, and -2sin x cos x for dx). So we have:
= Integral of [(2sin x cos x/u)(du/-2sin x cos x)] (simplify)
= Integral of -du/u
= -ln |u| + C (substitute 1 + cos2 x for u, and take off the absolute value symbol since the expression inside it is always positive)
= -ln (1 + cos2 x) + C
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