Got any ideas (what substitution should I use) to evaluate $$\int \frac{\sin^2(x)}{1+\sin^2(x)}dx~?$$
Answer
Write:
$$\frac {\sin^2 x} {1 + \sin^2 x} = \frac {1 +\sin^2 x - 1} {1 + \sin^2 x} = 1 - \frac 1 {1 + \sin^2 x} = 1 - \frac {\sec^2 x} {\sec^2 x + \tan^2 x} = 1 - \frac {\sec^2 x} {1 + 2\tan^2 x} $$
Now substitute $u = \tan x$ and the rest is easy.
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