Need some advice to solve this integral $\int\frac{\sin^2x}{1+\sin^2x}\mathrm dx$ I'm trying to use this subtitution $t=\tan(x/2)$. But I don´t get anywhere. I've tried $t=\tan(x)$ too.  Appreciate your help.  

$$\int\dfrac{\sin^2x}{1+\sin^2x}\mathrm dx$$
Note that $$\frac{\sin^2{x}}{1 + \sin^2{x}} = \frac{1 + \sin^2{x}}{1 + \sin^2{x}} - \frac{1}{1 + \sin^2{x}}$$

The first factor integrates to $x$. For the second integral, note that $$\frac{1}{1 + \sin^2{x}} = \frac{\sec^2{x}}{\tan^2{x} + \sec^2{x}} = \frac{\sec^2{x}}{2\tan^2{x} + 1}$$

Proceed using the substitution $u = \tan{x}$.
Write the ratio as

$$1- \frac{\cos^2 x}{1+\sin^2 x},$$

and recall that the derivative of $\arctan x$ is $1/(1+x^2)$ to obtain

$$x- 2^{-1/2} \arctan\bigl( \sqrt{2} \tan x\bigr)$$

(plus a constant)