\(\int \frac{sin^2x}{1 + sin^2x}dx = \int \frac{1 + sin^2x-1}{1 +sin^2x} dx\)
\(= \int 1- \frac 1{1 + sin^2 x} dx\)
\(= x - \int \frac 1{1 + sin^2x} dx\)
\(= x - \int \frac{sec^2x}{sec^2x + tan^2x}dx\)
\(= x - \int \frac{sec^2x}{1 + 2tan^2x} dx\)
Let \(tan \,x = t\)
then \(sec^2x \,dx = dt\)
\(\therefore \int \frac {sin^2x}{1 + sin^2x} dx = x - \int \frac {dt}{1 + 2t^2}\)
\(= x - \frac 12 \int \frac{dt}{\frac 12 + t^2}\)
\(=x - \frac 12 \,.\frac 1{\sqrt 2} tan^{-1} \left(\cfrac t{\frac 1{\sqrt2}}\right) \) \(\left(\because \int \frac 1{x^2 +a^2} dx = \frac 1a tan^{-1} \frac xa\right)\)
\(= x - \frac 1{2\sqrt 2} tan^{-1} (\sqrt 2 t)\)
\(= x - \frac1{2\sqrt 2} tan^{-1} (\sqrt 2 tan \,x) \) \((\because t = tan \,x)\)
