$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx}dx=$
$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx}\left(\ frac{1-sinx}{1-sinx}\right)dx$
$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sinx}{cos^2x}dx-\int_{0}^{\frac{\pi}{2}}\frac{sin^2x}{cos^2x}dx$
the first one substitute u=cosx the other tan^2x+1=sec^x .... tan^2x=sec^2-1 and now the integrate for sec^2x =tanx and integrate of -1 =-x
thats all
you know the answer right ??
but if you derive my answer you will get the integral I try it I think there is no problem if I multiply with 1-sinx
or if the integral have limits then we should look at the interval [0,pi/2]
I cant multiply with 1-sinx or 1-cosx or 1-tanx .....