∫−ππ2x(1+sinx)1+cos2xdx=

I=∫−ππ2xsinxdx1+cos2x  dropping odd term I=4∫0πxsinx1+cos2xdx,x→π−x I=4∫0π(π−x)sinx1+cos2xdx 2I=4π∫0πsinx1+cos2xdx=8π∫0π2sinx1+cos2xdx                                            =8πtan−1(−cosx)|0π2=8π(0+π4)                                            =2π2→I=π2