The correct option is D π2
Let I=2π∫0xsin8xsin8x+cos8xdx ...(1)I=2π∫0(2π−x)sin8(2π−x)sin8(2π−x)+cos8(2π−x)dx=2π∫0(2π−x)sin8xsin8x+cos8xdx ...(2)
Adding (1) & (2), we get :
⇒2I=2π2π∫0sin8xsin8x+cos8xdxI=π2π∫0sin8xsin8x+cos8xdx
I=4ππ2∫0sin8xsin8x+cos8xdx ...(3)
I=4ππ2∫0sin8(π2−x)sin8(π2−x)+cos8(π2−x)dxI=4ππ2∫0cos8xsin8x+cos8xdx ...(4)
Adding (3) and (4) , we get :
I=2ππ2∫01 dx=2π×π2=π2