∫π0sin2x(1+cosx)2dx=∫1−cos2x(1+cosx)2dx ∫sin2x(1+cosx)2dx=∫1−cos2x(1+cosx)2dx =2∫tan2x2dx2 2⎡⎢⎣∫π20(sec2x2−1)dx2⎤⎥⎦=2⎡⎢⎣tanx2∫π20−x2∫π20⎤⎥⎦ =2[(1−0)−(π3−0)] =2−π2
∫π20log sin x dx =