Question

积分∫xdx/[sin（x^2+1)]^2 dx=多少?

Answer 1

原式=1/2∫dx²/sin²(x²+1)

=1/2∫csc²(x²+1)d(x²+1)

=-1/2∫[-csc²(x²+1)]d(x²+1)

=-1/2*cot(x²+1)+C