Question

∫(0→π/2)xsinx^2dx

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Answer 1

若是 ∫ <0→π/2> xsin(x^2)dx
则 ∫ <0→π/2> xsin(x^2)dx = (1/2)∫ <0→π/2> sin(x^2)d(x^2)
= - (1/2)[cosx^2] <0→π/2> = (1/2)[1-cos(π^2/4)]

若是 ∫ <0→π/2> x(sinx)^2dx
则 ∫ <0→π/2> x(sinx)^2dx = (1/2) ∫ <0→π/2> x(1-cos2x)dx
= (1/2) ∫ <0→π/2> xdx - (1/2) ∫ <0→π/2> xcos2xdx
= (1/4) [x^2] <0→π/2> - (1/4) ∫ <0→π/2> xdsin2x
= π^2/16 - (1/4)[xsin2x] <0→π/2>+(1/4) ∫ <0→π/2> sin2xdx
= π^2/16 - 0 -(1/8) [cos2x]<0→π/2>
= π^2/16 +1/8

Answer 2

—1/3 原函数为1/3（cosx）的三次方