Question

大一！！！高数！！！！悬赏一百，求解一个定积分！！！谢谢大神！！！

Answer 1

根据公式：
∫(0,π)xf(sinx)dx
=π/2∫(0,π)f(sinx)dx
可得
原式
=π/2 ∫(0,π)sin^6xcos^4xdx
=π∫(0,π/2)sin^6x[1-sin^2x]^2dx
=π∫(0,π/2)sin^6x[1-2sin^2x+sin^4x]dx
=π∫(0,π/2)[sin^6x-2sin^8x+sin^10x]dx
=π×【5/6×3/4×1/2×π/2 -2×7/8×5/6×3/4×1/2×π/2+9/10×7/8×5/6×3/4×1/2×π/2】
=π×3π/512
=3π方/512

Answer 2

公式：∫[0→π] xf(sinx) dx = (π/2)∫[0→π] f(sinx) dx

∫[0→π] x(sinx)⁶(cosx)⁴ dx
由公式：
=(π/2)∫[0→π] (sinx)⁶(cosx)⁴ dx
=(π/2)∫[0→π/2] (sinx)⁶(cosx)⁴ dx + (π/2)∫[π/2→π] (sinx)⁶(cosx)⁴ dx
后一部分换元，令x=u+π/2，则dx=du，u：0→π/2
=(π/2)∫[0→π/2] (sinx)⁶(cosx)⁴ dx + (π/2)∫[0→π/2] (cosu)⁶(sinu)⁴ du
积分变量换回x
=(π/2)∫[0→π/2] (sinx)⁶(cosx)⁴ dx + (π/2)∫[0→π/2] (cosx)⁶(sinx)⁴ dx
=(π/2)∫[0→π/2] (sinx)⁴(cosx)⁴(sin²x+cos²x) dx
=(π/2)∫[0→π/2] (sinx)⁴(cosx)⁴dx
=(π/32)∫[0→π/2] (sin2x)⁴dx
=(π/32)∫[0→π/2] (1/4)(1-cos4x)² dx
=(π/128)∫[0→π/2] (1-2cos4x+cos²4x) dx
=(π/128)∫[0→π/2] [1-2cos4x+(1/2)(1+cos8x)] dx
=(π/128)∫[0→π/2] [3/2 - 2cos4x + cos8x] dx
=(π/128)(3/2)(π/2)
=3π²/512