Question

求xsin^6xcos^4xdx从0到π的定积分

Answer 1

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

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Answer 2

这个要用积化和差公式，原式=1/2(xsin10x+xsin2x),然后再分布积分，-xdcos10x-xdcos2x 缺系数你自己慢慢算可以了

Answer 3

见图