Question

求定积分，答案越详细越好，别把步骤省了，谢谢

Answer 1

1。（1）
令 y=π/2-x,则x=π/2-y
∫（π/2~0)f(cosx)dx=∫（0~π/2) f(cos(π/2-y))d(π/2-y)
=∫（0~π/2) -f(siny)dy
=-∫（0~π/2) f(siny)dy
=∫（π/2~0)f(siny)dy
=∫（π/2~0)f(sinx)dx
（2）
证明：令x=π-t,则x由0到π,t由π到0,dx=-dt
原式记为I
则I=-（积分区间π到0）∫(π-t)f(sin(π-t)dt
=-(积分区间π到0）∫(π-t)f(sin(t)dt
=(积分区间0到π）∫(π-t)f(sin(t)dt
=(积分区间0到π）∫πf(sin(t)dt-I
所以2I=(积分区间0到π）∫πf(sin(t)dt
即I=（π/2)∫f(sint)dt=(π/2)∫f(sinx)dx

2。
∫[0~π] (x sinx)/(1 + cos²x) dx
= ∫[0~π] (x sinx)/(2 - sin²x) dx,设f(x) = x/(2 - x²),则f(sinx) = sinx/(2 - sin²x)
= ∫[0~π] x f(sinx) dx
= (π/2)∫[0~π] f(sinx) dx
= (π/2)∫[0~π] sinx/(2 - sin²x) dx
= -(π/2)∫[0~π] 1/(1 + cos²x) d(cosx)
= -(π/2)arctan(cosx)_[0~π]
= -(π/2)[arctan(-1) - arctan(1)]
= -(π/2)(-π/4 - π/4)
= π²/4

追问

你好，第二小问，那个区间0~派，那个是这样表达吗？