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