设f(χ)为连续函数，证明：
(1)∫
0
π
χf(sinχ)dχ＝
f(sinχ)dχ＝π
f(sinχ)dχ；
(2)∫
0
2π
(｜sinχ｜)dχ＝4
f(sinχ)dχ＝π
f(sinχ)dχ；
(2)∫
0
2π
(｜sinχ｜)dχ＝4
【正确答案】正确答案：(1)令I＝∫
0
π
χf(sinχ)dχ，则 I＝∫
0
π
χf(sinχ)dχ
∫
π
0
(π－t)f(sint)(－dt)＝∫
0
π
(π－t)f(sint)dt ＝∫
0
π
(π－χ)f(sinχ)dχ＝π∫
0
π
f(sinχ)dχ－∫
0
π
χf(sinχ)dχ＝π∫
0
π
f(sinχ)dχ－I， 则I＝
(2)∫
0
2π
f(｜sinχ｜)dχ＝∫
-π
π
f(｜sinχ｜)dχ＝2∫
0
π
f(｜sinχ｜)dχ ＝2∫
0
π
f(sinχ)dχ＝4
∫
π
0
(π－t)f(sint)(－dt)＝∫
0
π
(π－t)f(sint)dt ＝∫
0
π
(π－χ)f(sinχ)dχ＝π∫
0
π
f(sinχ)dχ－∫
0
π
χf(sinχ)dχ＝π∫
0
π
f(sinχ)dχ－I， 则I＝
(2)∫
0
2π
f(｜sinχ｜)dχ＝∫
-π
π
f(｜sinχ｜)dχ＝2∫
0
π
f(｜sinχ｜)dχ ＝2∫
0
π
f(sinχ)dχ＝4
【答案解析】