解答题
13.设f(χ)为连续函数，证明：
(1)∫0πχf(sinχ)dχ＝
∫0πf(sinχ)dχ＝π
f(sinχ)dχ；
(2)∫02πf(｜sinχ｜)dχ＝4
(1)∫0πχf(sinχ)dχ＝
∫0πf(sinχ)dχ＝πf(sinχ)dχ；(2)∫02πf(｜sinχ｜)dχ＝4
【正确答案】(1)令I＝∫0πχf(sinχ)dχ，则
I＝∫0πχf(sinχ)dχ
∫0π(π－t)f(sint)(－dt)＝∫0π(π－t)F(sint)dt
＝∫0π(π－χ)f(sinχ)dχ＝π∫0π(sinχ)dχ－∫0πχf(sinχ)dχ－π∫0πf(sinχ)dχ－I，
则l＝∫0πχf(sinχ)dχ＝
∫0πf(sinχ)dχ＝π
f(sinχ)dχ．
(2)∫02πf(｜sinχ｜)dχ＝∫-ππf(｜sinx｜)dχ＝2∫0πf(｜sinχ｜)dχ
＝2∫0πf(sinχ)dχ＝4
I＝∫0πχf(sinχ)dχ
∫0π(π－t)f(sint)(－dt)＝∫0π(π－t)F(sint)dt＝∫0π(π－χ)f(sinχ)dχ＝π∫0π(sinχ)dχ－∫0πχf(sinχ)dχ－π∫0πf(sinχ)dχ－I，
则l＝∫0πχf(sinχ)dχ＝
∫0πf(sinχ)dχ＝πf(sinχ)dχ．(2)∫02πf(｜sinχ｜)dχ＝∫-ππf(｜sinx｜)dχ＝2∫0πf(｜sinχ｜)dχ
＝2∫0πf(sinχ)dχ＝4
【答案解析】