设f(χ)为连续函数,证明: (1)∫0πχf(sinχ)dχ=∫0πf(sinχ)dχ=πf(sinχ)dχ; (2)∫02πf(|sinχ|)dχ=4f(sinχ)dχ.

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问题 设f(χ)为连续函数,证明:
    (1)∫0πχf(sinχ)dχ=0πf(sinχ)dχ=πf(sinχ)dχ;
    (2)∫0f(|sinχ|)dχ=4f(sinχ)dχ.

选项

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)∫0f(|sinχ|)dχ=∫πf(|sinx|)dχ=2∫0πf(|sinχ|)dχ =2∫0πf(sinχ)dχ=4[*]f(sinχ)dχ.

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