设f(sin^2x)=x/sinx,且f∈c,求∫{[√x f(x)] / √(1-x) }dx
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设f(sin^2x)=x/sinx,且f∈c,求∫{[√x f(x)] / √(1-x) }dx
![设f(sin^2x)=x/sinx,且f∈c,求∫{[√x f(x)] / √(1-x) }dx](/uploads/image/z/5357825-17-5.jpg?t=%E8%AE%BEf%EF%BC%88sin%5E2x%29%3Dx%2Fsinx%2C%E4%B8%94f%E2%88%88c%2C%E6%B1%82%E2%88%AB%7B%5B%E2%88%9Ax+f%28x%29%5D+%2F+%E2%88%9A%281-x%29+%7Ddx)
设sinx=u x=arcsinu
f(u^2)=arcsinu/u
f(x)=arcsinx/√x
∫{[√x f(x)] / √(1-x) }dx
=∫{(√x*arcsinx/√x) / √(1-x) }dx
=∫{arcsinx/√(1-x)}dx
分部积分
设u=arcsinx dv=1/√(1-x)dx
du=1/√(1-x^2)dx v=-2√(1-x)
=-2arcsinx√(1-x)+∫{2√(1-x)/√(1-x^2)}dx
=-2arcsinx√(1-x)+∫{2/√1+x}dx
=-2arcsinx√(1-x)+4/3*(1+x)^(3/2)+c
f(u^2)=arcsinu/u
f(x)=arcsinx/√x
∫{[√x f(x)] / √(1-x) }dx
=∫{(√x*arcsinx/√x) / √(1-x) }dx
=∫{arcsinx/√(1-x)}dx
分部积分
设u=arcsinx dv=1/√(1-x)dx
du=1/√(1-x^2)dx v=-2√(1-x)
=-2arcsinx√(1-x)+∫{2√(1-x)/√(1-x^2)}dx
=-2arcsinx√(1-x)+∫{2/√1+x}dx
=-2arcsinx√(1-x)+4/3*(1+x)^(3/2)+c
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