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求∫sinx/(1%√sin2x)Dx

如图所示:

如图所示:

1-sin2x=sin^2(x)+cos^2(x)-2sinxcosx=(sinx-cosx)^2 ∫[√(1-sin2x) ]dx =∫|sinx-cosx|dx

令u=sin2x,则有sinx=u,x=arcsinu;因此:f(sin2x)=xsinx=f(u)=arcsinuu即:f(x)=arcsinxx.于是有:∫x1?xf(x)dx=

积化和差∫sinxsin2xsin3xdx=1/2∫(cosx-cos3x)sin3xdx=1/2∫cosxsin3xdx-1/2∫cos3xsin3xdx=1/4∫(sin2x+sin4x)dx-1/4∫sin6xdx=-1/8cos2x-1/16cos4x+1/24cos6x+C数学软件验算:

∫(lntanx/sin2x)dx =∫(lntanx)/2sinxcosx)dx =½∫(lntanx)cosx/(sinxcos²x)dx =½∫(lntanx)cosx/(sinx)dtanx =½∫(lntanx)/tanx)dtanx =½∫(lntanx)d(lntanx) =¼ [ln(tanx)]² + C

第一题,直接用万能公式法。即令u=tan(x/2)x=2arctanudx=2/(1+u^2)du,sinx=2u/(1+u^2),cos=(1-u^2)/(1+u^2)原式=∫(1+u^2)/4udu=(1/4)∫(u)^(-1)du+(1/4)∫udu=(1/4)lnu+(1/8)u^2+C=(1/4)ln[tan(x/2)]+(1/8)[tan(x/2)]^2+C第二题,原式=∫(1-sinx)/[...

答: ± (2/3)(sinx)^(3/2) + C 当x∈(2kπ,2kπ+π/2)时取 + 当x∈(2kπ+π/2,2kπ+π)时取 - ∫ √(sinx - sin³x) dx 定义域:x∈(2kπ,2kπ+π),k∈Z = ∫ √[sinx(1-sin²x)] dx = ∫ (sinx)^(1/2)*|cosx| dx 当x∈(2kπ,2kπ+π/2)时,cosx ≥ 0 = ∫ (sin...

-cosx+1/2cos2x

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