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∫(sinxsin2x)Dx的定积分

如图所示:

∫(sinxsin2x)dx =2∫sin²xcosxdx =2∫sin²xdsinx =2sin³x/3+C

∫(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

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...

1/[sin2x+2sinx] =1/[2sinxcosx+2sinx] =1/[2sinx(1+cosx)](上下都乘以sinx) =sinx/[2sinx*sinx*(1+cosx)] 所以 ∫dx/sin2x+2sinx =1/2∫sinx/[(1-(cosx)^2)(1+cosx)]dx =-1/2∫1/[(1-(cosx)^2)(1+cosx)]dcosx(凑微分法,记cosx=t) =-1/2∫1/[(1-t...

定积分∫0-π2(x+sinx)dx=(12x2-cosx)| 0-π2=(0-cos0)-[12×π24-cos(-π2)]=-1-π28.故答案为:-1-π28.

∫(lntanx)/(sin2x) dx =(1/2)∫(lntanx)/(sinxcosx) dx d(lntanx)=(sec²x/tanx)dx=dx/(sinxcosx) =(1/2)∫(lntanx)/(sinxcosx)·sinxcosxd(lntanx) =(1/2)∫(lntanx)d(lntanx) =(1/2)(lntanx)²/2+C =(1/4)(lntanx)²+C

因为sin2x = 2sinxcosx; ∫sin2xcosxdx = ∫2sinxcosxcosxdx = -2∫cosx^2dcosx = -2/3∫cosx^3

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