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∫(1/sin^3xCosx)Dx

如图

积分(sinx)^3d(sinx)=1/4(sinx)^4+C

∫ (cosx/sin^3x)dx =∫(sin^-3x)d(sinx) =(-1/2)sin^-2x+c

可以停止了,因为原函数不初等

这个是三角函数的不定积分,分母应先进性化简,计算步骤为: ∫1/(sinx+cosx)dx =∫dx/√2sin(x+π/4) =-(√2/2)∫dcos(x+π/4)/sin^2(x+π/4) =-(√2/4){∫dcos(x+π/4)/[1-cos(x+π/4)]+∫dcos(x+π/4)/[1+cos(x+π/4)]} =-(√2/4)ln{[1+cos(x+π/4)]/[1-cos...

解:∫sin^3xcos^2xdx =-∫sin^2xcos^2xdcosx =-∫(1-cos^2x)*cos^2xdcosx =-∫(cos^2x-cos^4x)dcosx =(1/5)*cos^5x-(1/3)*cos^3x

∫(sinx)^3·(cosx)^5dx =-∫(sinx)^2·(cosx)^5d(cosx) =∫[(cosx)^2-1](cosx)^5d(cosx) =∫(cosx)^7d(cosx)-∫(cosx)^5d(cosx) =(1/8)(cosx)^8-(1/6)(cosx)^6+C。

求不定积分∫dx/(sin³xcosx) 解:原式=∫(sin²x+cos²)dx/(sin³xcosx)=∫dx/(sinxcosx)+∫cosxdx/sin³x =∫d(2x)/sin(2x)+∫d(sinx)/sin³x=ln∣tanx∣-1/(2sin²x)+C

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