∫cosx^2/1+sinxcosx dx如何求解

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∫cosx^2/1+sinxcosx dx如何求解
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∫cosx^2/1+sinxcosx dx如何求解
∫cosx^2/1+sinxcosx dx如何求解

∫cosx^2/1+sinxcosx dx如何求解
∫ cos²x/(1 + sinxcosx) dx
= ∫ (1 + cos2x)/2 * 1/(2 + sin2x)/2 dx
= ∫ (1 + cos2x)/(2 + sin2x) dx
= (1/2)∫ dy/(2 + siny) + (1/2)∫ cosy/(2 + siny) dy
对于第一项,代u = tan(y/2),dy = 2du/(1 + u²),siny = 2u/(1 + u²)
= (1/2)∫ 1/[2 + 2u/(1 + u²)] * 2/(1 + u²) du + (1/2)∫ d(2 + siny)/(2 + siny)
= (1/2)∫ (1 + u²)/(2 + 2u² + 2u) * 2/(1 + u²) du + (1/2)ln(2 + siny)
= (1/2)∫ du/(u² + u + 1) + (1/2)ln(2 + siny)
= (1/2)∫ du/[(u + 1/2)² + 3/4] + (1/2)ln(2 + siny)
= (1/2) * (2/√3) * arctan[(u + 1/2)/(√3/2)] + (1/2)ln(2 + siny) + C
= (1/√3)arctan[(2u + 1)/√3] + (1/2)ln(2 + siny) + C
= (1/√3)arctan[(2tan(y/2) + 1)/√3] + (1/2)ln(2 + siny) + C
= (1/√3)arctan[(2tanx + 1)/√3] + (1/2)ln(2 + sin2x) + C

∫cosx^2/1+sinxcosx dx如何求解 sinxcosx/1+(cosx)^2的积分 求证:[(1-2sinxcosx)/sinx-cosx ]+cosx=sinx 求证(1-2sinxcosx)/(cosx^2-sinx^2)=(cosx^2-sinx^2)/(1+2sinxcosx) sin²x-2cosx=2,则sinxcosx+3sinx+cosx为?sin²x-2cosx=2,则sinxcosx+3sinx+cosx为A 3B -3C 1D -1谢啦~ ∫sinxcosx/(sinx+cosx)dx sinxcosx+cosx^2-2 化简 化简1+sinX+cosX+sinXcosX 化简：1/sinxcosx-cosx/sinx sinx+cosx+sinxcosx=1 求证(2sinxcosx)/(sinx+cosx-1)(sinx-cosx+1)=(1+cosx)/sinx 证明2sinxcosx/(sinx+cosx-1)(sinx-cosx+1)=1+cosx/sinx? 证明2SINXCOSX/(SINX+COSX-1)(SINX-COSX+1)=(I+COSX)/SINX 证明（1+SinX+CosX+2SinXCosX)/(1+SinX+CosX)=SinX+CosX 1+sinx+cosx+2sinxcosx/1+sinx+cosx=sinx+cosx证明 凑微分法求定积分∫（上限 派/2,下限0）sinxcosx/(1+cosx^2)dx sin^2xtanx+cos^2x/tanx+2sinxcosx-(1+cosx/sinxcosx)化简 解三角方程：2(sinx)^2-sinxcosx-(cosx)^2=1