求∫sinx dx/(sinx+cosx)的积分,x/2-ln|sinx+cosx|+c

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/28 22:52:24
求∫sinx dx/(sinx+cosx)的积分,x/2-ln|sinx+cosx|+c

求∫sinx dx/(sinx+cosx)的积分,x/2-ln|sinx+cosx|+c
求∫sinx dx/(sinx+cosx)的积分,x/2-ln|sinx+cosx|+c

求∫sinx dx/(sinx+cosx)的积分,x/2-ln|sinx+cosx|+c
sinx /(sinx+cosx)=(tanxcosx)/(tanxcosx+cosx)=tanx/(tanx+1)
令t=tanx,则dt=sec^2 xdx=(1+tan^2 x)dx=(1+t^2)dx,即dx=dt/(1+t^2),于是
∫sinx dx/(sinx+cosx)
=∫tdt/[(1+t)(1+t^2)]
=(1/2)∫[-1/(1+t)+(1+t)/(1+t^2)]dt
=(1/2)[∫-dt/(1+t)+∫(1+t)dt/(1+t^2)]
=(1/2)[-ln|1+t|+∫dt/(1+t^2)+∫tdt/(1+t^2)]
=(1/2)[-ln|1+t|+arctant+(1/2)ln(1+t^2)]+C
=(1/2)[-ln|1+tanx|+x+(1/2)ln(1+tan^2 x)]+C
=(1/2)[-ln|1+tanx|+x+ln|secx|]+C
=(x-ln|sinx+cosx|)/2+C

sinx /(sinx+cosx) = (1/2)【1 - (cosx-sinx) / (sinx+cosx) 】
原式 = ∫ (1/2)【1 - (cosx-sinx) / (sinx+cosx) 】dx
= x/2 - (1/2) ln| sinx+cosx | + C