求sinx/(1+sinx+cosx)的不定积分

老丫丫2022-10-04 11:39:541条回答

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sadfkhsdkfj 共回答了12个问题 | 采纳率100%: ∫ sinx/(1+sinx+cosx) dx
=∫sinx(sinx+cosx-1)/[(sinx+cosx+1)(sinx+cosx-1)] dx
=∫(sin^2x+sinxcosx-sinx)/[(sinx+cosx)^2-1] dx
=∫(sin^2x+sinxcosx-sinx)/(2sinxcosx) dx
=(1/2)∫sinx/cosx dx+(1/2)∫ dx-(1/2)∫1/cosx dx
=(-1/2)∫1/cosx d(cosx)+(1/2)∫ dx-(1/2)∫secx dx
=(1/2)x-(1/2)ln(cosx)+(1/2)x-(1/2)ln(secx+tanx)+C; 1年前