Question

计算不定积分

Answer 1

∫1/(1+sinx)dx
=∫(1-sinx)/[(1+sinx)(1-sinx)]dx -------------因为分母较复杂， 同时乘以(1-sinx)
=∫(1-sinx)/(1-sin^2 x)dx
=∫[(1-sinx)/cos^2 x]dx
=∫[(sin^2 x+cos^2 x)/cos^2 x]dx-∫[sinx/cos^2 x]d(x)
=x+∫[sin^2 x/cos^2 x]dx-∫[-1/cos^2 x]d(cosx)
=x-(1/cosx)+∫sinx[sinx/cos^2 x]dx
=x-(1/cosx)+∫sinxd[1/cosx]
=x-(1/cosx) +(sinx/cosx)-∫[1/cosx]dsinx ---------------分部积分公式
=x-(1/cosx)+tanx-∫[1/cosx]cosxdx
=x-secx+tanx-x+C
=tanx-secx+C

Answer 2