求下列不定积分

Question

如题所述

Answer 1

∫e^sinx * sin2x dx
=∫e^sinx * 2sinx * cosx*dx
=∫e^t * 2t dt 注：t = sinx，则 dt = cosx *dx
=2(t*e^t - ∫e^t dt)
=2t*e^t - 2e^t + C
=2sinx*e^(sinx) - 2e^(sinx) + C
∫(1 + sinx)/(1+cosx) *dx
=∫dx/(1+cosx) + ∫ sinx*dx/(1+ cosx)
=∫dx/{2[cos(x/2)]^2} - ∫d(1+cosx)/(1+cosx)
=∫[sec(x/2)]^2*d(x/2) - ln(1+cosx) + C1
=tan(x/2) + C2 - ln(1+cosx) + C1
=tan(x/2) - ln(1+cosx) + C 注：C= C1 + C2，C1、C2 和 C 都为常数