Question

第十四题怎么做？

Answer 1

令t=tan(x/2)，
则sinx=2t/(1+t²)，
cosx=(1-t²)/(1+t²)
dx=2/(1+t²)·dt

原式=∫1/[2sinx(1+cosx)]·dx
=∫(1+t²)²/(8t)·2/(1+t²)·dt
=1/4·∫(1+t²)/t·dt
=1/4·ln|t|+1/8·t²+C
=1/4·ln|tan(x/2)|+1/8·tan²(x/2)+C

Answer 2

1/[sin2x+2sinx]
=1/[2sinxcosx+2sinx]
=1/[2sinx(1+cosx)]（上下都乘以sinx）
=sinx/[2sinx*sinx*(1+cosx)]

所以
∫dx/sin2x+2sinx
=1/2∫sinx/[(1-(cosx)^2)(1+cosx)]dx
=-1/2∫1/[(1-(cosx)^2)(1+cosx)]dcosx(凑微分法，记cosx=t)
=-1/2∫1/[(1-t^2)(1+t)]dt
=-1/2{-1/4*ln(t-1)-1/2*1/(1+t)+1/4*ln(1+t)}+C
=1/8*(ln(cosx-1)+ln(cosx-1)*cosx+2-ln(1+cosx)-ln(1+cosx)*cosx)/(1+cosx)+C