计算∫1/(sinx+sin2x)dx,高手请进!

一楼的答案是错的吧,积分积错了
此题我只能想到2种解法了:
1.设u=tan(1/2*x),所以sinx=2u/(1+u^2),cosx=(1-u^2)/(1+u^2),dx=2/(1+u^2)du
代入化简得:
原式=∫1/u(3-u^2)du
=∫1/3*1/udu+∫2/3*1/(3-u^2)du
=-2/3*lnlu^2-3l+1/3*lnlul+C
=-2/3*lnltan(1/2*x)^2-3l+1/3*lnltan(1/2*x)l+C
2.分子分母同乘sinx,
所以原式=∫sinx/[(sinx+sin2x)sinx]dx
=-∫1/[(1-cosx^2)*(1+2cosx)]d(cosx)
设t=cosx
所以原式=-∫1/[(1-t^2)*(1+2t)]d(t)
=∫4/3*1/(1+2t)dt+∫1/6*1/(1-t)dt+∫(-1/2)*1/(1+t)dt
=2/3*lnl1+2tl-1/6*lnl1-tl-1/2lnl1+tl+C
=2/3*lnl1+2cosxl-1/6*lnl1-cosxl-1/2lnl1+cosxl+C