Question

1/(sinx+2)的不定积分如何求

Answer 1

简单计算一下即可，答案如图所示

Answer 2

1/(sinx+2)
=(1/2)/(0.5*sinx+1)dx
=1/(sin(x/2)cos(x/2)+1)d(x/2)
令t=x/2
原式=(1/sint*cost+1)dt
分子分母都除以(cost)^2
=(1/(cost)^2)/{[1/(cost)^2]+tant})dt
=1/{[1/(cost)^2]+tant}d(tant)
=[1/(1+(tantt)^2)+tant]d(tant)
令u=tant
={1/(1+u^2)+u}du
=1/[(0.5+u)^2+0.75]du
=[2/3^(1/2)]arctant[(2/3^(1/2))(u+1/2)]+c
=[2/3^(1/2)]arctant[(2/3^(1/2))(tant+1/2)]+c
=[2/3^(1/2)]arctant[(2/3^(1/2))(tant(x/2)+1/2)]+c
我已经验证过了，答案是正确的，不信你可以自己试一下