Question

根号下1+sinx dx 求积分

Answer 1

1+sinx=(sin(x/2)+cos(x/2))^2
即原式=∫（sin(x/2)+cos(x/2)）dx
=2∫sin(x/2)d(x/2)+2∫cos(x/2)d(x/2)
=2sin(x/2)-2cos(x/2)