计算不定积分：∫1/√(1+sinx)dx

答：
原积分
=∫1/√(1+cos(x-π/2)) dx
=∫1/√(1+2cos(x/2-π/4)^2-1) dx
=∫1/√(2cos(x/2-π/4)^2) dx
=1/√2 ∫1/cos(x/2-π/4) dx
=1/√2 ∫2sec(x/2-π/4) d(x/2-π/4)
=√2 ln|tan(x/2-π/4)+sec(x/2-π/4)| + C