∫1/(2+sinx)dx
做代换tgx/2=t
则sinx=2t/(1+t^2)
dx=d(2arctgt)=2dt/(1+t^2)
∫1/(2+sinx)dx
=∫[2/(1+t^2)]/[2+2t/(1+t^2)]dt
=∫2/(2(1+t^2)+2t)dt
=∫1/(t^2+t+1)dt
=∫1/[(t+1/2)^2+3/4]dt
=4/3∫1/{[(2t+1)/√3]^2+1}
=2/√3*arctg[(2t+1)/√3]+C
∫(下限为0 上限为π/2)1/(2+sinx)dx
做代换tgx/2=t
=∫(下限为0 上限为1)1/(t^2+t+1)dt
=2/√3*arctg[3/√3]-2/√3*arctg[(1/√3]
=2/√3(π/3-π/6)
=2/√3*(π/6)
=π/(3√3)
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