Integrate [sinx] / (2+sin 2x).

\(I=int frac{sinx}{2+sin2x}dx=frac{1}{2}int frac{sinx+cosx+sinx-cosx}{2+sin2x}dx=frac{1}{2}frac{sinx+cosx}{3-[sinx-cosx]^2}+frac{1}{2}frac{sinx+cosx}{1+[sinx+cosx]^2}\)

Now put sinx-cosx=t, then [cosx+sinx]dx=dt and put sinx+cosx=u,

then [sinx-cosx]dx=du
\(I=frac{1}{2}int frac{1}{^{sqrt{3}^2-t^2}}dt-frac{1}{2}int frac{1}{1+u^2}du=frac{1}{4sqrt{3}}lnfrac{sqrt{3}+t}{sqrt{3}-t}-frac{1}{2}tan^{-1}[u]+c\) \(I=frac{1}{4sqrt{3}}lnfrac{sqrt{3}+[sinx-cosx]}{sqrt{3}-[sinx-cosx]}-frac{1}{2}tan^{-1}[sinx+cosx]+c\)

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