Integral ∫π/20arctan2(6sinx3+cos2x)dx\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx

Is it possible to evaluate this integral in a closed form?
I=π/20arctan2(6sinx3+cos2x)dx

Answer

I will refer to the following result from my previous answer:

I(r,s)=π20arctan(rsinθ)arctan(ssinθ)dθ=πχ2(1+r21r×1+s21s),

where χ2 is the Legendre chi function. Using the addition formula for the arctangent, it follows that

arctan(6sinx3+cos2x)=arctan(32sinx112sin2x)=arctan(sinx)+arctan(12sinx). So it follows that

π20arctan2(6sinx3+cos2x)dx=I(1,1)+2I(1,12)+I(12,12),

which reduces to a combination of Legendre chi functions

π{χ2(322)+χ2(945)+2χ2((21)(52))}.

Attribution
Source : Link , Question Author : Laila Podlesny , Answer Author : Community

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