Integral \int_0^{\pi/2}\arctan^2(\frac{6\sin x}{3+\cos 2x})dx

Elise Winters

Elise Winters

Answered question

2022-04-30

Integral
0π2arctan2(6sinx3+cos2x)dx

Answer & Explanation

haillarip0c9

haillarip0c9

Beginner2022-05-01Added 23 answers

I'll mention the outcome below.
I(r,s)=0π2arctan(rsinθ)arcsin(ssinθ)dθ
=πχ2(1+r21r×1+s21s)
where χ2 that of Legendre's chi. By using the addition formula for the arctangent, it follows that
arctan(6sinx3+cos2x)=arctan(32sinx112sin2x)=arctan(sinx)+arctan(12sinx)
So it follows that
0π2arctan2(6sinx3+cos2x) dx =I(1,1)+2I(1,12)+I(12,12)
This is simplified to a Legendre's and chi function combination.
π[χ2(322)+χ(945)+2χ((21)(52))]

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