Is it possible to evaluate this integral in a closed form?

I=∫10arctan2x√1−x2dx

It also can be represented as

I=∫π/40ϕ2cosϕ√cos2ϕdϕ

**Answer**

Okay, finally I was able to prove it.

**Step 0. Observations.** In view of the following identity

∫π20arctan(rsinθ)dθ=2χ2(√1+r2−1r),

Vladimir’s result suggests that there may exists a general formula connecting

I(r,s)=∫π20arctan(rsinθ)arctan(ssinθ)dθ

and the Legendre chi function χ2. Indeed, inspired by Vladimir’s result, I conjectured that

I(r,s)=πχ2(√1+r2−1r⋅√1+s2−1s).

I succeeded in proving this identity, so I post a solution here.

**Step 1. Proof of the identity (1).** It is easy to check that the following identity holds:

arctan(ab)=∫∞1/badxa2+x2.

So it follows that

I(r,s)=∫∞1/r∫∞1/s∫π20sin2θ(x2+sin2θ)(y2+sin2θ)dθdydx=∫∞1/r∫∞1/s∫π201x2−y2(x2x2+sin2θ−y2y2+sin2θ)dθdydx=π2∫∞1/r∫∞1/s1x2−y2(x√x2+1−y√y2+1)dydx.

For the convenience of notation, we put

α=√r2+1−1randβ=√s2+1−1s.

Then it is easy to check that arsinh(1/r)=−logα and likewise for s and β. Thus with the substitution x↦sinhx and y↦sinhy, we have

I(r,s)=π2∫∞−logα∫∞−logβsinhxcoshy−sinhycoshxsinh2x−sinh2ydydx.

Applying the substitution e−x↦x and e−y↦y, it follows that

I(r,s)=π∫α0∫β0dydx1−x2y2=π∞∑n=0(∫α0x2ndx)(∫β0y2ndx)=π∞∑0(αβ)2n+1(2n+1)2=πχ2(αβ)

as desired, proving the identity (1).

**EDIT.** I found a much simpler and intuitive proof of (1). We first observe that (1) is equivalent to the following identity

∫π20arctan(2rsinθ1−r2)arctan(2ssinθ1−s2)dθ=πχ2(rs).

Now we first observe that from the addition formula for the hyperbolic tangent, we obtain the following formula

artanhx−artanhy=artanh(x−y1−xy)

which holds for sufficiently small x,y. Thus

arctan(2rsinθ1−r2)=1iartanh(2irsinθ1−r2)=artanh(reiθ)−artanh(re−iθ)i=2ℑartanh(reiθ)=2∞∑n=0sin(2n+1)θ2n+1r2n+1.

We readily check this holds for any |r|<1. Therefore

∫π20arctan(2rsinθ1−r2)arctan(2ssinθ1−s2)dθ=4∞∑m=0∞∑n=0r2m+1s2n+1(2m+1)(2n+1)∫π20sin(2m+1)θsin(2n+1)θdθ=2∞∑m=0∞∑n=0r2m+1s2n+1(2m+1)(2n+1)∫π20{cos(2m−2n)θ−cos(2m+2n+2)θ}dθ=π∞∑m=0∞∑n=0r2m+1s2n+1(2m+1)(2n+1)δm,n=πχ2(rs).

**Attribution***Source : Link , Question Author : Zakharia Stanley , Answer Author : Sangchul Lee*