# $\int_{0}^{\pi/2}\arctan(\sin(x))dx=\frac{\pi^2}{8}-\frac{\ln^2(\sqrt{2}-1)}{2}$

I am trying to evaluate the following integral, so far I tried 2 different ways, but could not finish the proof. Through the second method, it seems that I got closer

$$\int_{0}^{\pi/2}\arctan(\sin(x))dx=\frac{\pi^2}{8}-\frac{\ln^2(\sqrt{2}-1)}{2}$$

First Method

Consider the more general version with parameter k

$$I(k)=\int_{0}^{\pi/2}\arctan(k\sin(x))dx$$

$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin(x)}{1+k^2\sin^2(x)}dx$$

$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin(x)}{\cos^2(x)+\sin^2(x)+k^2\sin^2(x)}dx$$

$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin(x)}{\cos^2(x)+(1+k^2)\sin^2(x)}\frac{1}{\frac{\sin^2(x)}{\sin^2(x)}}dx$$

$$I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\csc(x)}{\cot^2(x)+(1+k^2)}dx$$

substitution $$\cot^2(x)=t$$ does not seem very helpful.

Second Method

let $$\sin(x)\longrightarrow x$$

$$\int_{0}^{\pi/2}\arctan(\sin(x))dx=\int_{0}^{1}\frac{\arctan(x)}{\sqrt{1-x^2}}dx$$

integrating by parts

$$\int_{0}^{1}\frac{\arctan(x)}{\sqrt{1-x^2}}dx=\arctan(x)\cdot\arcsin(x)|_{0}^{1}-\int_{0}^{1}\frac{\arcsin(x)}{1+x^2}dx$$

$$=\frac{\pi^2}{8}-\underbrace{\int_{0}^{1}\frac{\arcsin(x)}{1+x^2}dx}_{J}$$

using the expansion of $$\arcsin(x)$$

$$J=\int_{0}^{1}\frac{\arcsin(x)}{1+x^2}dx=\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(n!)^2}\frac{1}{2n+1}\int_{0}^{1}\frac{x^{2n+1}}{1+x^2}dx$$

Can someone indicate a method to help me finish at least one of the two methods? Thank you

• May 16 at 16:51
• @Robert Z thank you! May 16 at 16:53
• You can also write: $$\ln^2 (\sqrt{2}-1)=\mathrm{(sinh^{-1}1)^2}$$ May 16 at 16:59
• – FDP
May 17 at 13:15

$$I(k)=\int_{0}^{\pi/2}\arctan(k\sin x)dx,\>\>\>\>\> I^{\prime}(k)=\int_{0}^{\pi/2}\frac{\sin x}{1+k^2\sin^2x}dx$$ Evaluate $$I’(k)$$ with $$t=\cos x$$ \begin{align} I^{\prime}(k)=\int_{0}^{1}\frac{1}{1+k^2-k^2 t^2}dt =\frac{\text{arctanh}\frac{k}{\sqrt{1+k^2}}}{k\sqrt{1+k^2}} = \frac{\text{arcsinh}{\>k}}{k\sqrt{1+k^2}} \end{align} Then, $$I(0)=0$$, $$I(\infty) =\frac{\pi^2}4 = \int_0^\infty I’(k)dk$$
\begin{align} &\int_{0}^{\pi/2}\arctan(\sin x)dx =I(1)= \int_0^1 I’(k)dk \\ =&\int_{0}^{1}\frac{\text{arcsinh}{\>k}}{k\sqrt{1+k^2}}dk =-\frac12\int_{0}^{1}\frac{\text{arcsinh}{\>k}}{\text{arccsch}\>{k}}d\left(\text{arccsch}^2{k}\right) \\ =&- \frac12\text{arcsinh}\>k \>\text{arccsch}\>k\bigg|_0^1 +\frac12\int_0^1 \frac{\text{arcsinh}\>k+k\>\overset{k\to1/k}{\text{arccsch}{\>k}}}{k\sqrt{1+k^2}}dk\\ =& - \frac12\text{arcsinh}(1)\>\text{arccsch}(1) +\frac12\int_0^\infty I’(k)dk \\ = &- \frac12\text{arcsinh}^2(1)+\frac{\pi^2}8 \end{align}
Since $$|\sin x|\le 1$$, we may write $$\arctan(\sin x)=\sum_{n\ge0}\frac{(-1)^n}{2n+1}\sin(x)^{2n+1},$$ and thus $$J=\int_0^{\pi/2}\arctan(\sin x)dx=\sum_{n\ge0}\frac{(-1)^n}{2n+1}\int_0^{\pi/2}\sin(x)^{2n+1}dx.$$ The remaining integral is an easy use of the Beta function, and we have $$\int_0^{\pi/2}\sin(x)^{2n+1}dx=\frac{2^{2n}n!^2}{(2n+1)!},$$ so that $$J=\sum_{n\ge0}\frac{(-1)^n2^{2n}}{(2n+1)^2\binom{2n}{n}}.$$ Then this gives $$J=\sum_{n\ge0}\frac{(-1)^n2^{2n}}{(2n+1)^2\binom{2n}{n}}=\frac{\pi^2}{8}-\frac12\text{arcsinh}^2(1).$$