Old does not mean easy, like this integral.

Let us first substitute $x=\sin \theta$ . Our integral now becomes:

$\displaystyle\int_{0}^{\pi/4} \theta\cot \theta\text{ }\text{d}\theta=\mathfrak{I}\text{(say)}$

Now we shall use integration by parts:

$\mathfrak{I}=\left.\theta\ln \sin \theta\right|_{0}^{\pi/4}-\displaystyle\int_{0}^{\pi/4} \ln \sin \theta\text{ }\text{d}\theta$

$\therefore \mathfrak{I}=-\dfrac{\pi}{8}\ln 2-\displaystyle\int_{0}^{\pi/4} \ln \sin \theta\text{ }\text{d}\theta$

Now consider the Taylor expansion of $\bigg(-\ln (1-x)\bigg)$ :

$-\ln (1-x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+...$

Replacing $x$ by $e^{i2\theta}$ we get:

$-\ln (2\sin \theta)-i\bigg( \theta-\dfrac{\pi}{2}\bigg)=\displaystyle\sum_{r=1}^{\infty} \frac{e^{i2r\theta}}{r}$

Comparing the real parts on both sides we get:

$\ln \sin \theta=-\bigg( \displaystyle\sum_{r=1}^{\infty} \frac{\cos 2r\theta}{r}\bigg)-\ln 2$

Thus $\displaystyle\int_{0}^{\pi/4} \ln \sin \theta\text{ }\text{d}\theta=-\displaystyle\int_{0}^{\pi/4} \Bigg( \bigg( \displaystyle\sum_{r=1}^{\infty} \frac{\cos 2r\theta}{r}\bigg)+\ln 2\Bigg)\text{d}\theta$

$=-\Bigg( \bigg( \dfrac{1}{2}\displaystyle\sum_{r=1}^{\infty} \frac{\sin \dfrac{r\pi}{2}}{r^2}\bigg)+\dfrac{\pi}{4}\ln 2\Bigg)$

$=-\bigg( \dfrac{G}{2}+\dfrac{\pi}{4}\ln 2\bigg)$

Thus $\mathfrak{I}=\boxed{\dfrac{G}{2}+\dfrac{\pi}{8}\ln 2}$

Firstly, it is quite easy to show that:

$G = \displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)^2} = \displaystyle \int \limits_{0}^{1} \dfrac{1}{x} \int \limits_{0}^{x} \displaystyle \sum_{n = 0}^{\infty} (-1)^n t^{2n} dt$

$\Rightarrow G = \displaystyle \int \limits_{0}^{1} \dfrac{\arctan x}{x} dx$

Putting $x = \tan t,$

$G = \displaystyle \int_{0}^{\pi/4} 2t \csc 2t dt$

Applying by parts, we get :

$G = \displaystyle \int_{0}^{\pi/4} \ln(\cot t) dt$

Note: This identity(without proof) is given at the wiki page on Catalan constant.

Now, Let $\displaystyle \int_{0}^{\pi/4} \ln(\sin t) dt = I_{1}$ , and $\displaystyle \int_{0}^{\pi/4} \ln(\cos t) dt = I_{2}$ .

Then, $I_{1} + I_{2} = \displaystyle \int_{0}^{\pi/4} \ln\bigg(\dfrac{\sin 2t}{2}\bigg) dt = \dfrac{-\pi}{2} \ln 2$

Also, $I_{2} - I_{1} = \displaystyle \int_{0}^{\pi/4} \ln(\cot t) dt = G$

Hence, we get $I_{1} = \displaystyle \int_{0}^{\pi/4} \ln(\sin t) dt = -\dfrac{\pi}{4} \ln 2 - \dfrac{G}{2}$

Now, Putting $x= \sin t$ in $I = \displaystyle \int_{0}^{1/\sqrt{2}} \dfrac{\arcsin x}{x} dx$ , we get:

$I = \displaystyle \int_{0}^{\pi/4} t \cot t dt$

Now, applying by parts, we get:

$I = t \ln(\sin t) \bigg|_{0}^{\pi/4} - \displaystyle \int_{0}^{\pi/4} \ln(\sin t) dt$

= $\boxed{\dfrac{G}{2} + \dfrac{\pi}{8} \ln(2)}$

0.730181058376564+ determined (1/ 2) G + (Pi/ 8) Ln 2.