Another integration problem

First off all its, quite obvious to have substitution $\sin(x) \rightarrow x$ $I = \int_{0}^{1} \frac{\arcsin(x)}{1+x^2}$ Now using integration by parts, $I = \frac{\pi^2}{8} - \int_{0}^{1} \frac{\arctan(x)}{\sqrt{1-x^2}}$ Now, if nothing works and if you don't like series/complex numbers, then start thinking of Feynman way. Consider $f(a) = \int_{0}^{1} \frac{\arctan(ax)}{\sqrt{1-x^2}}$ After differentiating and integrating, $f'(a) = \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}}$ So we need to find out $f(1) = \int_{0}^{1} \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}}$ From now onwards i am giving sketch, as i think that one can carry forward easily as now part left is calculation. (Just one tricky part) Applying integration by parts, (Integrate $\frac{1}{a\sqrt{1+a^2}}$ Then in the new integral, substitute $\frac{1}{a} \rightarrow a$ You will get, $f(1) = \int_{1}^{\infty} \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}} - \ln^2(1+\sqrt{2})$ And, $\int_{1}^{\infty} \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}} = \int_{0}^{\infty} \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}} - \int_{0}^{1} \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}}$

Finally we get $f(1) = \frac{\pi^2}{8} - \frac{\ln^2(1+\sqrt{2})}{2}$ .

P.S : I have left many steps, they are left for reader -_-

Updated

Well, really nice problem.

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

Now we will consider the series $\displaystyle \sum_{n=0}^{\infty}{{(-1)}^{n}{a}^{n}} = \frac{1}{1+a}$

Substituting $\displaystyle a = sin(x)$ , we can write I as

$\displaystyle \int_{0}^{\pi/2}{xcos(x)\sum_{n=0}^{\infty}{{sin}^{2n}x{(-1)}^{n}}}$

$\displaystyle \sum_{n=0}^{\infty}{{(-1)}^{n}(\int_{0}^{\pi/2}{xcos(x){sin}^{2n}x}}$

Using IBP with $\displaystyle x = u, cos(x){sin}^{2n}x dx = dv \Rightarrow \frac{{sin}^{2n+1}x}{2n+1} = v$ ,

We will just do the integration first,

$\displaystyle \frac{x({sin}^{2n+1}x)}{2n+1} - \int{\frac{{sin}^{2n+1}x}{2n+1}}$ from $0$ to $\pi/2$

$\displaystyle \frac{\frac{\pi}{2}}{2n+1} - \frac{1}{2n+1}\int_{0}^{\pi/2}{{sin}^{2n+1}x}$

Now substituting in the sum,

$\displaystyle \sum_{n=0}^{\infty}{{(-1)}^{n}\frac{\frac{\pi}{2}}{2n+1}} - \sum_{n=0}^{\infty}{{(-1)}^{n}\frac{1}{2n+1}\int_{0}^{\pi/2}{{sin}^{2n+1}x}}$

First sum is just $\displaystyle \frac{\pi}{2}*{tan}^{-1}(1) = \frac{{\pi}^{2}}{8}$

Second will be $\displaystyle \int_{0}^{\pi/2}{\sum_{n=0}^{\infty}{\frac{{(-1)}^{n}{sin}^{2n+1}x}{2n+1}}}$

$\displaystyle \int_{0}^{\pi/2}{{tan}^{-1}(sinx)}$

Now using $a = sinx$ and IBP,

$\displaystyle = {sin}^{-1}a({tan}^{-1}a) - \int_{0}^{1}{\frac{{sin}^{-1}a}{1+{a}^{2}}}$

$\displaystyle = \frac{{\pi}^{2}}{8} - \int_{0}^{1}{\frac{{sin}^{-1}a}{1+{a}^{2}}}$

Hence, our answer suffices to $\displaystyle = \frac{{\pi}^{2}}{8} - \frac{{\pi}^{2}}{8} + \int_{0}^{1}{\frac{{sin}^{-1}a}{1+{a}^{2}}} = \int_{0}^{1}{\frac{{sin}^{-1}a}{1+{a}^{2}}}$

Any help with this integral would be appreciated! Thanks!

Integral's value is the answer - $\displaystyle \boxed{\frac{{log}^{2}(1 + \sqrt{2})}{2}}$

The answer is $= \dfrac { { log }^{ 2 }(1+\sqrt { 2 } ) }{ 2 }$