integration - Calculate $int_0^1frac{cos(arctan x)}{sqrt{x}}dx$

When I was considering derivatives and integrals (to try integration by parts) related with a closed form for $\zeta(3)$ (due to J. Jensen, see the Wikipedia), I've asked to me

Question. How it is possible to calculate $$\int_0^1\frac{\cos(\arctan x)}{\sqrt{x}}dx?$$
Can you justify a closed form for
$$\int\frac{\cos(\arctan x)}{\sqrt{x}}dx?$$
Many thanks.

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(The integral $\int \frac{\cos(\arctan x)}{1+\sqrt{x}}dx$ seems much more complicated).

Answer

$$I:=\displaystyle \int_0^1\frac{\cos\arctan x}{\sqrt{x}}\,\mathrm d x$$
$\displaystyle \cos\arctan x = \frac{1}{\sqrt{x^2+1}}$, so:
$$I=\int_0^1\frac{1}{\sqrt{x(x^2+1)}}\,\mathrm d x$$
Now substitute $x=\tan\frac\theta 2$:

$$I=\int_0^1 \frac{1}{\sqrt{x(x^2+1)}}\,\mathrm d x=\frac{\sqrt 2}{2} \int_0^{\frac\pi 2} \frac{\mathrm d \theta}{\sqrt{\sin\theta}}=\frac{\sqrt{2}}{4}\beta \left(\frac 14,\frac 12\right)=\boxed{\frac{\Gamma ^2 \left(\frac 14 \right)}{4\sqrt\pi}}$$
This is also equal to $K\left(\frac 12\right)$.