Evaluate this integral

Let arcsin(x) = y; Sin(y) = x; cos(pi/2 - y) = x; arcos(x) = pi/2 - y; Therefore arcSin(x) + arcCos(x) = pi/2. The expression evaluates to 2 *x /pi between 0 and 1 hence equal to 2/pi. So the answer is 2/pi.

Let $\begin{cases} \alpha = \arcsin x & \implies \sin \alpha = x \\ \beta = \arccos x & \implies \cos \beta = \sin \left(\dfrac \pi 2-\beta \right) = x \end{cases}$

$\implies \alpha = \dfrac \pi 2 - \beta \implies \alpha + \beta = \dfrac \pi 2$ .

Therefore, we have:

$\begin{aligned} I & = \int_0^1 \frac {dx}{\arcsin x + \arccos x} = \int_0^1 \frac 1{\alpha + \beta} \ dx = \int_0^1 \frac 1{\frac \pi 2} \ dx = \frac {2x}\pi \bigg|_0^1 = \boxed{\dfrac 2\pi} \end{aligned}$