Just...another integral

$\begin{aligned} I & = \int_0^1 \frac {\left(\sqrt x + \sqrt{1-x}\right)\sin \pi x}{1+\sqrt{2x}}dx & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^1 \left(\frac {\left(\sqrt x + \sqrt{1-x}\right)\sin \pi x}{1+\sqrt{2x}} + \frac {\left(\sqrt{1-x} + \sqrt x\right)\sin (\pi-\pi x)}{1+\sqrt{2(1-x)}} \right) dx \\ & = \frac 12 \int_0^1 \left(\sqrt x + \sqrt{1-x}\right)\left(\frac 1{1+\sqrt{2x}} + \frac 1{1+\sqrt{2(1-x)}} \right)\sin \pi x\ dx \\ & = \frac 12 \int_0^1 \left(\sqrt x + \sqrt{1-x}\right)\left(\frac {1-\sqrt{2x}}{1-2x} + \frac {1-\sqrt{2(1-x)}}{2x-1} \right) \sin \pi x\ dx \\ & = \frac 12 \int_0^1 \left(\sqrt x + \sqrt{1-x}\right)\left(\frac {\sqrt{2(1-x)}-\sqrt{2x}}{1-2x} \right) \sin \pi x\ dx \\ & = \frac 12 \int_0^1 \left(\frac {1-2x}{1-2x} \right)\sqrt 2 \sin \pi x\ dx \\ & = \frac 1{\sqrt 2} \int_0^1 \sin \pi x\ dx \\ & = \frac {\cos \pi x}{\sqrt 2 \pi} \bigg|_1^0 = \frac {\sqrt 2}\pi \approx \boxed{0.450} \end{aligned}$