The value of $$\int_0^{4042} \frac{\sqrt{x} d x}{\sqrt{x}+\sqrt{4042-x}}$$ is equal to

The value of $$\int_{-1 / 2}^{1 / 2} \cos ^{-1} x d x$$ is

Find the value of $$\int_0^1 \frac{\log (1+x)}{1+x^2} d x$$ is

The value of $$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x$$ is