$\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x=$

$\int_0^{32 \pi} \sqrt{1-\cos 4 x} d x=$

If $f(x)=\int \frac{\sin 2 x+2 \cos x}{4 \sin ^2 x+5 \sin x+1} d x$ and $f(0)=0$, then $f(\pi / 6)=$

$\int_{-2}^2 x^4\left(4-x^2\right)^{\frac{7}{2}} d x=$