How to find $\int_{0}^{4} \sqrt{16-x^2} dx$?

The easiest way to solve this integral is to notice that the curve $y=\sqrt{16-x^2}$ is one quarter of a circle, so the area under this curve will be one quarter the area of the circle. The circle has radius $4$, so area $\pi 4^2 = 16\pi$.

This means the integral must evaluate to $4\pi$.

With the suggested substitution,

$$\int_0^4\sqrt{16-x^2}dx=\int_0^{\frac\pi2}16\cos^2u\,du=8\int_0^{\pi/2}(\cos2u+1)\,du.$$

As $\cos2u$ runs from $1$ to $-1$ symmetrically, this contribution vanishes and $4\pi$ remain.

Does the answer have to be algebraically derived? If not, just draw it, recognize that it's a quarter of a circle with radius 4, thus the answer is $$\frac{\pi 4^2}{4} = 4\pi.$$

If you require something more analytic, you can work backward through double integrals: $$ \int_0^4 \sqrt{16-\cos^2 x} \operatorname{d} x = \int_0^4 \int_0^{\sqrt{16-\cos^2 x}} 1 \operatorname{d} y \operatorname{d} x.$$ From there, you again draw the integration region, and do a change of variables to polar coordinates: $$\begin{align} x &= r\cos\theta \\ y &= r\sin\theta \Rightarrow \\ I &= \int_0^{\pi/2} \int_0^4 r \operatorname{d} r \operatorname{d} \theta \\ & = \frac{\pi}{2} \left(\frac{4^2}{2}\right) = 4\pi \end{align}$$