How to find $\int_{0}^{4} \sqrt{16-x^2} dx$? What is the easiest way of solving this integral: 
$$\int_{0}^{4} \sqrt{16-x^2} dx$$ 
My idea was to substitute $x$ with $4\sin u$ and to get under the square root $\cos^{2}{u}$ so i can get rid of it, but then i get again $\cos^{2}u$. I suppose that could be solved using formula 
$$\cos^{2}{\frac{u}{2}} = \frac{1+\cos u}{2}$$ 
But then I got troubles with getting back substitution. 
Am I making somewhere mistake and if not how should I precede, or is there some easier way of solving it? 
 A: 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:
$$\begin{align}
x &= r\cos\theta \\
y &= r\sin\theta \Rightarrow \\
I &= \int_0^{\pi/2} \int_0^4 r \operatorname{d} r \operatorname{d} \theta \\
 & = \frac{\pi}{2} \left(\frac{4^2}{2}\right) = 4\pi
\end{align}$$
A: 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 $4\pi$.
A: 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.
A: First notice that $$\int_0^r\sqrt{r^2-x^2}dx=r^2\int_0^1\sqrt{1-x^2}dx.$$
Then by parts,
$$I=\int_0^1\sqrt{1-x^2}dx=\left.x\sqrt{1-x^2}\right|_0^1+\int_0^1\frac{x^2}{\sqrt{1-x^2}}dx.$$
But $x^2=1-(1-x^2)$ so that
$$I=\int_0^1\frac{1}{\sqrt{1-x^2}}dx-I.$$
The remaining integral is solved with an arc sine and yields $\pi/2$.
