u-substitution for a definite integral I want to solve the following integral:
$$\int_{-2}^2 \sqrt{4- x^2} dx$$
I am thinking of doing a u-substitution of the whole term inside the root. If I do this, I need to change the limits of the integral. Doing so, the new limits would be equal (0=0). Doesn't this mean that the value of the integral would be 0?
 A: An alternative way is to think this as surface of a semi circle with radius $2$. Then the answer is $2\pi$
A: The integral can be found with the substitution $x=\sin \theta$.
If we let $u=4-x^2$. Then $du=-2xdx$.
Note that $x=\sqrt{4-u}$ if $x\ge 0$ and $x=-\sqrt{4-u}$ if $x<0$.
So, $\displaystyle dx=\frac{du}{-2\sqrt{4-u}}$ if $x\ge 0$ and $\displaystyle dx=\frac{du}{2\sqrt{4-u}}$ if $x\le 0$.
So we actually do not have $\displaystyle \int_0^0 \frac{\sqrt{u}du}{-2\sqrt{4-u}}$.
\begin{align*}
\int_{-2}^2\sqrt{4-x^2}dx&=\int_{-2}^0\sqrt{4-x^2}dx+\int_0^2\sqrt{4-x^2}dx\\
&=\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}+\int_4^0 \frac{\sqrt{u}du}{-2\sqrt{4-u}}\\
&=\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}+\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}\\
&=\int_0^4 \frac{\sqrt{u}du}{\sqrt{4-u}}
\end{align*}
This does not help us to find the integral. But at least the integral is not $0$.
A: Why $0$? If you do the natural substitution ($x=2\sin u$ and $\mathrm dx=2\cos u\,\mathrm du$), then the limits will be $\pm\frac\pi2$.
