Proving an integral $\int \sqrt{a^2 - u^2} \, du$ How can I prove this?? Any hint or help would be great! Thanks :)
$$\int \sqrt{a^2-u^2} \mathrm{d}u = \frac{u}{2} \sqrt{a^2-u^2} +\frac{a^2}{2} \sin^{-1}(\frac{u}{a}) + C$$
 A: There are essentially three ways to go:
1) You introduce a $1$ and integrate by parts:
$$
\int 1\cdot\sqrt{a^2-u^2}\, du=u\sqrt{a^2-u^2}-\int u\frac{d}{du}\bigl(\sqrt{a^2-u^2}\bigr)\,du
$$
2) You change variables $u=a\sin t$.
3) You notice that your function describes a circle, draw a graph and find the primitive (connected with area) that way.
Can you proceed from here?
A: HINT: Because $a^2(\sin^2t+\cos^2t)=a^2$, substitution $u=a\sin t$ is a natural one.
A: Try manipulating the square root to get something of the form $\sqrt{1-t^2}$, then re-express (the square root) as  $\sqrt{1-\sin^2(\sin^{-1}t)}$.
Implementing this strategy would look like this:
$$\begin{array}{lll}
\int\sqrt{a^2-u^2}du&=&\rvert a \lvert\int\sqrt{1-\bigg(\frac{u}{\rvert a \lvert}\bigg)^2}du\\
&=&\rvert a \lvert\int\sqrt{1-\sin^2\bigg(\sin^{-1}\frac{u}{\rvert a \lvert}\bigg)}du\\
\end{array}$$
This leads naturally to the substitution
$$v = \sin^{-1}\frac{u}{\rvert a \lvert}$$
or alternatively
$$\rvert a \lvert \sin v = u$$
which should lead to this integral
$$a^2\int \cos^2vdv$$
