Can someone please explain why this is the first move when solving this integral. The integral which I am solving is $$\int_0^c\sqrt{c^2-x^2}~\mathrm dx$$ and the first thing which they suggest doing is setting $x=(\sqrt{a})/(\sqrt{b}) \sin u$. I understand everything after this but I do not comprehend where and why this step came to be. 
 A: Maybe it is better to divide the procedure into two steps.
First by letting $X=x/c$ (here we assume $c>0$) we obtain
$$\int_0^c\sqrt{c^2-x^2}\,dx=c^2\int_0^1\sqrt{1-X^2}\,dX$$
Now it seems quite natural to use the substitution $X=\sin u$.
A: You can integrate by geometry. $$ y = \sqrt{c^2-x^2}$$ is the top half of a circle of radius $c$ centered at the origin. Therefore the integral is a quarter of the area of a circle of radius $c$. 
I.e. $\frac{1}{4}\pi c^2$.
A: If you must do this by a method other than recognizing this as the area of one quarter of a circle, then the expression $\sqrt{c^2-x^2}$ suggests using the substitution
\begin{align}
x & = c\sin\theta \\
dx & = c\cos\theta\,d\theta \\[10pt]
\sqrt{c^2 -x^2} & = \sqrt{c^2 - c^2\sin^2\theta} = c\sqrt{1-\sin^2\theta} \\[5pt]
& = c\sqrt{\cos^2\theta}  = c \cos\theta.
\end{align}
As $x$ goes from $0$ to $c$, then $\sin\theta$ goes from $0$ to $1$, so $\theta$ goes from  $0$ to $\pi/2$. So we get
$$
\int_0^c \sqrt{c^2-x^2}\,dx = \int_0^{\pi/2} c^2\cos^2\theta\,d\theta.
$$
One way of evaluating the integral is by using the trigonometric identity
$$
\cos^2\theta = \frac 1 2 + \frac 1 2 \cos(2\theta).
$$
There is also a way to do it without finding any antiderivatives: Observe that by symmetry, since $\cos\theta = \sin(\frac\pi2 - \theta)$, we have
$$
\int_0^{\pi/2} \cos^2\theta\,d\theta = \int_0^{\pi/2} \sin^2\theta\,d\theta
$$
and then find the sum of those two integrals by using the identity $\cos^2\theta+\sin^2\theta=1$.
