Evaluate $\int \sqrt{ a^2 - x^2} dx$ Problem:

Evaluate the integral
$$
\int \sqrt{ a^2 - x^2 } dx.
$$

Solution:

Let us put
$$
x = a \sin \theta . \tag{0} 
$$
Then we have
$$
dx  = a \cos \theta d \theta, \tag{1} 
$$
and
$$
a^2 - x^2 = a^2 \left( 1 - \sin^2 \theta \right) = a^2 \cos^2 \theta. \tag{2} 
$$
So we have
$$
\begin{align}
& \ \ \ \int \sqrt{ a^2 - x^2 } dx \\
&= \int \sqrt{ a^2 - a^2 \sin^2 \theta } \, a \cos \theta d \theta \\
&= \int \sqrt{ a^2 \left( 1 - \sin^2 \theta \right) } \, a \cos \theta d \theta \\
&= \int \sqrt{ a^2 \cos^2 \theta }\, a \cos \theta d \theta \\
&= \int \big\lvert a \cos \theta \big\rvert \, a \cos \theta d \theta \\
& \mbox{[ note that for any real number $r$, we have $\lvert r \rvert = \sqrt{r^2}$ ]} \\
&= \int (a \cos \theta ) a \cos \theta d \theta \\
& \mbox{[ we assume that $0 <  \theta <  \frac{\pi}{2}$ so that $\cos \theta > 0$, } \\
& \qquad  \mbox{ and we also assume that $a > 0$ ]} \\
&= \int a^2 \cos^2 \theta d \theta \\
&= a^2 \int  \cos^2 \theta d \theta \\
&= a^2 \int \frac{2  \cos^2 \theta }{2} d \theta \\
&= a^2 \int  \frac{2  \cos^2 \theta -1 + 1 }{2} d \theta \\
&= a^2 \int \frac{ \cos 2 \theta + 1 }{2} d \theta \\
&= \frac{a^2}{2} \int (\cos 2 \theta + 1) d \theta \\
&= \frac{a^2}{2} \big( \int \cos 2 \theta d \theta + \int d \theta \big). 
\end{align}
$$


Thus we have
$$
\int \sqrt{ a^2 - x^2 } dx = \frac{a^2}{2} \big( \int \cos 2 \theta d \theta + \int d \theta \big). \tag{3}
$$


Now we evaluate $\int \cos 2 \theta d \theta$ as follows:


Let us put $2 \theta = \psi$. Then we have
$$
2 d \theta = d \psi,
$$
and hence
$$
d \theta = \frac{1}{2} d \psi.
$$
Then
$$
\begin{align}
\int \cos 2 \theta d \theta &= \int \cos \psi \frac{1}{2} d \psi \\
&= \frac12 \int \cos \psi d \psi \\
&= \frac12 \sin \psi \\
&= \frac12 \sin 2 \theta. \tag{4}  
\end{align}
$$


Now putting the value of $\int \cos 2 \theta d \theta$ from (4) into (3) we get
$$
\begin{align}
& \ \ \ \int \sqrt{ a^2 - x^2 } dx  \\ 
&= \frac{a^2}{2} \big( \int \cos 2 \theta d \theta + \int d \theta \big) \qquad \mbox{[ using (3) above ]} \\
&= \frac{a^2}{2} \left( \frac12 \sin 2 \theta \ + \theta  \right) + C \\ 
& \qquad \mbox{[ using (4) above; } \\ 
& \qquad \qquad \mbox{ note that $C$ is an arbitrary constant of integration ]}\\ 
&= \frac{a^2}{2} \left( \frac12 (2 \sin \theta \cos \theta ) + \theta \right) + C \\ 
&= \frac{a^2}{2} \big( \sin \theta \cos \theta  + \theta \big) + C \\ 
&= \frac{1}{2} \big( (a \sin \theta) (a \cos \theta) + a^2 \theta \big) +C \\
&= \frac12 \left( x \sqrt{a^2 - x^2} + a^2 \sin^{-1}  \frac{x}{a}  \right) + C \\ 
& \qquad \mbox{[ refer to (0) and (2) above; } \\ 
& \qquad  \qquad \mbox{ note that since $x = a \sin \theta$, therefore } \\ 
& \qquad \qquad \mbox{ we have $\sin \theta = \frac{x}{a}$ and so $\theta = \sin^{-1} \frac{x}{a}$. ] } \\
&= \frac{x}{2} \sqrt{ a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C.  
\end{align}
$$

Is this procedure correct, clear and rigorous enough?

You obtained the correct answer.  However, your assumption that $0 < \theta < \pi/2$ is not necessary.
Since you let $x = a\sin\theta$, where $a > 0$, then
$$\theta = \arcsin\left(\frac{x}{a}\right)$$
is defined to be the unique angle in the interval $[-\pi/2, \pi/2]$ such that $\sin\theta = x/a$.  Since $\theta \in [-\pi/2, \pi/2]$, $\cos\theta \geq 0$.  Hence, $|\cos\theta| = \cos\theta$.
