Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$ I need to solve the integral below, but I just can't figure how.

$$\int \sqrt{16-9x^2}\,dx$$

I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please explain to me how to finish it...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$

Integrate by parts

$$
\begin{eqnarray}
\color{blue}{ \int \sqrt{ 16 - 9 x^2 } d x } &=& x \sqrt{ 16 - 9 x^2 }
     + \int \frac{9 x^2}{ \sqrt{ 16 - 9 x^2 } } dx\\
&=& x \sqrt{ 16 - 9 x^2 }
     - \int \frac{16 - 9 x^2}{ \sqrt{ 16 - 9 x^2 } } dx
     + \int \frac{16}{ \sqrt{ 16 - 9 x^2 } } dx\\
&=& x \sqrt{ 16 - 9 x^2 }
     - \underbrace{ \color{blue}{ 
         \int \sqrt{ 16 - 9 x^2 } dx
       } }_{\displaystyle \text{This is the same!}}
     + \int \frac{16}{ \sqrt{ 16 - 9 x^2 } } dx.
\end{eqnarray}
$$

Rearange

$$
\begin{eqnarray}
\color{blue}{ \int \sqrt{ 16 - 9 x^2 } d x }
&=& \frac{1}{2} x \sqrt{ 16 - 9 x^2 }
     + \int \frac{8}{ \sqrt{ 16 - 9 x^2 } } dx\\
&=& \frac{1}{2} x \sqrt{ 16 - 9 x^2 }
     + \frac{8}{3} \underbrace{ \color{green}{ 
       \int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big)
     } }_{\displaystyle \text{This should be familiar!}^{(1)}}\\
\\
&=& \bbox[16px,border:2px solid #800000] { \color{#800000}{
     \frac{1}{2} x \sqrt{ 16 - 9 x^2 }
     + \frac{8}{3} {\sin^{-1}}\big( 3 x / 4\big) } }
\end{eqnarray}
$$

Footnotes
(1) In case
$$
\color{green}{ 
       \int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big)
     }
$$
is not familiar...

$$
\begin{eqnarray}
\color{green}{ 
       \int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big)
     } &\stackrel{\color{magenta}{3 x / 4 = \sin(\phi)}}=&
\int \frac{1}{\sqrt{1 - \sin^2(\phi)} } d\sin(\phi)\\
&=& \int \frac{\cos(\phi)}{\cos(\phi)} d\phi\\
&=& \phi\\
&\stackrel{\color{magenta}{\phi = {\sin^{-1}}\big( 3 x / 4\big)}}=&
{\sin^{-1}}\big( 3 x / 4\big).
\end{eqnarray}
$$

