I was asked to solve $\int{\sqrt{9-x^2}dx}$.
So I used trigonometric substitution and got $$ \frac{9}{2} \arcsin {\frac{x}{3}}+\frac{9}{4}\sin(2\arcsin{\frac{x}{3}}) +C $$
How can I find the $\frac{9}{4}\sin(2\arcsin(\frac{x}{3}))$ part?
The actual solution to the integral is $$ \frac{9}{2} \arcsin {\frac{x}{3}}+\frac{x\sqrt{9-x^2}}{2} +C $$
Hint:
$$\sin(2\arcsin(t))=2\sin(\arcsin(t))\cos(\arcsin(t))=2t\sqrt{1-t^2}.$$
Use $\sin2x=2\sin x\cos x$ and $\sin(\arcsin x)=x, \cos(\arcsin x)=\sqrt{1-x^2}$.
By simplicity, let us denote $\alpha = \arcsin(x/3)$. Then $\sin(2\arcsin(x/3)) = \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) = 2\frac{x}{3} \cos(\alpha)$. Thus: $$\frac{9}{4}\sin(2\arcsin(x/3)) = \frac{3}{2}x\cos(\arcsin(x/3))$$ Now, as you can see here, we have $\cos(\arcsin(x)) = \sqrt{1-x^{2}}$, so the above becomes: $$\frac{9}{4}\sin(2\arcsin(x/3)) = \frac{3}{2}x\sqrt{1-\frac{x^{2}}{9}} $$
$arcsin{\frac{x}{3}}=\alpha$
$sin\alpha=\frac{x}{3}$
$sin2\alpha=2\times\frac{x}{3}\times\frac{\sqrt{9-x^2}}{3}$,since $cos\alpha= \frac{\sqrt{9-x^2}}{3}$
$\frac{9}{4}sin(2arcsin\frac{x}{3})=\frac{9}{4}\times2 \times\frac{x}{3}\times\frac{\sqrt{9-x^2}}{3}= \frac{x\sqrt{9-x^2}}{2}$
Let $y=\arcsin(\frac{x}{3}),$ then $x=3\sin(y)$.
So since $\sin(2\arcsin(\frac{x}{3}))=\sin(2y)=2\sin(y)\cos(y)$ and $\sin^2{y}+\cos^2{y}=1$ you have
$$\cos^{2}(y)=1-\frac{x^2}{9}\implies\cos(y)=\sqrt{1-\frac{x^2}{9}}$$
$$\therefore \sin(2\arcsin(\frac{x}{3}))=2\sin(y)\cos(y)=\frac{2x}{3}\sqrt{\frac{9-x^2}{9}}=\frac{2}{9}x\sqrt{9-x^2}$$
Now multiply this by $\frac{9}{4}$ and you get the required result.
