How to solve this definite integral $ \int_0 ^a \sqrt {9-x^2} \ dx $ without using trigonometric substitution 
For what value(s) of $a$ is $$ \int_0 ^a \sqrt {9-x^2} \  dx = \pi \ \text{ ?} $$ 

I solved this question by using integral substitution, but I was told that it is better to obtain this solution by using the graphichal method. 
I tried it, but I wasn't able to have any useful insight. 
When I did it using trignomteric substituton, I ended up with the following equation for $a$, which I wasn't sure how to simplify (hence I need for a better method): 

$$ \Rightarrow \arcsin \left(\frac{a}{3}\right) + \frac {1}{2} \sin \left[ 2 \arcsin \left(\frac{a}{3} \right) \right] =  \frac {2}{9} \pi $$

 A: 
Graphically, the integral 
$$ \int_0 ^a \sqrt {9-x^2} \  dx = \pi $$
represents the shaded area in the diagram, which is the area sum of the circle sector of angle $x$ and the right triangle of base length $a=r\sin x$, i.e.
$$\pi = \frac12 x r^2 + \frac 12 r^2\sin x\cos x$$
where $r=3$ is the radius of the circle. Rewrite the equation as,
$$x + \frac12\sin 2x = \frac{2\pi}9$$
which does not have a close form solution and can only be solved numerically. But, a decent approximation can be obtained with $\sin 2x \approx 2x$ to obtain $x = \frac\pi9$. Thus, $a = 3\sin \frac\pi9 = 1.03$, compared with exact numerical result $1.07$
A: As you wrote it, considering the equation$$\arcsin \left(\frac{a}{3}\right) + \frac {1}{2} \sin \left[ 2 \arcsin \left(\frac{a}{3} \right) \right] =k \qquad \text{where} \qquad k=  \frac {2\pi}{9} $$ Composing  Taylor series around $a=0$, the lhs is
$$\frac{2 a}{3}-\frac{a^3}{81}-\frac{a^5}{4860}-\frac{a^7}{122472}+O\left(a^9\right)$$ Using series reversion
$$a=\frac{3 k}{2}+\frac{k^3}{16}+\frac{13 k^5}{1280}+\frac{493
   k^7}{215040}+O\left(k^9\right)$$
Plugging this result into the original equation and expanding again around $k=0$ gives for the lhs
$$k-\frac{37369 }{92897280}k^9+O\left(k^{11}\right)$$
Using $k=\frac {2\pi}{9}$, this gives $a=1.07033$ while the exact solution is
                                      $a=1.07036$
