My main objective is to rearrange the following to find $A=$.
I start with
$C=\int^{B}_{0}\sqrt {1+\dfrac {A^{2}}{B^{2}}\cos^{2}\left(\dfrac {x}{B}\right) }dx$
By substituting $y=x/B$ and using $\cos^2y=1-\sin^2y$
$C = B \sqrt{1 + \frac{A^2}{B^2}} \int_0^1 \sqrt{1 - \frac{A^2}{A^2 + B^2}\sin^2 y}\,dy = \sqrt{A^2 + B^2} E\left(1,\frac{A}{\sqrt{A^2 + B^2}}\right)$
This suggests that it would be easier to work with $k = A/\sqrt{A^2 + B^2}$ and $B$ rather than just $A$ and $B$. So
$C = \frac{B}{\sqrt{1 - k^2}} E(1,k)$
and knowing $k$ I can recover $A$ by using
$A = \frac{Bk}{\sqrt{1 - k^2}}$
So back to my main problem. I have two possible ways on how to find $k$ but I don't know how to start, any help will be greatly appreciated.
Method 1) Expand $\sqrt{1 - k^2 \sin^2 y}$ as a binomial series in $k^2 \sin^2y$ which can be integrated term by term (since the integral of $\sin^{2n} y$ is known) to end up with a polynomial in $k^2$; $k$ will have to be solved numerically.
Method 2) Find a root-finding algorithm to solve
$\frac{B}{\sqrt{1 - k^2}} E(1,k) - C = 0$
for $k$ and integrate numerically at each step of the algorithm.
Also, it might be important to mention that I can find values for $C$ given values for $A$ and $B$ but the whole point of this is to find values of $A$ given values for $C$ and $B$.