I need to prove the next equality,
$ \int \limits_{0}^{\frac{\pi}{2}}\int \limits_{0}^{\frac{\pi}{2}} \cos(2a \sin(x) \sin(y))dxdy = \left(\int \limits_{0}^{\frac{\pi}{2}}\cos(a \sin(z))dz \right)^2 $ for each $a \in \mathbb{R}$.
I am proving it as follows.
- I used a series expansion for the left integral:
$ \cos(2a \sin(x)\sin(y)) = 1 + \sum_{i = 1}^{\infty}\left( (-1)^{i}\frac{(2a)^{2i}}{(2i)!}\sin^{2i}(x)\sin^{2i}(y)\right) \qquad (1). $
After integrating $(1)$ will be as $ \frac{\pi^{2}}{4}\left(1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}(2i)!}{2^{2i}(i!)^{4}}\right)\right). $
- I used a series expansion for the right integral:
$ \int \limits_{0}^{\frac{\pi}{2}}\cos(a \sin(z))dz = \frac{\pi}{2}\left(1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}}{2^{2i}(i!)^{2}}\right)\right). $
How to prove an equality of the next expression,
$ \left(\frac{\pi}{2}\left( 1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}}{2^{2i}(i!)^{2}}\right)\right) \right)^{2} \qquad (2), $
$ \frac{\pi^{2}}{4}\left(1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}(2i)!}{2^{2i}(i!)^{4}}\right)\right), $
by gift to the square of $(2)$?