Find the area of the ellipsoid $(\frac{x}{a})^2 + (\frac{y}{a})^2 + (\frac{z}{b})^2 = 1$, using Surface Integral

Question

I need to find the area of the ellipsoid $(\frac{x}{a})^2 + (\frac{y}{a})^2 + (\frac{z}{b})^2 = 1$, using Surface Integral. I tried to use the method that I saw in another post on Math stackexchange. I parametrize the equation by letting $x = asin(u)cos(v)$, $y = a sin(u)sin(v)$, and $z = cos(u)$.
Then I have $\frac{\partial x}{\partial u} = a cos(v) cos(u)$, $\frac{\partial y}{\partial u} = a sin(v)cos(u)$, $\frac{\partial z}{\partial u} = -sin(u)$.
Then I let $$E = a^2cos^{2}(v)cos^2(u) + a^2 sin^2(v)cos^2(u) + sin^2(u)$$Then I have $\frac{\partial x}{\partial v} = -asin(u)sin(v)$, $\frac{\partial y}{\partial v} = a sin(u) cos(v)$, $\frac{\partial z}{\partial v} = 0$.
Then I have $$F = a^2sin^(u)sin^(v) + a^2 sin^2(u) cos^2(v)$$
Then I also have $$G = -a^2 sin(u)cos(u) sin(v)cos(v) + a^2 sin(v)cos(v)sin(u)cos(u) = 0$$Then $$E+F-G = a^2 cos^2(u) + sin^2 (u) + a^2 sin^2(u) = a^2 + sin^2(u)$$Then the area is $$ \int_{0}^{\pi} \int_{0}^{2\pi} \sqrt{a^2 + sin^2(u)} dvdu$$Then I'm stuck on how to calculate this integral, since my instructor hasn't talked anything about elliptic integral.