Proving an identity involving differentials of integrals 
Define $$ E(\theta,k) = \int_0^\theta \sqrt{1-k^2\sin{^2{x}}} dx$$ and $$F(\theta,k) = \int_0^\theta \frac{1}{\sqrt{1-k^2\sin{^2{x}}}} dx$$ We are to show $$\left(\frac{\partial E}{\partial k}\right)_\theta = \frac{E-F}{k} $$

Am I right in thinking: $$\left(\frac{\partial E}{\partial k}\right)_\theta =\int_0^\theta \frac{\partial}{\partial k}\sqrt{1-k^2\sin{^2{x}}} dx $$
Which gives $$k\left[\int_0^\theta \frac{\cos^2{x}}{\sqrt{1-k^2\sin{^2{x}}}} dx - \int_0^\theta \frac{1}{\sqrt{1-k^2\sin{^2{x}}}} dx\right]$$
which is $$k\left[\int_0^\theta \frac{\cos^2{x}}{\sqrt{1-k^2\sin{^2{x}}}} dx -F\right]$$
However, I can't seem to reduce the left integral to give what is required. Many thanks in advance.

The calculation of $\left(\frac{\partial E}{\partial k}\right)_{\theta}$ is fine, although you may not want to use $-\sin^2 x = \cos^2 x - 1$.  Instead, calculate $\frac{E-F}{k}$ directly by combining the integrals and simplifying.
