Solution
= Solution
The <fundamental theorem of calculus> and part (a) give
$$
I_\theta=E(\theta,k)\sqrt{1-k^2\sin^2\theta}=E(\theta,k)E_\theta(\theta,k)=\partial_\theta\left(\frac12E(\theta,k)^2\right).
$$
The difference is constant in $\theta$. Since $I(0,k)=E(0,k)=0$, the constant is zero, proving
$$
\boxed{I(\theta,k)=\frac12E(\theta,k)^2.}
$$
This is simply the <chain rule> integrated, with $E$ itself as the antiderivative of the square-root factor.