Can this integral be reduced to an elliptic integral, or to anything else that is easier to evaluate? New user here......I have run into a problem where I am trying to evaluate the following integral (if at all possible, analytically):
$ I = \int_{0}^{\pi/2}\sin(x) \sqrt{1 + k^2 \sin^2(x)}\, dx  $ 
To me, it seems that this has to be related to one or other of the elliptic integrals, somehow, but I am not being able to see the connection.
Any help would be appreciated, thank you !
 A: Hints:
It can be expressed with elementary functions, due to the $\sin x$ factor in front of the root. We can suppose $k>0$.


*

*First step: rewrite the integral as
\begin{align}
\int_{0}^{\pi/2}\mkern-9mu\sin x\,\sqrt{\mathstrut 1 + k^2 \sin^2\!x}\, \mathrm dx=\int_{0}^{\pi/2}\mkern-9mu\sin x \,\sqrt{\mathstrut1 + k^2-k^2 \cos^2\!x}\, \mathrm dx \\
=\sqrt{\mathstrut1+k^2}\int_{0}^{\pi/2}\mkern-9mu\sin x \,\sqrt{1 -\frac{k^2}{1 + k^2}\cos^2\!x}\, \mathrm dx 
\end{align} use the substitution

*Second step: use the substitution
$$\begin{cases}\cos\theta=\frac k{\sqrt{1+k^2}}\,\cos x,\quad \theta\in [0,\pi]\iff \theta=\arccos\biggl(\frac k{\sqrt{1+k^2}}\,\cos x\biggr),\\
\sin\theta\, \mathrm d\theta=\frac k{\sqrt{1+k^2}}\,\sin x\,\mathrm d x, \\
x=0\longleftrightarrow \theta =\arccos\frac k{\sqrt{1+k^2}}, \quad  x=\frac\pi 2\longleftrightarrow \theta=\frac\pi 2.
\end{cases} $$
Denote $\theta_0=\arccos\frac k{\sqrt{1+k^2}}$. The integral becomes
$$I=\frac{1+k^2}k \int_0^{\theta_0}\sin\theta\sqrt{1-\cos^2\!\theta\,}\, \mathrm d\theta =\sin^2\!\theta\, \mathrm d\theta =\frac{1+k^2}k\int_0^{\theta_0}\frac{1-\cos2\theta}2\, \mathrm d\theta.
$$

