Can we find $\int\sqrt{1+a^2\sin^2ax}\,dx$? I do not think trigonometic subsitutions work here, $\displaystyle\int\sqrt{1+a^2\sin^2ax}\,dx$ where $a$ is a real number. (clear when $a=0$)
I was wondering if there is a way to find the antiderivative or not.
 A: You can find an antiderivative, but it is not in terms of elementary functions.
Hint 1: Use elliptic integrals, specifically use the one defined as:
$$E(\varphi | k^2):= \int_0^{\varphi} \sqrt{1-k^2\sin^2(\theta)} d\theta $$
This is known as the incomplete elliptic integral of the second kind.
Hint 2: To use the above hint, make a substitution of the form $u=ax$ in your integral. This is to get it into the form so that you can apply the elliptic integral. So your integrand will be:
$$\cfrac{\sqrt{1+a^2\sin^2(u)}}{a} $$
Do you think you can take it from there?
Edit: Just to be clear, the integral defined above by $E(\varphi|k^2)$ is also an indefinite integral (just like the one you want to solve). The variable $\theta$ is just a dummy variable.
A: Use $u=ax$, which means $dx=\dfrac{1}{a}\,\mathrm{d}u$
$$\int\sqrt{1+a^2\sin^2ax}\,dx=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{a}}}
{\displaystyle\int}\sqrt{a^2\sin^2\left(u\right)+1}\,\mathrm{d}u$$
Solving for
$${\displaystyle\int}\sqrt{a^2\sin^2\left(u\right)+1}\,\mathrm{d}u=\operatorname{E}\left(u\,\middle|\,-a^2\right)$$
And plugging them in
$$=\dfrac{\operatorname{E}\left(ax\,\middle|\,-a^2\right)}{a}+C$$
