Integral of $\sqrt{4-\sin^2(x)}$

Question

So we have to find the antiderivative of $\sqrt{4-\sin^2x}$.
What I did was to put $\sin x=t;\cos xdx=dt$.
But now I am not able to calculate the integral of $\sqrt\frac{4-x^2}{1-x^2}$.
So how to initiate the solution of the problem? Please guide.
WolframAlpha says something elliptic, which I dont know yet. So plase help.

Answer 1

If you consider the ellipse of equation $$ \frac{x^2}{3}+\frac{y^2}{4}=1 $$ and try to compute the length of an arc of it starting from $(\sqrt{3},0)$ counterclockwise, you can use the parametrization $$ x=\sqrt{3}\cos t,\qquad y=2\sin t $$ that gives an integral of the form $$ \int_0^{t_0}\sqrt{3\sin^2t+4\cos^2t}\,dt= \int_0^{t_0}\sqrt{4-\sin^2t}\,dt $$ where $t_0$ represents the parameter corresponding to the terminal point of the arc.

Such integrals are not computable in terms of elementary functions and are known as elliptic integrals.