Integrate $$I=\int \frac{dx}{\sqrt{4-\sin^2 x}}$$

we can write it as

$$I=\sqrt{2} \times \int \frac{dx}{\sqrt{7+\cos 2x}}$$

Put $\tan x=t$ we get

$$I=\int \frac{dt}{\sqrt{1+t^2}\sqrt{8+6t^2}}$$

is it possible to continue?


Nope, this is an elliptic integral as per Botond's comment. The best you can do is express it as a elliptic integral of the first kind:

$$\frac{1}{2}\int (1 - \frac{1}{4} \sin^2 x)^{-1/2} \, \mathrm{d}x = \frac{1}{2}F\left(x \, \bigg | \,\frac{1}{4}\right)$$


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