= Solution
Use the <integration by substitution> $x=\sin\theta$, with $-\pi/6\leq\theta\leq\pi/6$. On this interval $\cos\theta>0$, so it cancels the positive denominator <square root>. The transformed <integral> is
$$
\frac12\int_{-\pi/6}^{\pi/6}\sqrt{1-\frac49\sin^2\theta}\,d\theta.
$$
Its integrand is even. Therefore the answer in the PDF's modulus-amplitude convention for the <elliptic integral of the second kind> is
$$
\boxed{E\left(\frac23,\frac\pi6\right)}.
$$
The modulus is $2/3$, not the squared parameter $4/9$.
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