$\int \frac{\sin 2x \, dx}{\sqrt{9 - \cos^4 x}}$ equals

The integral to evaluate: $\int \frac{\sin 2x \, dx}{\sqrt{9 - \cos^4 x}}$

Using the double angle formula:

$\sin 2x = 2\sin x \cos x$

The integral becomes:

$\int \frac{2\sin x \cos x \, dx}{\sqrt{9 - \cos^4 x}}$

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Let $u = \cos^2 x$

$\frac{du}{dx} = 2\cos x \cdot (-\sin x)$

$\frac{du}{dx} = -2\sin x \cos x$

$du = -2\sin x \cos x \, dx$

$2\sin x \cos x \, dx = -du$

Also, $\cos^4 x = (\cos^2 x)^2 = u^2$

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Substituting into the integral:

$\int \frac{-du}{\sqrt{9 - u^2}}$

$= -\int \frac{du}{\sqrt{9 - u^2}}$

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Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$

Here $a^2 = 9$, so $a = 3$

$-\int \frac{du}{\sqrt{9 - u^2}} = -\sin^{-1}\left(\frac{u}{3}\right) + C$

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Substituting back $u = \cos^2 x$:

$-\sin^{-1}\left(\frac{\cos^2 x}{3}\right) + C$