For $x \in \left(0, \frac{\pi}{2}\right)$, $\int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx$ is equal to

For $x \in \left(0, \frac{\pi}{2}\right)$, find $\int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx$

Using the double angle formula $\sin 2x = 2\sin x \cos x$:

$\int \frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} dx$

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Let $t = \sin x - \cos x$

Differentiating:

$\frac{dt}{dx} = \cos x + \sin x$

$dt = (\sin x + \cos x)dx$

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To express the denominator in terms of $t$:

$t^2 = (\sin x - \cos x)^2$

$t^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x$

$t^2 = 1 - 2\sin x \cos x$

Therefore:

$2\sin x \cos x = 1 - t^2$

$\sqrt{2\sin x \cos x} = \sqrt{1 - t^2}$

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The integral becomes:

$\int \frac{dt}{\sqrt{1-t^2}}$

$= \sin^{-1}(t) + C$

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Substituting back $t = \sin x - \cos x$:

$\sin^{-1}(\sin x - \cos x) + C$