$\int \frac{dx}{2\sin^2 x + 5\cos^2 x}$ is equal to

$\int \frac{dx}{2\sin^2 x + 5\cos^2 x}$

Divide both numerator and denominator by $\cos^2 x$:

$\int \frac{dx}{2\sin^2 x + 5\cos^2 x}$

$= \int \frac{\frac{1}{\cos^2 x} \cdot dx}{\frac{2\sin^2 x}{\cos^2 x} + \frac{5\cos^2 x}{\cos^2 x}}$

$= \int \frac{\sec^2 x \, dx}{2\tan^2 x + 5}$

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Let $t = \tan x$

Then $dt = \sec^2 x \, dx$

The integral becomes:

$\int \frac{dt}{2t^2 + 5}$

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$\int \frac{dt}{2t^2 + 5}$

$= \int \frac{dt}{2(t^2 + \frac{5}{2})}$

$= \frac{1}{2} \int \frac{dt}{t^2 + \frac{5}{2}}$

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Using $\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$

Here $a^2 = \frac{5}{2}$, so $a = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}$

$\frac{1}{2} \int \frac{dt}{t^2 + \frac{5}{2}}$

$= \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{10}}{2}} \tan^{-1}\left(\frac{t}{\frac{\sqrt{10}}{2}}\right) + C$

$= \frac{1}{2} \cdot \frac{2}{\sqrt{10}} \tan^{-1}\left(\frac{2t}{\sqrt{10}}\right) + C$

$= \frac{1}{\sqrt{10}} \tan^{-1}\left(\frac{\sqrt{2} \cdot t}{\sqrt{5}}\right) + C$

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Substituting $t = \tan x$ back:

$\frac{1}{\sqrt{10}} \tan^{-1}\left(\frac{\sqrt{2}\tan x}{\sqrt{5}}\right) + C$