Evaluate: \(\int\frac{2}{3 + 2\sin^2 x + 5\cos^2 x}\,dx\)

Answer: \(\frac{1}{\sqrt{10}}\tan^{-1}\left(\frac{\sqrt{5}\tan x}{2\sqrt{2}}\right) + c\)

Divide numerator and denominator by \(\cos^2 x\):

\(I = \int\frac{2\sec^2 x}{3\sec^2 x + 2\tan^2 x + 5}\,dx = \int\frac{2\sec^2 x}{3 + 3\tan^2 x + 2\tan^2 x + 5}\,dx = \int\frac{2\sec^2 x}{5\tan^2 x + 8}\,dx\).

Put \(\tan x = t\), \(\sec^2 x\,dx = dt\):

\(I = \int\frac{2\,dt}{5t^2 + 8} = \frac{2}{5}\int\frac{dt}{t^2 + \left(\sqrt{\frac{8}{5}}\right)^2} = \frac{2}{5}\cdot\frac{1}{\sqrt{8/5}}\tan^{-1}\left(\frac{t}{\sqrt{8/5}}\right) + c\).

Since \(\frac{2}{5}\cdot\frac{\sqrt{5}}{2\sqrt{2}} = \frac{1}{\sqrt{10}}\) and \(\frac{t}{\sqrt{8/5}} = \frac{\sqrt{5}\,t}{2\sqrt{2}}\),

\(I = \frac{1}{\sqrt{10}}\tan^{-1}\left(\frac{\sqrt{5}\tan x}{2\sqrt{2}}\right) + c\).