Integrale definito

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{sin^2x+2cos^2}{ sin(2x)} \, dx =$

$ = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{sin^2x}{ 2sinx \cdot cosx} \, dx +2\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{cos^2x}{ 2sinx \cdot cosx} \, dx  =$

$ = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{sinx}{ 2 cosx} \, dx +2\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{cosx}{ 2sinx} \, dx  =$

$ = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{2} tan x \, dx +2\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{2} cot x \, dx  =$

Facciamo appello ai soliti integrali noti, nel senso che se esplicitamente richiesto sappiamo come risolverli

$= \left. -\frac{1}{2}ln(cos x) \right|_{\frac{\pi}{6}}^{\frac{\pi}{4}} + \left. \ln(sin x) \right|_{\frac{\pi}{6}}^{\frac{\pi}{4}}=$

$= \left. \frac{1}{2}ln(tan x) + \frac{1}{2}ln(sin x)\right|_{\frac{\pi}{6}}^{\frac{\pi}{4}} = $

$ = \frac{1}{2}ln(\frac{\sqrt{2}}{2}) - \frac{1}{2}ln(\frac{\sqrt{3}}{3}) - \frac{1}{2}ln(\frac{1}{2}) = $

$ = \frac{1}{2}ln(\frac{\sqrt{2}}{2}) + \frac{1}{2}ln(\sqrt{3}) + \frac{1}{2}ln(2) = $

$ = \frac{1}{2}ln(\frac{\sqrt{6}}{2}) + \frac{1}{2}ln(2) = $

$ = \frac{1}{2}ln(\sqrt{6}) -\frac{1}{2}ln(2) + \frac{1}{2}ln(2) = $

$ = \frac{1}{2}ln(\sqrt{6}) = $

$ = \frac{1}{4}ln 6 $