Integrali

Poniamo $x = \frac{\sqrt{2}}{\sqrt{5}} sin t $ da cui discendono:

• $dx = \frac{\sqrt{2}}{\sqrt{5}} cos t \, dt $
• $t = arcsin (\frac{\sqrt{5}}{\sqrt{2}} x)  $
• $ cost = \sqrt{1-sin^2t} = \frac{1}{\sqrt{2}} \sqrt{2-5x^2} $
• $ (2-3x) = (2-3\frac{\sqrt{2}}{\sqrt{5}} sin t) = \frac{\sqrt{1}}{\sqrt{5}} (2\sqrt{5}-3\sqrt{2}sin t) $

per cui

$ \int \frac{2-3x}{\sqrt{2-5x^2}} \, dx = $

$ = \int \frac { (2\sqrt{5}-3\sqrt{2}sin t)}{\sqrt{5}\sqrt{2}\sqrt{1-sin^2 t}} \cdot \frac{\sqrt{2}}{\sqrt{5}} cos(t) \, dt = $ 

$ = \int \frac{ (2\sqrt{5}-3\sqrt{2}sin t)}{5cos(t)} \cdot cos(t) \, dt = $

$ = \int \frac{ (2\sqrt{5}-3\sqrt{2}sin t)}{5} \, dt = $ 

$ = \int \frac{2}{\sqrt{5}} \, dt - \int \frac{3 \sqrt{2}}{5}sin t \, dt = $

$ = \frac{2}{\sqrt{5}} t + \frac{3 \sqrt{2}}{5} cos(t) + c $

$ = \frac{2}{\sqrt{5}} arcsin (\frac{\sqrt{5}}{\sqrt{2}} x) + \frac{3 \sqrt{2}}{5} \frac{1}{\sqrt{2}} \sqrt{2-5x^2} + c = $ 

$ = \frac{2}{\sqrt{5}} arcsin (\frac{\sqrt{5}}{\sqrt{2}} x) + \frac{3}{5} \sqrt{2-5x^2} + c $