Utente:F l a n k e r/Sandbox

${\displaystyle \int _{-2}^{2}{\sqrt {9t^{4}+16t^{2}}}\,dt=\int _{-2}^{2}t\;{\sqrt {9t^{2}+16}}\,dt}$

dato che:

${\displaystyle d\left(9t^{2}+16\right)=18t\,dt}$

allora:

${\displaystyle \int _{-2}^{2}{\frac {1}{18}}{\sqrt {9t^{2}+16}}\ d\left(9t^{2}+16\right)}$

che è come se fosse l'integrale di:

${\displaystyle \int {\frac {1}{18}}{\sqrt {x}}\,d\left(x\right)}$

quindi:

${\displaystyle \int _{-2}^{2}{\frac {1}{18}}{\sqrt {9t^{2}+16}}\ d\left(9t^{2}+16\right)={\frac {1}{18}}\cdot {\frac {2}{3}}\left[9t^{2}+16\right]_{-2}^{2}}$