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:

${\displaystyle \int _{0}^{2}{\frac {1}{18}}{\sqrt {9t^{2}+16}}\ d\left(9t^{2}+16\right)\qquad \mathrm {per} \ t>0}$

${\displaystyle -\int _{-2}^{0}{\frac {1}{18}}{\sqrt {9t^{2}+16}}\ d\left(9t^{2}+16\right)\qquad \mathrm {per} \ t<0}$

che è come se fosse l'integrale di:

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

quindi:

${\displaystyle \int _{-2}^{2}{\frac {1}{18}}{\sqrt {9t^{2}+16}}\ d\left(9t^{2}+16\right)={\frac {1}{18}}\cdot {\frac {2}{3}}\left[\left(9t^{2}+16\right)^{\frac {3}{2}}\right]_{0}^{2}-{\frac {1}{18}}\cdot {\frac {2}{3}}\left[\left(9t^{2}+16\right)^{\frac {3}{2}}\right]_{-2}^{0}={\frac {1}{27}}\left[\left(9t^{2}+16\right)^{\frac {3}{2}}\right]_{0}^{2}-{\frac {1}{27}}\left[\left(9t^{2}+16\right)^{\frac {3}{2}}\right]_{-2}^{0}=...}$