Question

∫_0^3▒〖(2+3/(√9-x^2 ))dx〗

Answer 1

$\displaystyle \int\limits_{0}^{3}2+\frac{3}{\sqrt{9-x^2}}dx=\int\limits_{0}^{3}2dx+\int\limits_{0}^{3}\frac{3}{\sqrt{9-x^2}}dx$
fazemos a substituição trigonométrica:
$x=3\cos\phi\implies dx=-3\sin \phi d\phi$
a integral ficará:
$\displaystyle \int\limits_{0}^{3}2dx+\int\limits_{0}^{3}\frac{3}{\sqrt{9-x^2}}dx=\int\limits_{0}^{3}2dx+\int\limits_{0}^{3}\frac{3}{\sqrt{9-9\cos^2\phi}}(-3\cos\phi d\phi)$
agora é só resolver a fração e integrar:
$\displaystyle \int\limits_{0}^{3}2dx-\int\limits_{0}^{3}\frac{9\cos\phi d\phi}{\sqrt{9-9\sin^2\phi}}= \int\limits_{0}^{3}2dx+\int\limits_{3}^{0}\frac{9\cos\phi d\phi}{\sqrt{9-9\sin^2\phi}}~~~(1)\\\\\\  \int\limits_{0}^{3}2dx+\int\limits_{3}^{0}\frac{9\cos\phi d\phi}{\sqrt{9-9\sin^2\phi}}= \int\limits_{0}^{3}2dx+9\int\limits_{3}^{0}\frac{\cos\phi d\phi}{\sqrt9\sqrt{1-\sin^2\phi}}~(2)\\\\\\ \int\limits_{0}^{3}2dx+9\int\limits_{3}^{0}\frac{\cos\phi d\phi}{3\sqrt{1-\sin^2\phi}}=\int\limits_{0}^{3}2dx+\frac{9}{3}\int\limits_{3}^{0}\frac{\cos\phi d\phi}{\cos\phi}~~(3)\\\\\\\int\limits_{0}^{3}2dx+\frac{9}{3}\int\limits_{3}^{0}d\phi=\left[\frac{}{}2x\right]^{3}_{0}+3\left[\frac{}{}\phi\right]_{3}^{0}~~~(4)$
Note que $\displaystyle \phi=\arcsin\left(\frac{x}{3}\right)$
então
$\displaystyle\left[\frac{}{}2x\right]^{3}_{0}+3\left[\frac{}{}\phi\right]_{3}^{0}=(2\cdot3+0)+3\left(\arcsin\left(\frac{0}{3}\right)-\arcsin(1)\right)~~(5)\\\\\\\int\limits_{0}^{3}2dx+\int\limits_{3}^{0}\frac{9}{\sqrt{9-x^2}}dx=6+3\pi-\frac{3\pi}{2}=\frac{12+6\pi-3\pi}{2}=\boxed{\frac{12+3\pi}{2}}$