Question

Evaluate the integral
${\int }_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}}\frac{dx}{\sqrt{4-9x^2}}$

• A. $\frac{\pi }{36}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{7\pi }{30}$

Answer 1

The correct option is A $\frac{\pi }{36}$

${\int }_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}}\frac{dx}{2\sqrt{1-\frac{a{x}^2}{2}}}=\frac{1}{2}{\int }_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}}\frac{dx}{\sqrt{1-(\frac{3x}{2}{)}^2}}$

$={[\frac{2}{3}\times \frac{1}{2}si{n}^{-1}\left(\frac{3x}{2}\right)]}_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}}$

$=\frac{2}{3}\times \frac{1}{2}[si{n}^{-1}\left(\frac{3\sqrt{3}}{3\left(2\right)}\right)-si{n}^{-1}\left(\frac{3\sqrt{2}}{3\times 2}\right)]$

$=\frac{1}{3}[si{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)-sin\left(\frac{1}{\sqrt{2}}\right)]$

$=\frac{1}{3}[\frac{\pi }{3}-\frac{\pi }{4}]$

$=\frac{1}{3}\left(\frac{\pi }{12}\right)$

$=\frac{\pi }{36}$