Question

Choose the correct answer.
${\int }_0^{\frac{2}{3}}\frac{1}{4+9x^2}dx$
$\left(a\right)\frac{\pi }{6}\left(b\right)\frac{\pi }{12}\left(c\right)\frac{\pi }{24}\left(d\right)\frac{\pi }{4}$

Answer 1

${\int }_0^{\frac{2}{3}}\frac{1}{4+9x^2}dx=\frac{1}{9}{\int }_0^{\frac{2}{3}}\frac{1}{\frac{4}{9}+x^2}dx=\frac{1}{9}{\int }_0^{\frac{2}{3}}\frac{1}{(\frac{2}{3}{)}^2+x^2}=\frac{1}{9}.\frac{1}{\frac{2}{3}}{\left[ta{n}^{-1}\left(\frac{x}{\frac{2}{3}}\right)\right]}_0^{\frac{2}{3}}[\because \int \frac{dx}{a+x^2}=\frac{1}{a}ta{n}^{-1}\frac{x}{a}]=\frac{1}{6}{\left[ta{n}^{-1}\right(\frac{3x}{2}\left)\right]}_0^{\frac{2}{3}}=\frac{1}{6}[ta{n}^{-1}(\frac{3}{2}.\frac{2}{3})-ta{n}^{-1}0]=\frac{1}{6}(ta{n}^{-1}1-0)=\frac{1}{6}.\frac{\pi }{4}=\frac{\pi }{24}.$
Hence, option (c) is the correct.