Question

${\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}$ equals

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{24}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (C)

$\frac{\pi }{24}$

$\int \frac{dx}{4+9x^2}=\int \frac{dx}{(2{)}^2+(3x{)}^2}$
Put $3x=t\Rightarrow 3dx=dt$
$\therefore \int \frac{dx}{(2{)}^2+(3x{)}^2}=\frac{1}{3}\int \frac{dt}{(2{)}^2+t^2}$
$=\frac{1}{3}\left[\frac{1}{2}{\tan }^{-1}\frac{t}{2}\right]$
$=\frac{1}{6}{\tan }^{-1}\left(\frac{3x}{2}\right)$
$=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
${\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}=F\left(\frac{2}{3}\right)-F\left(0\right)$
$=\frac{1}{6}{\tan }^{-1}(\frac{3}{2}\cdot \frac{2}{3})-\frac{1}{6}{\tan }^{-1}0$
$=\frac{1}{6}{\tan }^{-1}1-0$
$=\frac{1}{6}\times \frac{\pi }{4}=\frac{\pi }{24}$