$\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$ equals
A. $\frac{\pi}{6}$
B. $\frac{\pi}{12}$
C. $\frac{\pi}{24}$
D. $\frac{\pi}{4}$
$\int \frac{d x}{4+9 x^{2}}=\int \frac{d x}{(2)^{2}+(3 x)^{2}}$
Put $3 x=t \Rightarrow 3 d x=d t$
$\therefore \int \frac{d x}{(2)^{2}+(3 x)^{2}}=\frac{1}{3} \int \frac{d t}{(2)^{2}+t^{2}}$
$=\frac{1}{3}\left[\frac{1}{2} \tan ^{-1} \frac{t}{2}\right]$
$=\frac{1}{6} \tan ^{-1}\left(\frac{3 x}{2}\right)$
$=\mathrm{F}(x)$
By second fundamental theorem of calculus, we obtain
$\int_{0}^{2} \frac{d x}{4+9 x^{2}}=\mathrm{F}\left(\frac{2}{3}\right)-\mathrm{F}(0)$
$=\frac{1}{6} \tan ^{-1}\left(\frac{3}{2} \cdot \frac{2}{3}\right)-\frac{1}{6} \tan ^{-1} 0$
$=\frac{1}{6} \tan ^{-1} 1-0$
$=\frac{1}{6} \times \frac{\pi}{4}$
$=\frac{\pi}{24}$
Hence, the correct answer is C.
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