int_0^{2/3} frac{dx}{4 + 9x^2} = | vedclass

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(B) To evaluate the integral $I = \int_0^{2/3} \frac{dx}{4 + 9x^2}$,we first rewrite the denominator: $I = \int_0^{2/3} \frac{dx}{2^2 + (3x)^2}$. Let

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$\frac{\pi}{24}$

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(B) To evaluate the integral $I = \int_0^{2/3} \frac{dx}{4 + 9x^2}$,we first rewrite the denominator: $I = \int_0^{2/3} \frac{dx}{2^2 + (3x)^2}$. Let $3x = u$,then $3dx = du$,which implies $dx = \frac{du}{3}$. When $x = 0$,$u = 0$. When $x = 2/3$,$u = 2$. Substituting these into the integral: $I = \int_0^2 \frac{1}{2^2 + u^2} \cdot \frac{du}{3} = \frac{1}{3} \int_0^2 \frac{du}{2^2 + u^2}$. Using the standard formula $\int \frac{du}{a^2 + u^2} = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + C$: $I = \frac{1}{3} \left[ \frac{1}{2} 	an^{-1}(\frac{u}{2}) ight]_0^2 = \frac{1}{6} [	an^{-1}(1) - 	an^{-1}(0)]$. Since $	an^{-1}(1) = \frac{\pi}{4}$ and $	an^{-1}(0) = 0$: $I = \frac{1}{6} 	imes \frac{\pi}{4} = \frac{\pi}{24}$.

$\int_0^{2/3} \frac{dx}{4 + 9x^2} = $

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