Evaluate the definite integral int_{0}^{frac{ | vedclass

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(B) We need to evaluate the integral $I = \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$.First,rewrite the integrand as $\int \frac{d x}{(2)^{2}+(3 x)^

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$\frac{\pi}{24}$

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(B) We need to evaluate the integral $I = \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$.First,rewrite the integrand as $\int \frac{d x}{(2)^{2}+(3 x)^{2}}$.Let $3x = t$,then $3 dx = dt$,which implies $dx = \frac{1}{3} dt$.The indefinite integral becomes $\int \frac{1}{3} \frac{dt}{(2)^{2}+t^{2}} = \frac{1}{3} \cdot \frac{1}{2} 	an^{-1}(\frac{t}{2}) = \frac{1}{6} 	an^{-1}(\frac{3x}{2})$.Applying the limits from $0$ to $\frac{2}{3}$:$I = \left[ \frac{1}{6} 	an^{-1}(\frac{3x}{2}) ight]_{0}^{\frac{2}{3}}$$I = \frac{1}{6} 	an^{-1}(\frac{3}{2} \cdot \frac{2}{3}) - \frac{1}{6} 	an^{-1}(0)$$I = \frac{1}{6} 	an^{-1}(1) - 0$Since $	an^{-1}(1) = \frac{\pi}{4}$,we have $I = \frac{1}{6} \cdot \frac{\pi}{4} = \frac{\pi}{24}$.Thus,the correct option is $B$.

Evaluate the definite integral $\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$.

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