If int_{0}^{k} frac{dx}{2 + 18x^2} = frac{pi} | vedclass

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(C) Given the integral $\int_{0}^{k} \frac{dx}{2 + 18x^2} = \frac{\pi}{24}$.Factor out $2$ from the denominator: $\frac{1}{2} \int_{0}^{k} \frac{dx}{1

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$\frac{1}{3}$

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(C) Given the integral $\int_{0}^{k} \frac{dx}{2 + 18x^2} = \frac{\pi}{24}$.Factor out $2$ from the denominator: $\frac{1}{2} \int_{0}^{k} \frac{dx}{1 + 9x^2} = \frac{\pi}{24}$.Multiply both sides by $2$: $\int_{0}^{k} \frac{dx}{1 + (3x)^2} = \frac{\pi}{12}$.Using the formula $\int \frac{dx}{1 + a^2x^2} = \frac{1}{a} 	an^{-1}(ax) + C$,we get:$\left[ \frac{1}{3} 	an^{-1}(3x) ight]_{0}^{k} = \frac{\pi}{12}$.$\frac{1}{3} 	an^{-1}(3k) - \frac{1}{3} 	an^{-1}(0) = \frac{\pi}{12}$.Since $	an^{-1}(0) = 0$,we have $\frac{1}{3} 	an^{-1}(3k) = \frac{\pi}{12}$.$	an^{-1}(3k) = \frac{\pi}{4}$.$3k = 	an(\frac{\pi}{4}) = 1$.Therefore,$k = \frac{1}{3}$.

If $\int_{0}^{k} \frac{dx}{2 + 18x^2} = \frac{\pi}{24}$,then the value of $k$ is

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