If int_0^k frac{dx}{2 + 8x^2} = frac{pi}{16}, | vedclass

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(B) Given the integral $\int_0^k \frac{dx}{2 + 8x^2} = \frac{\pi}{16}$.Factor out $2$ from the denominator: $\frac{1}{2} \int_0^k \frac{dx}{1 + 4x^2}

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$\frac{1}{2}$

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(B) Given the integral $\int_0^k \frac{dx}{2 + 8x^2} = \frac{\pi}{16}$.Factor out $2$ from the denominator: $\frac{1}{2} \int_0^k \frac{dx}{1 + 4x^2} = \frac{1}{2} \int_0^k \frac{dx}{1 + (2x)^2}$.Let $t = 2x$,then $dt = 2dx$ or $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=k, t=2k$.The integral becomes $\frac{1}{2} \int_0^{2k} \frac{dt/2}{1 + t^2} = \frac{1}{4} \int_0^{2k} \frac{dt}{1 + t^2}$.Evaluating the integral: $\frac{1}{4} [	an^{-1} t]_0^{2k} = \frac{1}{4} 	an^{-1}(2k)$.Equating to the given value: $\frac{1}{4} 	an^{-1}(2k) = \frac{\pi}{16}$.$	an^{-1}(2k) = \frac{\pi}{4}$.$2k = 	an(\frac{\pi}{4}) = 1$.Therefore,$k = \frac{1}{2}$.

If $\int_0^k \frac{dx}{2 + 8x^2} = \frac{\pi}{16}$,then $k = $

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