If int_{0}^{a} frac{dx}{1+4x^{2}} = frac{pi}{ | vedclass

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(A) Given the integral $\int_{0}^{a} \frac{dx}{1+(2x)^{2}} = \frac{\pi}{8}$.Let $2x = t$,then $2dx = dt$ or $dx = \frac{dt}{2}$.When $x = 0, t = 0$ an

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$\frac{1}{2}$

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(A) Given the integral $\int_{0}^{a} \frac{dx}{1+(2x)^{2}} = \frac{\pi}{8}$.Let $2x = t$,then $2dx = dt$ or $dx = \frac{dt}{2}$.When $x = 0, t = 0$ and when $x = a, t = 2a$.The integral becomes $\int_{0}^{2a} \frac{1}{1+t^{2}} \cdot \frac{dt}{2} = \frac{\pi}{8}$.$\frac{1}{2} [	an^{-1}(t)]_{0}^{2a} = \frac{\pi}{8}$.$	an^{-1}(2a) - 	an^{-1}(0) = \frac{\pi}{4}$.$	an^{-1}(2a) = \frac{\pi}{4}$.$2a = 	an(\frac{\pi}{4}) = 1$.$a = \frac{1}{2}$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

If $\int_{0}^{a} \frac{dx}{1+4x^{2}} = \frac{\pi}{8}$,then $a =$

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