If int_{0}^{a} frac{dx}{4 + x^2} = frac{pi}{8 | vedclass

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(B) Given the integral: $\int_{0}^{a} \frac{dx}{4 + x^2} = \frac{\pi}{8}$ We know that $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} 	an^{-1} \left( \frac

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$2$

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(B) Given the integral: $\int_{0}^{a} \frac{dx}{4 + x^2} = \frac{\pi}{8}$ We know that $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} 	an^{-1} \left( \frac{x}{k} ight) + C$. Here,$k^2 = 4$,so $k = 2$. Applying the limits from $0$ to $a$: $\left[ \frac{1}{2} 	an^{-1} \left( \frac{x}{2} ight) ight]_{0}^{a} = \frac{\pi}{8}$ $\frac{1}{2} 	an^{-1} \left( \frac{a}{2} ight) - \frac{1}{2} 	an^{-1} (0) = \frac{\pi}{8}$ Since $	an^{-1}(0) = 0$,we have: $\frac{1}{2} 	an^{-1} \left( \frac{a}{2} ight) = \frac{\pi}{8}$ $	an^{-1} \left( \frac{a}{2} ight) = \frac{\pi}{4}$ Taking the tangent of both sides: $\frac{a}{2} = 	an \left( \frac{\pi}{4} ight)$ $\frac{a}{2} = 1$ $a = 2$

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

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