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

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Aicharya · Accepted Answer

We have $\int_{0}^{a} \frac{d x}{1+4 x^{2}}=\frac{\pi}{8}$ $	herefore \frac{\pi}{8}=\frac{1}{4} \int_{0}^{a} \frac{d x^{2}+\left(\frac{1}{2}ight)^{2}}{2}=\frac{1}{4} \frac{1}{\left(\frac{1}{2}ight)}\left[	an ^{-1} \frac{x}{\left(\frac{1}{2}ight)}ight]_{0}^{a}$ $\quad=\frac{1}{2}\left[	an ^{-1} 2 a-	an ^{-1} 0ight]$ $	herefore \frac{\pi}{8}=\frac{1}{2} 	an ^{-1} 2 a \Rightarrow \frac{\pi}{4}=	an ^{-1} 2 a \Rightarrow 2 a=	an \frac{\pi}{4}=1 \Rightarrow a=\frac{1}{2}$

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

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