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If $\int_{o}^{a} \frac{d x}{1+4 x^{2}}=\frac{\pi}{8}$, then $a=$
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Verified Answer
The correct answer is:
$\frac{1}{2}$
We have $\int_{0}^{a} \frac{d x}{1+4 x^{2}}=\frac{\pi}{8}$
$\therefore \frac{\pi}{8}=\frac{1}{4} \int_{0}^{a} \frac{d x^{2}+\left(\frac{1}{2}\right)^{2}}{2}=\frac{1}{4} \frac{1}{\left(\frac{1}{2}\right)}\left[\tan ^{-1} \frac{x}{\left(\frac{1}{2}\right)}\right]_{0}^{a}$
$\quad=\frac{1}{2}\left[\tan ^{-1} 2 a-\tan ^{-1} 0\right]$
$\therefore \frac{\pi}{8}=\frac{1}{2} \tan ^{-1} 2 a \Rightarrow \frac{\pi}{4}=\tan ^{-1} 2 a \Rightarrow 2 a=\tan \frac{\pi}{4}=1 \Rightarrow a=\frac{1}{2}$
$\therefore \frac{\pi}{8}=\frac{1}{4} \int_{0}^{a} \frac{d x^{2}+\left(\frac{1}{2}\right)^{2}}{2}=\frac{1}{4} \frac{1}{\left(\frac{1}{2}\right)}\left[\tan ^{-1} \frac{x}{\left(\frac{1}{2}\right)}\right]_{0}^{a}$
$\quad=\frac{1}{2}\left[\tan ^{-1} 2 a-\tan ^{-1} 0\right]$
$\therefore \frac{\pi}{8}=\frac{1}{2} \tan ^{-1} 2 a \Rightarrow \frac{\pi}{4}=\tan ^{-1} 2 a \Rightarrow 2 a=\tan \frac{\pi}{4}=1 \Rightarrow a=\frac{1}{2}$
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