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If $\int\limits_{0}^{a}\frac{1}{1+4x^2}dx=\frac{π}{8}$, then a equal |
$\frac{π}{2}$ $\frac{π}{4}$ 1 $\frac{1}{2}$ |
$\frac{1}{2}$ |
We have, $\int\limits_{0}^{a}\frac{1}{1+4x^2}dx=\frac{π}{8}$ $⇒\int\limits_{0}^{a}\frac{1}{1^2+(2x)^2}dx=\frac{π}{8}⇒\frac{1}{2}[\tan^{-1}2x]_{0}^{a}=\frac{π}{8}$ $⇒\tan^{-1}2a=\frac{π}{4}⇒2a=\tan\frac{π}{4}⇒a=\frac{1}{2}$ |
