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