If $\int_0^a {\frac{1}{{1 + 4{x^2}}}} dx = \frac{\pi }{8},$ then $a =$..........

Let $I = \int_0^a {\frac{1}{{1 + 4{x^2}}}} dx = \frac{\pi }{8}$

Now, $\int_0^a {\frac{1}{{4\left( {\frac{1}{4} + {x^2}} \right)}}} dx = \frac{2}{4}\left[ {{{\tan }^{ - 1}}2x} \right]_0^a$

$= \frac{1}{2}{\tan ^{ - 1}}2a - 0 = \pi /8$

$\frac{1}{2}{\tan ^{ - 1}}2a = \frac{\pi }{8}$

$\Rightarrow$ ${\tan ^{ - 1}}2a = \pi /4$

$\Rightarrow$ $2a = 1$

therefore,$a = \frac{1}{2}$