Question

If $\underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}dx=\frac{\pi }{8},$ then the value of $4a$ is

Answer 1

$\underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}dx=\frac{\pi }{8}\Rightarrow \underset{0}{\overset{a}{\int }}\frac{1}{4}\cdot \frac{1}{{\left(\frac{1}{2}\right)}^2+(x{)}^2}dx=\frac{\pi }{8}\Rightarrow {\left[\frac{1}{2}{\tan }^{-1}\right(2x\left)\right]}_0^a=\frac{\pi }{8}\{\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\}\Rightarrow {\tan }^{-1}\left(2a\right)=\frac{\pi }{4}\Rightarrow a=\frac{1}{2}\therefore 4a=2$