Question

If ${\int }_0^a\frac{1}{4+x^2}dx=\frac{\pi }{8}$ find the value of a

Answer 1

${\int }_0^a\frac{1}{4+x^2}dx={\int }_0^a\frac{1}{2^2+x^2}dx\Rightarrow \int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$

(Formula)

${\left[\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right]}_0^a\Rightarrow \frac{\pi }{8}=\frac{1}{2}{\tan }^{-1}\frac{a}{2}+\frac{1}{2}{\tan }^{-1}0$

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

$\Rightarrow \tan \left(\frac{\pi }{4}\right)=\frac{a}{2}$

$a=2$

Hence the correct value of $a$ is $2$