Question

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

Answer 1

${\int }_0^a\frac{1}{4+x^2}dx=\frac{\pi }{8}\Rightarrow \frac{1}{2}{\left(ta{n}^{-1}\frac{x}{2}\right)}_0^a=\frac{\pi }{8}$

$\Rightarrow ta{n}^{-1}\frac{a}{2}=\frac{\pi }{4}\Rightarrow \frac{a}{2}=tan\frac{\pi }{4}\Rightarrow a=2$