Question

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

Answer 1

Let $I={\int }_0^a\frac{1}{4+x^2}dx$

Put $x=2u\Rightarrow dx=2u$

$={\int }_0^a\frac{1}{4+(2u{)}^2}2du$

$={\int }_0^a\frac{1}{2u^2+2}du$

$=\frac{1}{2}{\int }_0^a\frac{1}{u^2+1}du$

$=\frac{1}{2}[ta{n}^{-1}u{]}_0^a+C$

Substitute back $u=\frac{x}{2}$

$=\frac{1}{2}[ta{n}^{-1}\left(\frac{x}{2}\right){]}_0^a+C$

$\frac{1}{2}\left[ta{n}^{-1}\left(\frac{a}{2}\right)\right]=\frac{\pi }{8}$

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

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

$\Rightarrow \frac{a}{2}=1$

Therefore, $a=2.$