Question

If ${\int }_0^a\frac{1}{4+x^2}dx=\frac{\mathrm{\pi }}{8}$, find the value of a. [CBSE 2014]

Answer 1

$$\begin{gathered} {\int }_0^a\frac{1}{4+x^2}dx=\frac{\mathrm{\pi }}{8} \\ \Rightarrow {\overline{)\frac{1}{2}{\tan }^{-1}\frac{x}{2}}}_0^a=\frac{\mathrm{\pi }}{8}\left[\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\right] \\ \Rightarrow \frac{1}{2}\left({\tan }^{-1}\frac{a}{2}-{\tan }^{-1}0\right)=\frac{\mathrm{\pi }}{8} \\ \Rightarrow {\tan }^{-1}\frac{a}{2}-0=\frac{\mathrm{\pi }}{4} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\tan }^{-1}\frac{a}{2}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{a}{2}=\tan \frac{\mathrm{\pi }}{4}=1 \\ \Rightarrow a=2 \end{gathered}$$

Thus, the value of a is 2.