Question

If $\underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}dx=\frac{\mathrm{\pi }}{8},$ then a = _______________.

Answer 1

$$\begin{gathered} \mathrm{Given}:\underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}dx=\frac{\mathrm{\pi }}{8}...\left(1\right) \\ \mathrm{Let}I=\underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}dx \\ \Rightarrow I=\frac{1}{4}\underset{0}{\overset{a}{\int }}\frac{1}{{\displaystyle \frac{1}{4}}+x^2}dx \\ \Rightarrow I=\frac{1}{4}\underset{0}{\overset{a}{\int }}\frac{1}{{\displaystyle {\left(\frac{1}{2}\right)}^2+x^2}}dx \\ \Rightarrow I=\frac{1}{4}{\left\{\frac{1}{{\displaystyle \frac{1}{2}}}{\tan }^{-1}\frac{x}{{\displaystyle \frac{1}{2}}}\right\}}_0^a\left(\because \int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c\right) \\ \Rightarrow I=\frac{1}{4}{\left\{2{\tan }^{-1}2x\right\}}_0^a \\ \Rightarrow I=\frac{1}{4}{\left\{2{\tan }^{-1}2a-2{\tan }^{-1}0\right\}}_0^a \\ \Rightarrow I=\frac{1}{2}{\tan }^{-1}2a \\ \mathrm{Since},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}dx=\frac{\mathrm{\pi }}{8} \\ \mathrm{Therefore},\frac{1}{2}{\tan }^{-1}2a=\frac{\mathrm{\pi }}{8} \\ \Rightarrow {\tan }^{-1}2a=\frac{\mathrm{\pi }}{4} \\ \Rightarrow 2a=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow 2a=1 \\ \Rightarrow a=\frac{1}{2} \end{gathered}$$

​
Hence, a = $\underline{\frac{1}{2}}$.