Question

If $\underset{0}{\overset{k}{\int }}\frac{1}{2+8x^2}dx=\frac{\mathrm{\pi }}{16},$ find the value of k.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {\int }_0^k\frac{1}{2+8x^2}\mathit{}\mathit{d}x=\frac{\mathrm{\pi }}{16} \\ \Rightarrow \frac{1}{8}{\int }_0^k\frac{1}{{\displaystyle \frac{1}{4}}+x^2}\mathit{d}x=\frac{\mathrm{\pi }}{16} \\ \Rightarrow \frac{1}{4}{\left[{\tan }^{-1}2x\right]}_0^k=\frac{\mathrm{\pi }}{16} \\ \Rightarrow {\tan }^{-1}2k=\frac{\mathrm{\pi }}{4} \\ \Rightarrow 2k=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow 2k=1 \\ \Rightarrow k=\frac{1}{2} \end{gathered}$$