Question

${\int }_0^k\frac{1}{2+8x^2}dx=\frac{\pi }{16}$ , find the value of k

Answer 1

$\begin{array}{c}{\int }_0^k\frac{1}{2+8k^2}dx=\pi /16\\ \Rightarrow \frac{1}{8}{\int }_0^K\frac{1}{\frac{1}{4}+K^2}dx=\pi /16\\ \Rightarrow \frac{1}{8}{\int }_0^k\frac{dk}{{\left(\frac{1}{2}\right)}^2+k^2}=\pi /16\\ \Rightarrow \frac{1}{8}{\left[\frac{1}{2}{\tan }^{-1}2k\right]}^k+c\\ {\Rightarrow }^{1}16{{\left[{\tan }^{-1}2k\right]}_0}^k=\frac{1}{16}\times \pi =\frac{\pi }{16}Ans.\end{array}$