Question

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

Answer 1

${\int }_0^k\left(\frac{1}{2(1+4x^2)}\right)dx=\left(\frac{\pi }{16}\right)or\left(\frac{1}{8}\right){\int }_0^k\left(\frac{1}{\left(\frac{1}{4}\right)+x^2}\right)dx=\left(\frac{\pi }{16}\right)={\int }_0^k\left(\frac{1}{\left(\right(\frac{1}{2}){)}^2+x^2}\right)dx=\left(\frac{\pi }{2}\right)$

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

$=2(ta{n}^{-1}2k-ta{n}^{-1}0)=\left(\frac{\pi }{2}\right)=ta{n}^{-1}2k=\left(\frac{\pi }{4}\right)2k=tan\left(\frac{\pi }{4}\right)$

$2k=1\therefore k=\left(\frac{1}{2}\right)$