Question

Evaluate : $\int\limits^k_0 {\frac{1}{2+8x^{2}}} \, dx=\frac{\pi}{16}$; Find the value of 'k'.

Answer 1

➡️Hope it will help you‼️‼️

Answer 2

Answer:

$k = \dfrac 12 tan\dfrac {\pi}{16}$

Step-by-step explanation:

$\int\limits^k_0 \dfrac 1{2 + 8x^2} \, dx = \dfrac {\pi}{16}\\\\ \int\limits^k_0 \dfrac 1{(\sqrt{2} )^2 + (2\sqrt{2} x)^2} \, dx = \dfrac {\pi}{16}$

By the identity of $tan^{-1} \dfrac xa$

$\int \dfrac 1{a^2 + x^2} = tan^{-1} \dfrac xa +c$

$\left [tan^{-1} \dfrac {2\sqrt{2}x}{\sqrt{2}}\right ]_{0}^{k} = \dfrac {\pi}{16}\\\left [tan^{-1} 2x \right ]_{0}^{k} = \dfrac {\pi}{16}\\tan^{-1} 2k - tan^{-1}0 = \dfrac {\pi}{16}\\\\2k = tan\dfrac {\pi}{16}\\ k = \dfrac 12tan\dfrac {\pi}{16}$