If $\int_0^{\mathrm{k}} \frac{d x}{2+8 x^2}=\frac{\pi}{16}$, then value of $\mathrm{k}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \int_0^k \frac{d x}{2+8 x^2}=\frac{\pi}{16} \Rightarrow \frac{1}{2} \int_0^k \frac{d x}{1+(2 x)^2}=\frac{\pi}{16} \ & \Rightarrow \frac{1}{2}\left[\frac{	an ^{-1}(2 x)}{2}ight]_0^k=\frac{\pi}{16} \ & \Rightarrow 	an ^{-1}(2 k)=\frac{\pi}{16} \ & \Rightarrow 2 k=1 \ & \Rightarrow k=\frac{1}{2}\end{aligned}$

If $\int_0^{\mathrm{k}} \frac{d x}{2+8 x^2}=\frac{\pi}{16}$, then value of $\mathrm{k}$ is

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