∫0k dx2+8x2=π16 then k=

G.I =12∫0K dx1+4x2=12∫0k dx1+(2x)2=12⋅12⋅tan−1⁡(2x]0k⇒14tan−1⁡(2k)=π16tan−1⁡(2k)=π42k=1k=12

∫0k dx2+8x2=π16 then k=

G.I =12∫0K dx1+4x2=12∫0k dx1+(2x)2=12⋅12⋅tan−1⁡(2x]0k⇒14tan−1⁡(2k)=π16tan−1⁡(2k)=π42k=1k=12

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$\int_{0}^{k} \frac{d x}{2 + 8 x^{2}} = \frac{π}{16} then k =$

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$\frac{π}{2}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} G.I = \frac{1}{2} \int_{0}^{K} \frac{d x}{1 + 4 x^{2}} = \frac{1}{2} \int_{0}^{k} \frac{d x}{1 + (2 x)^{2}} \\ = \frac{1}{2} \cdot \frac{1}{2} \cdot [\tan^{- 1} (2 x]_{0}^{k} \\ \Rightarrow \frac{1}{4} [\tan^{- 1} (2 k)] = \frac{π}{16} \end{array}$