Question

\[\int\limits_0^k \frac{1}{2 + 8 x^2} dx = \frac{\pi}{16},\] find the value of k.

Answer 1

\[\text{We have}, \]
\[ \int_0^k \frac{1}{2 + 8 x^2} d x = \frac{\pi}{16}\]
\[ \Rightarrow \frac{1}{8} \int_0^k \frac{1}{\frac{1}{4} + x^2} d x = \frac{\pi}{16}\]
\[ \Rightarrow \frac{1}{4} \left[ \tan^{- 1} 2x \right]_0^k = \frac{\pi}{16}\]
\[ \Rightarrow \tan^{- 1} 2k = \frac{\pi}{4}\]
\[ \Rightarrow 2k = \tan\frac{\pi}{4}\]
\[ \Rightarrow 2k = 1\]
\[ \Rightarrow k = \frac{1}{2}\]