Question

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 

Options

• \[\frac{\pi}{2}\]
• \[\frac{1}{2}\]
• \[\frac{\pi}{4}\]

• 1

Answer 1

\[\frac{1}{2}\]

\[\int_0^\alpha \frac{1}{1 + 4 x^2} d x = \frac{\pi}{8}\]

\[ \Rightarrow \int_0^\alpha \frac{1}{1 + \left( 2x \right)^2} d x = \frac{\pi}{8}\]

\[ \Rightarrow \frac{1}{2} \left[ \tan^{- 1} 2x \right]_0^\alpha = \frac{\pi}{8}\]

\[ \Rightarrow \frac{1}{2} \tan^{- 1} 2\alpha = \frac{\pi}{8}\]

\[ \Rightarrow 2\alpha = \tan\frac{\pi}{4}\]

\[ \Rightarrow 2\alpha = 1\]

\[ \therefore \alpha = \frac{1}{2}\]