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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
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Solution
\[\int_0^\alpha \frac{1}{1 + 4 x^2} d x = \frac{\pi}{8}\]
\[ \Rightarrow \int_0^\alpha \frac{1}{1 + \left( 2x \right)^2} d x = \frac{\pi}{8}\]
\[ \Rightarrow \frac{1}{2} \left[ \tan^{- 1} 2x \right]_0^\alpha = \frac{\pi}{8}\]
\[ \Rightarrow \frac{1}{2} \tan^{- 1} 2\alpha = \frac{\pi}{8}\]
\[ \Rightarrow 2\alpha = \tan\frac{\pi}{4}\]
\[ \Rightarrow 2\alpha = 1\]
\[ \therefore \alpha = \frac{1}{2}\]
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