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प्रश्न
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
उत्तर
\[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\]
\[ \Rightarrow \frac{1}{2} \tan^{- 1} \left.\frac{x}{2}\right|_0^a = \frac{\pi}{8} ................\left[ \int\frac{1}{a^2 + x^2}dx = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right]\]
\[ \Rightarrow \frac{1}{2}\left( \tan^{- 1} \frac{a}{2} - \tan^{- 1} 0 \right) = \frac{\pi}{8}\]
\[ \Rightarrow \tan^{- 1} \frac{a}{2} - 0 = \frac{\pi}{4}\]
\[\Rightarrow \tan^{- 1} \frac{a}{2} = \frac{\pi}{4}\]
\[ \Rightarrow \frac{a}{2} = \tan\frac{\pi}{4} = 1\]
\[ \Rightarrow a = 2\]
Thus, the value of a is 2.
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