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प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
उत्तर
\[Let\ I = \int_\frac{\pi}{4}^\frac{\pi}{2} \cot x\ d\ x\ . Then, \]
\[I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \cot x\frac{- (cosec x + \cot x)}{cosec x + \cot x} dx\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - \cot^2 x}{cosec x + \cot x} dx\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - {cosec}^2 x + 1}{cosec x + \cot x} dx \left[ \because {cosec}^2 x = 1 + \cot^2 x \right]\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - {cosec}^2 x}{cosec x + \cot x} dx - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{1}{cosec x + \cot x}dx\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - {cosec}^2 x}{cosec x + \cot x} dx - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{\sin x}{1 + \cos x}dx\]
\[ \Rightarrow I = - \left[ \log \left( cosec x + \cot x \right) \right]_\frac{\pi}{4}^\frac{\pi}{2} + \left[ \log \left( 1 + \cos x \right) \right]_\frac{\pi}{4}^\frac{\pi}{2} \]
\[ \Rightarrow I = - \log \left( 1 + \infty \right) + \log \left( \sqrt{2} + 1 \right) + \log \left( 1 + 0 \right) - \log \left( 1 + \frac{1}{\sqrt{2}} \right)\]
\[ \Rightarrow I = \log \left( \sqrt{2} + 1 \right) - \log \left( \frac{\sqrt{2} + 1}{\sqrt{2}} \right)\]
\[ \Rightarrow I = \log \left( \frac{\sqrt{2}\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} + 1 \right)} \right)\]
\[ \Rightarrow I = \log\sqrt{2}\]
\[ \Rightarrow I = \frac{1}{2}\log 2\]
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