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प्रश्न
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
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उत्तर
We have,
\[I = \int_0^\frac{\pi}{2} \frac{\cos^2 x}{\sin x + \cos x} d x ...............(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^2 \left( \frac{\pi}{2} - x \right)}{\sin\left( \frac{\pi}{2} - x \right) + \cos\left( \frac{\pi}{2} - x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\sin^2 x}{\cos x + \sin x} dx .................(2)\]
Adding (1) and (2)
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{\cos^2 x}{\sin x + \cos x} + \frac{\sin^2 x}{\cos x + \sin x} \right]dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{1}{\sin x + \cos x} \right]dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{1}{\frac{2\tan\frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} \right]dx\]
\[= - \int_0^\frac{\pi}{2} \frac{1 + \tan^2 \frac{x}{2}}{\tan^2 \frac{x}{2} - 2\tan\frac{x}{2} - 1} dx\]
\[ = - \int_0^\frac{\pi}{2} \frac{\sec^2 \frac{x}{2}}{\tan^2 \frac{x}{2} - 2\tan\frac{x}{2} - 1} dx\]
\[\text{Putting }\tan\frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2}dx = dt\]
\[ \Rightarrow \sec^2 \frac{x}{2}dx = 2dt\]
\[\text{When }x \to 0; t \to 0\]
\[\text{and }x \to \frac{\pi}{2}; t \to 1\]
\[\therefore 2I = - 2 \int_0^1 \frac{dt}{t^2 - 2t - 1}\]
\[ \Rightarrow I = - \int_0^1 \frac{dt}{\left( t - 1 \right)^2 - \left( \sqrt{2} \right)^2}\]
\[ = - \frac{1}{2\sqrt{2}} \left[ \log\left| \frac{t - 1 - \sqrt{2}}{t - 1 + \sqrt{2}} \right| \right]_0^1 \]
\[ = - \frac{1}{2\sqrt{2}}\left[ \log\left| - 1 \right| - \log\left| \frac{- 1 - \sqrt{2}}{- 1 + \sqrt{2}} \right| \right]\]
\[ = - \frac{1}{2\sqrt{2}}\left[ \log 1 - \log\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right]\]
\[= - \frac{1}{2\sqrt{2}}\left[ - \log\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right]\]
\[ = \frac{1}{2\sqrt{2}}\log\left[ \frac{\left( \sqrt{2} + 1 \right)\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)} \right]\]
\[ = \frac{1}{2\sqrt{2}}\log\left[ \frac{\left( \sqrt{2} + 1 \right)^2}{\left( 2 - 1 \right)} \right]\]
\[ = \frac{1}{2\sqrt{2}}\log \left( \sqrt{2} + 1 \right)^2 \]
\[ = \frac{1}{2\sqrt{2}} \times 2 \log\left( \sqrt{2} + 1 \right)\]
\[ = \frac{1}{\sqrt{2}}\log\left( \sqrt{2} + 1 \right)\]
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