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