Question

\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]

Answer 1

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)

\[2I = \int_0^\frac{\pi}{2} \left[ \frac{\sin^2 x}{\sin x + \cos x} + \frac{\cos^2 x}{\cos x + \sin x} \right] d x\]

\[ = \int_0^\frac{\pi}{2} \left[ \frac{1}{\sin x + \cos x} \right] dx\]

\[ = \int_0^\frac{\pi}{2} \left[ \frac{1 + \tan^2 \frac{x}{2}}{2\tan\frac{x}{2} + 1 - \tan^2 \frac{x}{2}} \right] dx\]

\[ = \int_0^\frac{\pi}{2} \frac{\sec^2 \frac{x}{2}}{2\tan\frac{x}{2} + 1 - \tan^2 \frac{x}{2}} dx\]

\[\text{Putting }\tan\frac{x}{2} = t\]

\[ \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2}dx = dt\]

\[ \Rightarrow \sec^2 \frac{x}{2}dx = 2 dt\]

\[\text{When }x \to 0; t \to 0\]

\[\text{and }x \to \frac{\pi}{2}; t \to 1\]

\[ \therefore 2I = \int_0^1 \frac{2dt}{2t + 1 - t^2} dx\]

\[ = 2 \int_0^1 \frac{dt}{\left( \sqrt{2} \right)^2 - \left( t - 1 \right)^2}\]

\[ = \frac{2}{2\sqrt{2}} \left[ \log\left| \frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} \right| \right]_0^1 \]

\[ = \frac{1}{\sqrt{2}}\left[ \log\left( \frac{\sqrt{2}}{\sqrt{2}} \right) - log\left| \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right| \right] \]

\[ = \frac{1}{\sqrt{2}}\left[ 0 - \log\left| \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right| \right]\]

\[ = - \frac{1}{\sqrt{2}}\log\left| \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right|\]

\[ = \frac{1}{\sqrt{2}}\log\left| \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right|\]

\[ = \frac{1}{\sqrt{2}}\log\left[ \frac{\left( \sqrt{2} + 1 \right)\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)} \right]\]

\[2I = \frac{1}{\sqrt{2}}\log\left[ \frac{\left( \sqrt{2} + 1 \right)^2}{2 - 1} \right]\]

\[2I = \frac{2}{\sqrt{2}}\log\left( \sqrt{2} + 1 \right)\]

\[\text{Hence }I = \frac{1}{\sqrt{2}}\log\left( \sqrt{2} + 1 \right)\]