Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{{\sin }^{2}x}{\sin x+\cos x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{2}x}{\sin x+\cos x}\mathit{d}x\mathit{}.....\left(1\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)}{\sin \left(\frac{\mathrm{\pi }}{2}-x\right)+\cos \left(\frac{\mathrm{\pi }}{2}-x\right)}\mathit{d}x \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^{2}x}{\cos x+\sin x}dx.....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{{\sin }^{2}x}{\sin x+\cos x}+\frac{{\cos }^{2}x}{\cos x+\sin x}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{\sin x+\cos x}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{2\tan {\displaystyle \frac{x}{2}}+1-{\tan }^2{\displaystyle \frac{x}{2}}}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{2\tan {\displaystyle \frac{x}{2}}+1-{\tan }^2{\displaystyle \frac{{\displaystyle x}}{2}}}dx \\ \mathrm{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 \\ \mathrm{When}x\to 0;t\to 0 \\ \mathrm{and}x\to \frac{\mathrm{\pi }}{2};t\to 1 \\ \therefore 2I={\int }_0^1\frac{{\displaystyle 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) \\ \mathrm{Hence}I=\frac{1}{\sqrt{2}}\log \left(\sqrt{2}+1\right) \end{gathered}$$