Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^{2}x}{\sin x+\cos x}dx.....\left(1\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{2}x}{\cos x+\sin x}dx.....\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{{\cos }^{2}x}{\sin x+\cos x}+\frac{{\sin }^{2}x}{\cos x+\sin x}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{\sin x+\cos x}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{{\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}+\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}\right]dx \end{gathered}$$

$$\begin{gathered} =-{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{{\tan }^2{\displaystyle \frac{x}{2}}-2\tan {\displaystyle \frac{x}{2}}-1}dx \\ =-{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{{\tan }^2{\displaystyle \frac{x}{2}}-2\tan {\displaystyle \frac{x}{2}}-1}dx \\ \mathrm{Putting}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt \\ \Rightarrow {\sec }^2\frac{x}{2}dx=2dt \\ \mathrm{When}x\to 0;t\to 0 \\ \mathrm{and}x\to \frac{\mathrm{\pi }}{2};t\to 1 \end{gathered}$$

$$\begin{gathered} \therefore 2I=-2{\int }_0^1\frac{dt}{t^2-2t-1} \\ \Rightarrow I=-{\int }_0^1\frac{dt}{{\left(t-1\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =-\frac{1}{2\sqrt{2}}{\left[\log \left|\frac{t-1-\sqrt{2}}{t-1+\sqrt{2}}\right|\right]}_0^1 \\ =-\frac{1}{2\sqrt{2}}\left[\log \left|-1\right|-\log \left|\frac{-1-\sqrt{2}}{-1+\sqrt{2}}\right|\right] \\ =-\frac{1}{2\sqrt{2}}\left[\log 1-\log \frac{\sqrt{2}+1}{\sqrt{2}-1}\right] \end{gathered}$$

$$\begin{gathered} =-\frac{1}{2\sqrt{2}}\left[-\log \frac{\sqrt{2}+1}{\sqrt{2}-1}\right] \\ =\frac{1}{2\sqrt{2}}\log \left[\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\right] \\ =\frac{1}{2\sqrt{2}}\log \left[\frac{{\left(\sqrt{2}+1\right)}^2}{\left(2-1\right)}\right] \\ =\frac{1}{2\sqrt{2}}\log {\left(\sqrt{2}+1\right)}^2 \\ =\frac{1}{2\sqrt{2}}\times 2\log \left(\sqrt{2}+1\right) \\ =\frac{1}{\sqrt{2}}\log \left(\sqrt{2}+1\right) \end{gathered}$$