Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{5\cos x+3\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{5\cos x+3\sin x}\mathit{d}x\mathit{.}\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{5\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}\right)+{\displaystyle 3\left(\frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}\right)}}\mathit{d}x\left[\because \sin A=\left(\frac{2\tan {\displaystyle \frac{A}{2}}}{1+{\tan }^2{\displaystyle \frac{A}{2}}}\right),\cos A=\left(\frac{1-{\tan }^2{\displaystyle \frac{A}{2}}}{1+{\tan }^2{\displaystyle \frac{A}{2}}}\right)\right] \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{5-5{\tan }^2{\displaystyle \frac{x}{2}}+6\tan {\displaystyle \frac{x}{2}}}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{5-5{\tan }^2{\displaystyle \frac{x}{2}}+6\tan {\displaystyle \frac{x}{2}}}dx \\ \mathrm{Let}\tan \frac{x}{2}=t.\mathrm{Then},\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt \\ \mathrm{Also},x=0,t=0\mathrm{and}x=\frac{\mathrm{\pi }}{2},t=1 \\ \therefore I={\int }_0^1\frac{2dt}{5-5t^2+6t} \\ \Rightarrow I=\frac{1}{5}{\int }_0^1\frac{2dt}{1-t^2+{\displaystyle \frac{6}{5}}t+{\displaystyle \frac{36}{100}}-{\displaystyle \frac{36}{100}}} \\ =\frac{2}{5}{\int }_0^1\frac{dt}{-{\left(t-{\displaystyle \frac{6}{10}}\right)}^2+{\displaystyle \frac{136}{100}}} \\ =\frac{2}{5}\times \frac{10}{\sqrt{136}}{\left[-\log \left(\frac{t-{\displaystyle \frac{6}{10}}-{\displaystyle \frac{\sqrt{136}}{10}}}{t-{\displaystyle \frac{6}{10}}+{\displaystyle \frac{\sqrt{136}}{10}}}\right)\right]}_0^1 \\ =\frac{1}{\sqrt{34}}\left[-\log \left(\frac{4-2\sqrt{34}}{4+2\sqrt{34}}\right)+\log \left(\frac{-6-2\sqrt{34}}{-6+2\sqrt{34}}\right)\right] \\ =\frac{1}{\sqrt{34}}\log \left(\frac{6+2\sqrt{34}}{6-2\sqrt{34}}\times \frac{4+2\sqrt{34}}{4-2\sqrt{34}}\right) \\ =\frac{1}{\sqrt{34}}\log \left(\frac{160+20\sqrt{34}}{160-20\sqrt{34}}\right) \\ =\frac{1}{\sqrt{34}}\log \left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right) \end{gathered}$$