Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}\frac{1}{5+3\cos x}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I={\int }_0^{\mathrm{\pi }}\frac{1}{5+3\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}\right)}\mathit{d}x \\ \Rightarrow I={\int }_0^{\mathrm{\pi }}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{5+5{\tan }^2{\displaystyle \frac{x}{2}}-3{\tan }^2{\displaystyle \frac{x}{2}}}dx \\ \Rightarrow I={\int }_0^{\mathrm{\pi }}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{5+2{\tan }^2{\displaystyle \frac{x}{2}}}dx \\ \mathrm{Let}\tan \frac{x}{2}=t.\mathrm{Then},\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt \\ \mathrm{When}x=0,t=0\mathrm{and}x=\mathrm{\pi },t=\infty \\ \therefore I={\int }_0^{\infty }\frac{dt}{5+2t^2} \\ \Rightarrow I=\frac{1}{2}{\int }_0^{\infty }\frac{dt}{{\displaystyle \frac{5}{2}}+t^2} \\ \Rightarrow I=\frac{1}{2}{\left[{\tan }^{-1}\frac{\sqrt{5}t}{\sqrt{2}}\right]}_0^{\infty } \\ \Rightarrow I=\frac{1}{2}\left(\frac{\mathrm{\pi }}{2}-0\right) \\ \Rightarrow I=\frac{\mathrm{\pi }}{4} \end{gathered}$$