Question

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

(a) π/4
(b) π/8
(c) π/2
(d) 0

Answer 1

(a) π/4


$$\begin{gathered} {\int }_0^{\mathrm{\pi }}\frac{1}{5+3\cos x}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{1}{5+3{\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2\frac{x}{2}}}}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{1+{\tan }^2\frac{x}{2}}{5+5{\tan }^2\frac{x}{2}+3-3{\tan }^2\frac{x}{2}}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{se{c}^2{\displaystyle \frac{x}{2}}}{8+2{\tan }^2\frac{x}{2}}dx \\ \mathrm{Let}\tan \frac{x}{2}=t,\mathrm{then}se{c}^2\frac{x}{2}dx=2dt \\ \mathrm{When}x=0,t=0,x=\mathrm{\pi },\mathrm{t}=\infty \\ \mathrm{Therefore}\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ \frac{1}{2}{\int }_0^{\infty }\frac{\mathrm{dt}}{4+{\mathrm{t}}^2} \\ =\frac{1}{2}{\left[{\tan }^{-1}\frac{\mathrm{t}}{2}\right]}_0^{\infty } \\ =\frac{1}{2}\left(\frac{\mathrm{\pi }}{2}-0\right)=\frac{\mathrm{\pi }}{4} \end{gathered}$$