Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{dx}{4\cos x+2\sin x}$

Answer 1

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{4\cos x+2\sin x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{4-4{\tan }^2\frac{x}{2}+4\tan {\displaystyle \frac{x}{2}}}dx \\ \mathrm{Let}\tan \frac{x}{2}=t,\mathrm{then}\frac{1}{2}se{c}^2\frac{x}{2}dx=dt \\ \mathrm{When}x=0,t=0,x=\frac{\mathrm{\pi }}{2},t=1 \\ =\frac{-1}{4}{\int }_0^1\frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2-{\displaystyle \frac{5}{4}}} \\ =\frac{-1}{4}\times \frac{-4}{\sqrt{5}}{\left[\log \frac{2t-1-\sqrt{5}}{2t-1+\sqrt{5}}\right]}_0^1 \\ =\frac{1}{\sqrt{5}}\log \frac{\sqrt{5}+1}{\sqrt{5}-1} \end{gathered}$$