Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{2\cos x+4\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{2\cos x+4\sin x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{2-2{\tan }^2{\displaystyle \frac{x}{2}}+8\tan \frac{x}{2}}dx \\ \mathrm{Putting}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt \\ \mathrm{When}x\to 0;t\to 0 \\ \mathrm{and}x\to \frac{\mathrm{\pi }}{2};t\to 1 \\ \therefore I=2{\int }_0^1\frac{dt}{2-2t^2+8t} \\ =-\frac{2}{2}{\int }_0^1\frac{dt}{t^2-4t-1} \\ =-{\int }_0^1\frac{dt}{{\left(t-2\right)}^2-5} \\ ={\int }_0^1\frac{dt}{{\left(\sqrt{5}\right)}^2-{\left(t-2\right)}^2} \\ =\frac{1}{2\sqrt{5}}{\left[\log \left|\frac{\sqrt{5}+t-2}{\sqrt{5}-t+2}\right|\right]}_0^1 \\ =\frac{1}{2\sqrt{5}}\left[\log \frac{\sqrt{5}-1}{\sqrt{5}+1}-\log \frac{\sqrt{5}-2}{\sqrt{5}+2}\right] \\ =\frac{1}{2\sqrt{5}}\log \left[\frac{\sqrt{5}-1}{\sqrt{5}+1}\times \frac{\sqrt{5}+2}{\sqrt{5}-2}\right] \\ =\frac{1}{2\sqrt{5}}\log \left[\frac{5+2\sqrt{5}-\sqrt{5}-2}{5-2\sqrt{5}+\sqrt{5}-2}\right] \\ =\frac{1}{2\sqrt{5}}\log \left[\frac{\sqrt{5}+3}{-\sqrt{5}+3}\right] \end{gathered}$$
$$\begin{gathered} I=\frac{1}{2\sqrt{5}}\log \left(\frac{3+\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}\right) \\ I=\frac{1}{2\sqrt{5}}log{\left(\frac{3+\sqrt{5}}{2}\right)}^2 \\ I=\frac{2}{2\sqrt{5}}log\left(\frac{3+\sqrt{5}}{2}\right) \\ I=\frac{1}{\sqrt{5}}log\left(\frac{3+\sqrt{5}}{2}\right) \end{gathered}$$