Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{5+4\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{5+4\sin x}\mathit{}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}\mathit{,} \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{5+4\left({\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}\right)}\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{5\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)+8\tan {\displaystyle \frac{x}{2}}}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sec }^2{\displaystyle \frac{x}{2}}}{5{\tan }^2{\displaystyle \frac{x}{2}}+8\tan {\displaystyle \frac{x}{2}}+5}dx \\ \mathrm{Let}\tan \frac{x}{2}=t.\mathrm{Then},\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt \\ \mathrm{When}x=0,t=0\mathrm{and}x=\frac{\mathrm{\pi }}{2},\mathit{}t=1 \\ \therefore I=2{\int }_0^1\frac{1}{5t^2+8t+5}dt \\ \Rightarrow I=2{\int }_0^1\frac{1}{{\left(\sqrt{5}t\right)}^2+8t+5+{\left({\displaystyle \frac{4}{\sqrt{5}}}\right)}^2-{\left({\displaystyle \frac{4}{\sqrt{5}}}\right)}^2}dt \\ \Rightarrow I=2{\int }_0^1\frac{1}{{\left(\sqrt{5}t+{\displaystyle \frac{4}{\sqrt{5}}}\right)}^2+{\displaystyle \frac{9}{5}}}dt \\ \Rightarrow I=\frac{2}{3}{\left[{\tan }^{-1}\left(\frac{\sqrt{5}t+{\displaystyle \frac{4}{\sqrt{5}}}}{{\displaystyle \frac{3}{\sqrt{5}}}}\right)\right]}_0^1 \\ \Rightarrow I=\frac{2}{3}\left[{\tan }^{-1}3-{\tan }^{-1}\frac{4}{3}\right] \\ \Rightarrow I=\frac{2}{3}\left[{\tan }^{-1}\left(\frac{3-{\displaystyle \frac{4}{3}}}{1+3\times {\displaystyle \frac{4}{3}}}\right)\right] \\ \Rightarrow I=\frac{2}{3}{\tan }^{-1}\frac{1}{3} \end{gathered}$$