Question

If $\int \frac{1}{5+4\sin x}dx=A{\tan }^{-1}\left(B\tan \frac{x}{2}+\frac{4}{3}\right)+C,$ then

(a) A = $\frac{2}{3}$, B = $\frac{5}{3}$

(b) A = $\frac{1}{3}$, B = $\frac{2}{3}$

(c) A = $-\frac{2}{3}$, B = $\frac{5}{3}$

(d) A = $\frac{1}{3}$, B = $-\frac{5}{3}$

Answer 1

(a) A = $\frac{2}{3}$, B = $\frac{5}{3}$


$$\begin{gathered} \int \frac{1}{5+4\sin x}dx=A{\tan }^{-1}\left(B\tan \frac{x}{2}+\frac{4}{3}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}..\left(1\right) \\ \mathrm{Considering}\mathrm{the}\mathrm{LHS}\mathrm{of}\mathrm{eq}\left(1\right) \\ \mathrm{Putting}\sin x=\frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}} \\ \Rightarrow \int \frac{1}{5+{\displaystyle \frac{8\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}dx \\ \Rightarrow \int \frac{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)}{5\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)+8\tan {\displaystyle \frac{x}{2}}}dx \\ \Rightarrow \int \frac{{\sec }^2{\displaystyle \frac{x}{2}}}{5{\tan }^2{\displaystyle \frac{x}{2}}+8\tan {\displaystyle \frac{x}{2}}+5}dx...\left(2\right) \\ \mathrm{Let}\tan \frac{x}{2}=t \\ \Rightarrow {\sec }^2\frac{x}{2}\times \frac{1}{2}dx=dt \\ \Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2dt \\ \therefore \mathrm{Putting}\tan \frac{x}{2}=t\mathrm{and}{\sec }^2\left(\frac{x}{2}\right)dx=2dt\mathrm{we}\mathrm{get}, \\ \int \frac{2dt}{5t^2+8t+5} \\ \Rightarrow \frac{2}{5}\int \frac{dt}{t^2+{\displaystyle \frac{8}{5}}t+1} \\ \Rightarrow \frac{2}{5}\int \frac{dt}{t^2+{\displaystyle \frac{8}{5}}t+{\left({\displaystyle \frac{4}{5}}\right)}^2-{\left({\displaystyle \frac{4}{5}}\right)}^2+1} \\ \Rightarrow \frac{2}{5}\int \frac{dt}{{\left(t+{\displaystyle \frac{4}{5}}\right)}^2+1-{\displaystyle \frac{16}{25}}} \\ \Rightarrow \frac{2}{5}\int \frac{dt}{{\left(t+{\displaystyle \frac{4}{5}}\right)}^2+{\left({\displaystyle \frac{3}{5}}\right)}^2} \\ \Rightarrow \frac{2}{5}\times \frac{5}{3}{\tan }^{-1}\left(\frac{t+{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{3}{5}}}\right)+C \\ \Rightarrow \frac{2}{3}{\tan }^{-1}\left(\frac{5t+4}{3}\right)+C \\ \Rightarrow \frac{2}{3}{\tan }^{-1}\left(\frac{5}{3}\tan \frac{x}{2}+\frac{4}{3}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}t\mathit{=}\tan \frac{x}{\mathit{2}}\right)\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}...\left(3\right) \\ \mathrm{Comparing}\mathrm{eq}\left(3\right)\mathrm{with}\mathrm{the}\mathrm{RHS}\mathrm{of}\mathrm{eq}\left(1\right)\mathrm{we}\mathrm{get}, \\ \therefore \mathit{}A=\frac{2}{3},\mathit{}B=\frac{5}{3} \end{gathered}$$