Question

$\int \frac{1}{13+3\cos x+4\sin x}dx$

Answer 1

$$\begin{gathered} LetI=\int \frac{1}{13+3\cos x+4\sin x}dx \\ \mathrm{Putting}\cos x=\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}\mathrm{and}\sin x=\frac{2\tan \left({\displaystyle \frac{x}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{x}{2}}\right)} \\ \therefore I=\int \frac{1}{13+3\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}\right)+4\times 2{\displaystyle \frac{\tan \left({\displaystyle \frac{x}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{x}{2}}\right)}}}dx \\ =\int \frac{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)}{13\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)+3-3{\tan }^2{\displaystyle \frac{x}{2}}+8\tan \left({\displaystyle \frac{x}{2}}\right)}dx \\ =\int \frac{{\sec }^2{\displaystyle \frac{x}{2}}}{13{\tan }^2{\displaystyle \frac{x}{2}}-3{\tan }^2{\displaystyle \frac{x}{2}}+16+8\tan \left({\displaystyle \frac{x}{2}}\right)}dx \\ =\int \frac{{\sec }^2{\displaystyle \frac{x}{2}}}{10{\tan }^2\left({\displaystyle \frac{x}{2}}\right)+8\tan \left({\displaystyle \frac{x}{2}}\right)+16}dx \\ \mathrm{Let}\tan \left(\frac{x}{2}\right)=t \\ \Rightarrow \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=dt \\ \Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2dt \\ \therefore I=\int \frac{2dt}{10t^2+8t+16} \\ =\int \frac{dt}{5t^2+4t+8} \\ =\frac{1}{5}\int \frac{dt}{t^2+{\displaystyle \frac{4}{5}}t+{\displaystyle \frac{8}{5}}} \\ =\frac{1}{5}\int \frac{dt}{t^2+{\displaystyle \frac{4}{5}}t+{\left({\displaystyle \frac{2}{5}}\right)}^2-{\left({\displaystyle \frac{2}{5}}\right)}^2+{\displaystyle \frac{8}{5}}} \\ =\frac{1}{5}\int \frac{dt}{{\left(t+{\displaystyle \frac{2}{5}}\right)}^2-{\displaystyle \frac{4}{25}}+{\displaystyle \frac{8}{5}}} \\ =\frac{1}{5}\int \frac{dt}{{\left(t+{\displaystyle \frac{2}{5}}\right)}^2+{\displaystyle \frac{-4+40}{25}}} \\ =\frac{1}{5}\int \frac{dt}{{\left(t+{\displaystyle \frac{2}{5}}\right)}^2+{\left({\displaystyle \frac{6}{5}}\right)}^2} \\ =\frac{1}{5}\times \frac{5}{6}{\tan }^{-1}\left(\frac{t+{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{6}{5}}}\right)+C \\ =\frac{1}{6}{\tan }^{-1}\left(\frac{5t+2}{6}\right)+C \\ =\frac{1}{6}{\tan }^{-1}\left(\frac{5\tan {\displaystyle \frac{x}{2}}+2}{6}\right)+C \end{gathered}$$