Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{\sin x+\sqrt{3}\cos x}dx \\ \mathrm{Putting}\sin x=\frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}\mathrm{and}\cos x=\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}} \\ \Rightarrow I=\int \frac{1}{{\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}+{\displaystyle \sqrt{3}\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}}dx \\ =\int \frac{1+{\tan }^2{\displaystyle \frac{x}{2}}}{2\tan {\displaystyle \frac{x}{2}}+\sqrt{3}-\sqrt{3}{\tan }^2{\displaystyle \frac{x}{2}}}dx \\ =\int \frac{{\sec }^2{\displaystyle \frac{x}{2}}}{-\sqrt{3}{\tan }^2{\displaystyle \frac{x}{2}}+2\tan {\displaystyle \frac{x}{2}}+\sqrt{3}}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Let}\tan \frac{x}{2}=t \\ \Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt \\ \Rightarrow {\sec }^2\frac{x}{2}dx=2dt \\ \therefore I=2\int \frac{dt}{-\sqrt{3}{t}^2+2t+\sqrt{3}} \\ =-\frac{2}{\sqrt{3}}\int \frac{dt}{t^2-{\displaystyle \frac{2}{\sqrt{3}}}t-1} \\ =-\frac{2}{\sqrt{3}}\int \frac{dt}{t^2-{\displaystyle \frac{2}{\sqrt{3}}}t+{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2-{\displaystyle {\left(\frac{1}{\sqrt{3}}\right)}^2}-1} \\ =-\frac{2}{\sqrt{3}}\int \frac{dt}{{\left(t-{\displaystyle \frac{1}{\sqrt{3}}}\right)}^2-{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^2} \\ =-\frac{2}{\sqrt{3}}\times \frac{1}{2{\displaystyle \frac{2}{\sqrt{3}}}}\log \left|\frac{t-{\displaystyle \frac{1}{\sqrt{3}}}-{\displaystyle \frac{2}{\sqrt{3}}}}{t-{\displaystyle \frac{1}{\sqrt{3}}}+{\displaystyle \frac{2}{\sqrt{3}}}}\right|+C \end{gathered}$$

$$\begin{gathered} =-\frac{1}{2}\log \left|\frac{t-{\displaystyle \frac{3}{\sqrt{3}}}}{t+{\displaystyle \frac{1}{\sqrt{3}}}}\right|+C \\ =-\frac{1}{2}\log \left|\frac{\sqrt{3}t-3}{\sqrt{3}t+1}\right|+C \\ =\frac{1}{2}\log \left|\frac{\sqrt{3}t+1}{\sqrt{3}t-3}\right|+C \\ =\frac{1}{2}\log \left|\frac{\sqrt{3}\tan {\displaystyle \frac{x}{2}}+1}{\sqrt{3}\tan {\displaystyle \frac{x}{2}}-3}\right|+C \\ \mathrm{or},\frac{1}{2}\log \left|\frac{1+\sqrt{3}\tan {\displaystyle \frac{x}{2}}}{3-\sqrt{3}\tan {\displaystyle \frac{x}{2}}}\right|+C \end{gathered}$$