Question

\[\int\frac{1}{7 + 5 \cos x} dx =\]

Options

• \[\frac{1}{\sqrt{6}} \tan^{- 1} \left( \frac{1}{\sqrt{6}}\tan\frac{x}{2} \right) + C\]

• \[\frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}}\tan\frac{x}{2} \right) + C\]

• \[\frac{1}{4} \tan^{- 1} \left( \tan\frac{x}{2} \right) + C\]
• \[\frac{1}{7} \tan^{- 1} \left( \tan\frac{x}{2} \right) + C\]

Answer 1

\[\frac{1}{\sqrt{6}} \tan^{- 1} \left( \frac{1}{\sqrt{6}}\tan\frac{x}{2} \right) + C\]

 

 

\[\text{Let }I = \int\frac{dx}{7 + 5 \cos x}\]

\[\text{Putting }\cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \therefore I = \int\frac{dx}{7 + 5 \times \left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)}\]
\[ = \int\frac{\left( 1 + \tan^2 \frac{x}{2} \right) dx}{7\left( 1 + \tan^2 \frac{x}{2} \right) + 5 - 5 \tan^2 \frac{x}{2}}\]
\[ = \int\frac{\sec^2 \frac{x}{2} dx}{2 \tan^2 \frac{x}{2} + 12}\]
\[ = \frac{1}{2}\int\frac{\sec^2 \frac{x}{2}dx}{\tan^2 \frac{x}{2} + \left( \sqrt{6} \right)^2}\]
\[\text{Let }\tan \frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) dx = dt\]
\[ \Rightarrow \sec^2 \left( \frac{x}{2} \right) dx = 2 dt\]
\[ \therefore I = \frac{1}{2}\int\frac{2 dt}{t^2 + \left( \sqrt{6} \right)^2}\]
\[ = \frac{1}{\sqrt{6}} \tan^{- 1} \left( \frac{t}{\sqrt{6}} \right) + C ................\left( \because \int\frac{1}{a^2 + x^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right)\]
\[ = \frac{1}{\sqrt{6}} \tan^{- 1} \left( \frac{\tan \frac{x}{2}}{\sqrt{6}} \right) + C .............\left( \because t = \tan \frac{x}{2} \right)\]