Question

\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]

Options

• log (3 + 4 cos² x) + C

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

Answer 1

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

 

 

\[\text{Let }I = \int\frac{\sin x}{3 + 4 \cos^2 x}dx\]

\[\text{Putting }\cos x = t\]

\[ \Rightarrow - \sin x dx = dt\]

\[ \therefore I = \int\frac{- dt}{3 + 4 t^2}\]

\[ = \frac{1}{4}\int\frac{- dt}{t^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]

\[ = \frac{- 1}{4} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} \left( \frac{t \times 2}{\sqrt{3}} \right) + C .............\left( \because \int\frac{1}{x^2 + a^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right)\]

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

\[ = - \frac{1}{2\sqrt{3}} \tan^{- 1} \left( \frac{2 \cos x}{\sqrt{3}} \right) + C .............\left( \because t = \cos x \right)\]