Question

\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]

Answer 1

\[\text{  Let  I } = \int\frac{1}{\sin x \left( 2 + 3 \cos x \right)}\text{ dx}\]

\[ = \int\frac{\sin x}{\sin^2 x \left( 2 + 3 \cos x \right)}dx\]

\[ = \int\frac{\sin x}{\left( 1 - \cos^2 x \right) \left( 2 + 3 \cos x \right)} dx\]

\[ = \int\frac{\sin x}{\left( 1 - \cos x \right) \left( 1 + \cos x \right) \left( 2 + 3 \cos x \right)} dx\]

\[\text{ Putting   cos x = t }\]

\[ \Rightarrow - \text{ sin  x  dx  = dt}\]

\[ \therefore I = \int\frac{- 1}{\left( 1 - t \right) \left( 1 + t \right) \left( 2 + 3t \right)}dt\]

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

\[\text{ Let }\frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 3t + 2 \right)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{3t + 2}\]

\[ \Rightarrow \frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 3t + 2 \right)} = \frac{A \left( t + 1 \right) \left( 3t + 2 \right) + B \left( t - 1 \right) \left( 3t + 2 \right) + C \left( t + 1 \right) \left( t - 1 \right)}{\left( t - 1 \right) \left( t + 1 \right) \left( 3t + 2 \right)}\]

\[ \Rightarrow 1 = A \left( t + 1 \right) \left( 3t + 2 \right) + B \left( t - 1 \right) \left( 3t + 2 \right) + C \left( t + 1 \right) \left( t - 1 \right)\]

\[\text{ Putting  t + 1 = 0 or t = - 1}\]

\[ \Rightarrow 1 = A \times 0 + B \left( - 1 - 1 \right) \left( 3 \times - 1 + 2 \right) + C \times 0\]

\[ \therefore B = \frac{1}{2}\]

\[\text{ Now , putting t - 1 = 0 or t = 1 }\]

\[ \Rightarrow 1 = A \left( 1 + 1 \right) \left( 3 + 2 \right) + B \times 0 + C \times 0\]

\[ \therefore A = \frac{1}{10}\]

\[\text{ Now, putting 3t + 2 = 0 or t} = \frac{- 2}{3}\]

\[ \Rightarrow 1 = A \times 0 + B \times 0 + C \left( - \frac{2}{3} + 1 \right) \left( - \frac{2}{3} - 1 \right)\]

\[ \Rightarrow 1 = C \left( \frac{1}{3} \right) \left( \frac{- 5}{3} \right)\]

\[ \therefore C = \frac{- 9}{5}\]

\[ \therefore I = \int\frac{1}{10 \left( t - 1 \right)}dt + \frac{1}{2}\int\frac{1}{t + 1}dt - \frac{9}{5}\int\frac{1}{3t + 2}dt\]

\[ = \frac{1}{10} \text{ ln }\left| t - 1 \right| + \frac{1}{2} \text{ ln }\left| t + 1 \right| - \frac{9}{5} \text{ ln }\frac{\left| 3t + 2 \right|}{3} + C\]

\[ = \frac{1}{10} \text{ ln} \left| t - 1 \right| + \frac{1}{2} \text{ log } \left| t + 1 \right| - \frac{3}{5} \text{ ln} \left| 3t + 2 \right| + C\]

\[ = \frac{1}{10} + \text{ ln } \left| \cos x - 1 \right| + \frac{1}{2} \text{ ln }\left| \cos x + 1 \right| - \frac{3}{5} \text{ ln } \left| 3 \cos x + 2 \right| + C.......... \left[ \because t = \cos x \right]\]