Question

\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\frac{1}{\sin x + \sin 2x}dx\]

\[ = \int\frac{1}{\sin x + 2 \sin x \cos x}dx\]

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

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

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

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

\[\text{ Putting  cos  x } = t\]

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

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

\[\therefore I = - \int\frac{1}{\left( 1 - t \right) \left( 1 + t \right) \left( 1 + 2t \right)}dt\]
\[ = \int\frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 2t + 1 \right)}dt\]
\[ \therefore \frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 2t + 1 \right)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{2t + 1}\]
\[ \Rightarrow 1 = A \left( t + 1 \right) \left( 2t + 1 \right) + B \left( t - 1 \right) \left( 2t + 1 \right) + C \left( t - 1 \right) \left( t + 1 \right)\]
\[\text{ Putting  t + 1 = 0 or t = - 1}\]
\[ \Rightarrow 1 = A \times 0 + B \left( - 1 - 1 \right) \left( - 2 + 1 \right) + C \times 0\]
\[ \Rightarrow 1 = B \left( 2 \right)\]
\[ \therefore B = \frac{1}{2}\]
\[\text{ Now, putting t - 1 = 0 or t = 1 }\]
\[ \Rightarrow 1 = A \left( 2 \right) \left( 3 \right) + B \times 0 + C \times 0\]
\[ \therefore A = \frac{1}{6}\]
\[\text{ Now, putting 2t + 1 = 0 or t} = - \frac{1}{2}\]
\[ \Rightarrow 1 = A \times 0 + B \times 0 + C \left( - \frac{1}{2} - 1 \right) \left( - \frac{1}{2} + 1 \right)\]
\[ \Rightarrow 1 = C \left( - \frac{3}{2} \right) \left( \frac{1}{2} \right)\]
\[ \therefore C = - \frac{4}{3}\]
\[ \therefore I = \frac{1}{6}\int\frac{1}{t - 1}dt + \frac{1}{2}\int\frac{1}{t + 1}dt - \frac{4}{3}\int\frac{1}{2t + 1}dt\]
\[ = \frac{1}{6} \text{ ln }\left| t - 1 \right| + \frac{1}{2} \text{ log } \left| t + 1 \right| - \frac{4}{3} \text{ ln} \frac{\left| 2t + 1 \right|}{2} + C\]
\[ = \frac{1}{6} \text{ ln} \left| t - 1 \right| + \frac{1}{2} \text{ ln} \left| t + 1 \right| - \frac{2}{3} \text{ ln } \left| 2t + 1 \right| + C\]
\[ = \frac{1}{6}\text{ ln } \left| \cos x - 1 \right| + \frac{1}{2} \text{ ln} \left| \cos x + 1 \right| - \frac{2}{3} \text{ ln }\left| 2 \cos x + 1 \right| + C \left[ \because t = \cos x \right]\]