Question

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

Answer 1

We have,
\[I = \int\frac{dx}{\sin x + \sin 2x}\]
\[ = \int\frac{dx}{\sin x + 2 \sin x \cos x}\]
\[ = \int\frac{dx}{\sin x \left( 1 + 2 \cos x \right)}\]
\[ = \int\frac{\sin x dx}{\sin^2 x \left( 1 + 2 \cos x \right)}\]
\[ = \int\frac{\sin x dx}{\left( 1 - \cos^2 x \right) \left( 1 + 2 \cos x \right)}\]
\[ = \int\frac{\sin x dx}{\left( 1 - \cos x \right) \left( 1 + \cos x \right) \left( 1 + 2 \cos x \right)}\]
\[\text{Putting }\cos x = t\]
\[ \Rightarrow - \sin x dx = dt\]
\[ \Rightarrow \sin x dx = - dt\]
\[ \therefore I = \int\frac{- dt}{\left( 1 - t \right) \left( 1 + t \right) \left( 1 + 2t \right)}\]
\[ = \int\frac{dt}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)}\]
\[\text{Let }\frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{1 + 2t}\]
\[ \Rightarrow \frac{1}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)} = \frac{A \left( t + 1 \right) \left( 1 + 2t \right) + B \left( t - 1 \right) \left( 1 + 2t \right) + C \left( t - 1 \right) \left( t + 1 \right)}{\left( t - 1 \right) \left( t + 1 \right) \left( 1 + 2t \right)}\]
\[ \Rightarrow 1 = A \left( t + 1 \right) \left( 1 + 2t \right) + B \left( t - 1 \right) \left( 1 + 2t \right) + C \left( t - 1 \right) \left( t + 1 \right)\]
\[\text{Putting t + 1 = 0}\]
\[ \Rightarrow t = - 1\]
\[1 = B \left( - 1 - 1 \right) \left( 1 - 2 \right)\]
\[ \Rightarrow 1 = B \left( - 2 \right) \left( - 1 \right)\]
\[ \Rightarrow B = \frac{1}{2}\]
\[\text{Putting t - 1 = 0}\]
\[ \Rightarrow t = 1\]
\[1 = A \left( 1 + 1 \right) \left( 1 + 2 \right)\]
\[ \Rightarrow 1 = A\left( 2 \right)\left( 3 \right)\]
\[ \Rightarrow A = \frac{1}{6}\]
\[\text{Putting 1 + 2t = 0}\]
\[t = - \frac{1}{2}\]
\[ \Rightarrow 1 = A \times 0 + B \times 0 + C \left( - \frac{1}{2} - 1 \right) \left( - \frac{1}{2} + 1 \right)\]
\[1 = C \left( - \frac{3}{2} \right) \left( \frac{1}{2} \right)\]
\[C = - \frac{4}{3}\]
Then,
\[I = \frac{1}{6}\int\frac{dt}{t - 1} + \frac{1}{2}\int\frac{dt}{t + 1} - \frac{4}{3}\int\frac{dt}{1 + 2t}\]
\[ = \frac{1}{6} \log \left| t - 1 \right| + \frac{1}{2} \log \left| t + 1 \right| - \frac{4}{3} \times \frac{\log \left| 1 + 2t \right|}{2} + C\]
\[ = \frac{1}{6} \log \left| t - 1 \right| + \frac{1}{2} \log \left| t + 1 \right| - \frac{2}{3}\log \left| 1 + 2t \right| + C\]
\[ = \frac{1}{6} \log \left| \cos x - 1 \right| + \frac{1}{2} \log \left| \cos x + 1 \right| - \frac{2}{3} \log \left| 1 + 2 \cos x \right| + C\]