$\int \frac{1}{\sin x + \sin 2 x} d x$

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Solution

$We have, I = \int \frac{d x}{\sin x + \sin 2 x} = \int \frac{d x}{\sin x + 2 \sin x \cos x} = \int \frac{d x}{\sin x (1 + 2 \cos x)} = \int \frac{\sin x d x}{\sin^{2} x (1 + 2 \cos x)} = \int \frac{\sin x d x}{(1 - \cos^{2} x) (1 + 2 \cos x)} = \int \frac{\sin x d x}{(1 - \cos x) (1 + \cos x) (1 + 2 \cos x)} Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = \int \frac{- d t}{(1 - t) (1 + t) (1 + 2 t)} = \int \frac{d t}{(t - 1) (t + 1) (1 + 2 t)} Let \frac{1}{(t - 1) (t + 1) (1 + 2 t)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{1 + 2 t} \Rightarrow \frac{1}{(t - 1) (t + 1) (1 + 2 t)} = \frac{A (t + 1) (1 + 2 t) + B (t - 1) (1 + 2 t) + C (t - 1) (t + 1)}{(t - 1) (t + 1) (1 + 2 t)} \Rightarrow 1 = A (t + 1) (1 + 2 t) + B (t - 1) (1 + 2 t) + C (t - 1) (t + 1) Putting t + 1 = 0 \Rightarrow t = - 1 1 = B (- 1 - 1) (1 - 2) \Rightarrow 1 = B (- 2) (- 1) \Rightarrow B = \frac{1}{2} Putting t - 1 = 0 \Rightarrow t = 1 1 = A (1 + 1) (1 + 2) \Rightarrow 1 = A (2) (3) \Rightarrow A = \frac{1}{6} Putting 1 + 2 t = 0 t = - \frac{1}{2} \Rightarrow 1 = A \times 0 + B \times 0 + C (- \frac{1}{2} - 1) (- \frac{1}{2} + 1) 1 = C (- \frac{3}{2}) (\frac{1}{2}) C = - \frac{4}{3} Then, I = \frac{1}{6} \int \frac{d t}{t - 1} + \frac{1}{2} \int \frac{d t}{t + 1} - \frac{4}{3} \int \frac{d t}{1 + 2 t} = \frac{1}{6} \log |t - 1| + \frac{1}{2} \log |t + 1| - \frac{4}{3} \times \frac{\log |1 + 2 t|}{2} + C = \frac{1}{6} \log |t - 1| + \frac{1}{2} \log |t + 1| - \frac{2}{3} \log |1 + 2 t| + C = \frac{1}{6} \log |\cos x - 1| + \frac{1}{2} \log |\cos x + 1| - \frac{2}{3} \log |1 + 2 \cos x| + C$

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