Question:
The evaluation of the indefinite integral $\int \frac{dx}{\sin x + \sin 2x}$ is:
The evaluation of the indefinite integral $\int \frac{dx}{\sin x + \sin 2x}$ is:
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Always convert trigonometric sums into product form before substitution.
Updated On: Jun 10, 2026
- $\frac{1}{4}\log|1-\cos x| + \frac{1}{3}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$
- $\frac{1}{3}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{1}{3}\log|1+2\cos x| + c$
- $\frac{1}{6}\log|1-\cos x| + \frac{1}{2}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$
- $\frac{1}{6}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{2}{3}\log|1+2\cos x| + c$
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The Correct Option is C
Solution and Explanation
Concept: The integral involves trigonometric expressions where transforming to cosine form and using substitution reduces it to partial fractions.
Step 1: Simplify denominator \[ \sin 2x = 2\sin x \cos x \] \[ \sin x + \sin 2x = \sin x(1 + 2\cos x) \] Thus, \[ I = \int \frac{dx}{\sin x(1 + 2\cos x)} \]
Step 2: Multiply and substitute Multiply numerator and denominator by \(\sin x\): \[ I = \int \frac{\sin x\,dx}{\sin^2 x(1+2\cos x)} \] \[ \sin^2 x = 1 - \cos^2 x \] So, \[ I = \int \frac{\sin x\,dx}{(1-\cos^2 x)(1+2\cos x)} \] Let \( t = \cos x \Rightarrow dt = -\sin x dx \)
Step 3: Convert \[ I = \int \frac{-dt}{(1-t^2)(1+2t)} \] \[ = \int \frac{dt}{(t-1)(t+1)(2t+1)} \]
Step 4: Partial fractions \[ \frac{1}{(t-1)(t+1)(2t+1)} = \frac{A}{t-1} + \frac{B}{t+1} + \frac{C}{2t+1} \] Solving: \[ A=\frac{1}{6},\quad B=\frac{1}{2},\quad C=-\frac{4}{3} \]
Step 5: Integration \[ I = \frac{1}{6}\log|t-1| + \frac{1}{2}\log|t+1| - \frac{2}{3}\log|2t+1| \] Substitute \( t=\cos x \).
Step 1: Simplify denominator \[ \sin 2x = 2\sin x \cos x \] \[ \sin x + \sin 2x = \sin x(1 + 2\cos x) \] Thus, \[ I = \int \frac{dx}{\sin x(1 + 2\cos x)} \]
Step 2: Multiply and substitute Multiply numerator and denominator by \(\sin x\): \[ I = \int \frac{\sin x\,dx}{\sin^2 x(1+2\cos x)} \] \[ \sin^2 x = 1 - \cos^2 x \] So, \[ I = \int \frac{\sin x\,dx}{(1-\cos^2 x)(1+2\cos x)} \] Let \( t = \cos x \Rightarrow dt = -\sin x dx \)
Step 3: Convert \[ I = \int \frac{-dt}{(1-t^2)(1+2t)} \] \[ = \int \frac{dt}{(t-1)(t+1)(2t+1)} \]
Step 4: Partial fractions \[ \frac{1}{(t-1)(t+1)(2t+1)} = \frac{A}{t-1} + \frac{B}{t+1} + \frac{C}{2t+1} \] Solving: \[ A=\frac{1}{6},\quad B=\frac{1}{2},\quad C=-\frac{4}{3} \]
Step 5: Integration \[ I = \frac{1}{6}\log|t-1| + \frac{1}{2}\log|t+1| - \frac{2}{3}\log|2t+1| \] Substitute \( t=\cos x \).
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