AP EAMCET · Maths · Indefinite Integration
\(\int \frac{d x}{\sin x+\sin 2 x}=\)
- A \(\begin{array}{r}\frac{1}{2} \log _e|1+\cos x|+\frac{1}{6} \log _e|1-\cos x|-\frac{2}{3} \log _e \mid \\ 1+2 \cos x \mid+c\end{array}\)
- B \(\begin{array}{r}\frac{1}{2} \log _e|1+\cos x|-\frac{2}{3} \log _e|1-\cos x|+\frac{1}{2} \log _e \mid \\ 1+2 \cos x \mid+c\end{array}\)
- C \(\begin{array}{r}\frac{1}{2} \log _e|1+\sin x|-\frac{1}{3} \log _e|1-\sin x|-\frac{1}{3} \log _e \mid \\ 1+\cos x \mid+c\end{array}\)
- D \(\begin{array}{r}\frac{1}{3} \log _e|1-\sin x|+\frac{1}{2} \log _e|1+\cos x|-\frac{2}{3} \log _e \mid \\ 1-2 \cos x \mid+c\end{array}\)
Answer & Solution
Correct Answer
(A) \(\begin{array}{r}\frac{1}{2} \log _e|1+\cos x|+\frac{1}{6} \log _e|1-\cos x|-\frac{2}{3} \log _e \mid \\ 1+2 \cos x \mid+c\end{array}\)
Step-by-step Solution
Detailed explanation
Given, \[ \begin{aligned} & \int \frac{d x}{\sin x+\sin 2 x}=\int \frac{d x}{\sin x+2 \sin x \cos x} \\ & =\int \frac{d x}{\sin x(1+2 \cos x)} \\ & =\int \frac{\sin x d x}{\sin ^2 x(1+2 \cos x)} \\ & =\int \frac{-\sin x d x}{\left(\cos ^2 x-1\right)(1+2 \cos x)} \end{aligned} \]…
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