AP EAMCET · Maths · Indefinite Integration
\[
\int \frac{d x}{\sin x+\sin 2 x}=
\]
- A \(\frac{1}{6} \log (1-\cos x)+\frac{1}{2} \log (1+\cos x)+\)
\[
\frac{2}{3} \log |1+2 \cos x|+c
\] - B \[
\begin{aligned}
& \frac{1}{6} \log (1-\cos x)-\frac{1}{2} \log (1+\cos x)- \\
& \frac{2}{3} \log |1+2 \cos x|+c
\end{aligned}
\] - C \[
\begin{aligned}
& \frac{1}{6} \log (1-\cos x)+\frac{1}{2} \log (1+\cos x)- \\
& \frac{2}{3} \log |1+2 \cos x|+c
\end{aligned}
\] - D \(\frac{1}{6} \log [(1-\cos x)(1+\cos x)|1+2 \cos x|]+c\)
Answer & Solution
Correct Answer
(C) \[
\begin{aligned}
& \frac{1}{6} \log (1-\cos x)+\frac{1}{2} \log (1+\cos x)- \\
& \frac{2}{3} \log |1+2 \cos x|+c
\end{aligned}
\]
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { } I=\int \frac{d x}{\sin x+\sin 2 x} \\ & =\int \frac{d x}{\sin x(1+2 \cos x)}=\int \frac{\sin x d x}{\sin ^2 x(1+2 \cos x)} \\ & =\int \frac{\sin x d x}{(1-\cos x)(1+\cos x)(1+2 \cos x)} \end{aligned} \] Let \(\cos x=t \Rightarrow-\sin x d x=d t\)…
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