If \(\int \frac{\sin \theta}{\sin 3 \theta} d \theta=\frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan \theta}\right|+c\), then \(k=\)

(A)
\(\int \frac{\sin \theta}{\sin 3 \theta} d \theta =\int \frac{\sin \theta}{3 \sin \theta-4 \sin ^{2} \theta} d \theta=\) \(\int \frac{\sin \theta}{\sin \theta\left(3-4 \sin ^{2} \theta\right)} d \theta \)
\( =\int \frac{1}{3-4 \sin ^{2} \theta} d \theta\)
Dividing both numerator and denominator by \(\cos ^{2} \theta\), we get
\(I=\int \frac{\sec ^{2} \theta}{3 \sec ^{2} \theta-4 \tan ^{2} \theta} d \theta=\) \(\int \frac{\sec ^{2} \theta}{3\left(1+\tan ^{2} \theta\right)-4 \tan ^{2} \theta} d \theta\)
\(I=\int \frac{\sec ^{2} \theta}{3-\tan ^{2} \theta} d \theta\)
Put \(\tan \theta=t \Rightarrow \sec ^{2} \theta d \theta=d t\)
\(I=\int \frac{1}{(\sqrt{3})^{2}-t^{2}} d t=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\tan \theta}{\sqrt{3}-\tan \theta}\right|\)
Comparing with given data, we get \(\mathrm{k}=\sqrt{3}\)