(A)
$\begin{aligned}
\int \frac{\sin \theta}{\sin 3 \theta} d \theta &=\int \frac{\sin \theta}{3 \sin \theta-4 \sin ^{2} \theta} d \theta=\int \frac{\sin \theta}{\sin \theta\left(3-4 \sin ^{2} \theta\right)} d \theta \\
&=\int \frac{1}{3-4 \sin ^{2} \theta} d \theta
\end{aligned}$
Dividing both numerator and denominator by $\cos ^{2} \theta$, we get
$\begin{aligned}
I &=\int \frac{\sec ^{2} \theta}{3 \sec ^{2} \theta-4 \tan ^{2} \theta} d \theta=\int \frac{\sec ^{2} \theta}{3\left(1+\tan ^{2} \theta\right)-4 \tan ^{2} \theta} d \theta \\
I &=\int \frac{\sec ^{2} \theta}{3-\tan ^{2} \theta} d \theta
\end{aligned}$
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 $\mathrm{k}=\sqrt{3}$