If $\int \frac{\sin heta}{\sin 3 heta} d heta=\frac{1}{2 k} \log \left|\frac{k+ an heta}{k- an heta} ight|+c$, then $k=$

Question

If $\int \frac{\sin 	heta}{\sin 3 	heta} d 	heta=\frac{1}{2 k} \log \left|\frac{k+	an 	heta}{k-	an 	heta}ight|+c$, then $k=$

Aicharya · Accepted Answer

$\begin{aligned} \int \frac{\sin 	heta}{\sin 3 	heta} d 	heta &=\int \frac{\sin 	heta}{3 \sin 	heta-4 \sin ^{2} 	heta} d 	heta=\int \frac{\sin 	heta}{\sin 	heta\left(3-4 \sin ^{2} 	hetaight)} d 	heta \ &=\int \frac{1}{3-4 \sin ^{2} 	heta} d 	heta \end{aligned}$ Dividing both numerator and denominator by $\cos ^{2} 	heta$, we get $\begin{aligned} I &=\int \frac{\sec ^{2} 	heta}{3 \sec ^{2} 	heta-4 	an ^{2} 	heta} d 	heta=\int \frac{\sec ^{2} 	heta}{3\left(1+	an ^{2} 	hetaight)-4 	an ^{2} 	heta} d 	heta \ I &=\int \frac{\sec ^{2} 	heta}{3-	an ^{2} 	heta} d 	heta \end{aligned}$ Put $	an 	heta=t \Rightarrow \sec ^{2} 	heta d 	heta=d t$ $I=\int \frac{1}{(\sqrt{3})^{2}-t^{2}} d t=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}ight|=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+	an 	heta}{\sqrt{3}-	an 	heta}ight|$ Comparing with given data, we get $\mathrm{k}=\sqrt{3}$

If $\int \frac{\sin \theta}{\sin 3 \theta} d \theta=\frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan…

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