If int frac{sin theta}{sin 3 theta} d theta = | vedclass

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(A) We have the integral $I = \int \frac{\sin 	heta}{\sin 3 	heta} d 	heta$. Using the identity $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$,w

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$\sqrt{3}$

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(A) We have the integral $I = \int \frac{\sin 	heta}{\sin 3 	heta} d 	heta$. Using the identity $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$,we get: $I = \int \frac{\sin 	heta}{\sin 	heta(3 - 4 \sin^2 	heta)} d 	heta = \int \frac{1}{3 - 4 \sin^2 	heta} d 	heta$. Dividing the numerator and denominator by $\cos^2 	heta$: $I = \int \frac{\sec^2 	heta}{3 \sec^2 	heta - 4 	an^2 	heta} d 	heta = \int \frac{\sec^2 	heta}{3(1 + 	an^2 	heta) - 4 	an^2 	heta} d 	heta = \int \frac{\sec^2 	heta}{3 - 	an^2 	heta} d 	heta$. Let $	an 	heta = t$,then $\sec^2 	heta d 	heta = dt$. $I = \int \frac{1}{(\sqrt{3})^2 - t^2} dt$. Using the formula $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight| + c$: $I = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + t}{\sqrt{3} - t} ight| + c = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + 	an 	heta}{\sqrt{3} - 	an 	heta} ight| + c$. Comparing this with the given expression $\frac{1}{2 k} \log \left|\frac{k+	an 	heta}{k-	an 	heta}ight|+c$,we find $k = \sqrt{3}$.

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

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