int {frac{{sin 2theta dtheta}}{{(1 - {{sin }^ | vedclass

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Question

(C) Let $I = \int \frac{\sin 2	heta d	heta}{\cos^2 	heta \cos 3	heta} = \int \frac{2 \sin 	heta \cos 	heta d	heta}{\cos^2 	heta (4 \cos^3 	he

Vedclass · Accepted Answer

$\frac{2}{3}\ln \left| {{{\left( {\frac{{\cos 	heta + \cos \frac{\pi }{6}}}{{\cos 	heta - \cos \frac{\pi }{6}}}} ight)}^{	an \frac{\pi }{6}}}{e^{\sec (\pi - 	heta )}}} ight| + C$

Vedclass · Answer

(C) Let $I = \int \frac{\sin 2	heta d	heta}{\cos^2 	heta \cos 3	heta} = \int \frac{2 \sin 	heta \cos 	heta d	heta}{\cos^2 	heta (4 \cos^3 	heta - 3 \cos 	heta)} = \int \frac{2 \sin 	heta d	heta}{\cos^2 	heta (4 \cos^2 	heta - 3)}$.Let $\cos 	heta = x$,then $-\sin 	heta d	heta = dx$.$I = -2 \int \frac{dx}{x^2(4x^2 - 3)} = -\frac{2}{3} \int \frac{4x^2 - (4x^2 - 3)}{x^2(4x^2 - 3)} dx = -\frac{2}{3} \int \left( \frac{1}{x^2 - 3/4} - \frac{1}{x^2} ight) dx$.$I = -\frac{2}{3} \left( \frac{1}{2(\sqrt{3}/2)} \ln \left| \frac{x - \sqrt{3}/2}{x + \sqrt{3}/2} ight| + \frac{1}{x} ight) + C = -\frac{2}{3} \left( \frac{1}{\sqrt{3}} \ln \left| \frac{2x - \sqrt{3}}{2x + \sqrt{3}} ight| + \frac{1}{x} ight) + C$.Since $	an(\pi/6) = 1/\sqrt{3}$ and $\cos(\pi/6) = \sqrt{3}/2$,we have $I = \frac{2}{3} \ln \left| \left( \frac{2x + \sqrt{3}}{2x - \sqrt{3}} ight)^{	an(\pi/6)} \cdot e^{-1/x} ight| + C$.Substituting $x = \cos 	heta$,$I = \frac{2}{3} \ln \left| \left( \frac{\cos 	heta + \cos(\pi/6)}{\cos 	heta - \cos(\pi/6)} ight)^{	an(\pi/6)} \cdot e^{-\sec 	heta} ight| + C$.Since $-\sec 	heta = \sec(\pi - 	heta)$,the result matches option $C$.

$\int {\frac{{\sin 2\theta d\theta}}{{(1 - {{\sin }^2}\theta )\cos 3\theta }}} $ is equal to (where $C$ is the constant of integration).

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