Question

If $\int \frac{{\sec }^2\theta \sin 2\theta d\theta }{\cos 3\theta }=f\left(\theta \right)+C$ for some arbitrary constant $C,$ then $f\left(\theta \right)$ is equal to

• A. $\frac{2}{3}\ln \left|{\left(\frac{\cos \theta +\cos \frac{\pi }{6}}{\cos \theta -\cos \frac{\pi }{6}}\right)}^{\tan \frac{\pi }{6}}{e}^{\sec \theta }\right|$
• B. $\frac{2}{3}\ln \left|{\left(\frac{\cos \theta -\cos \frac{\pi }{6}}{\cos \theta +\cos \frac{\pi }{6}}\right)}^{\tan \frac{\pi }{6}}{e}^{\cos \theta }\right|$
• C. $\frac{2}{3}\ln \left|{\left(\frac{\cos \theta +\cos \frac{\pi }{6}}{\cos \theta -\cos \frac{\pi }{6}}\right)}^{\tan \frac{\pi }{6}}{e}^{\sec (\pi -\theta )}\right|$
• D. $\frac{2}{3}\ln \left|{\left(\frac{\cos \theta +\cos \frac{\pi }{6}}{\cos \theta -\cos \frac{\pi }{6}}\right)}^{\tan \frac{\pi }{6}}{e}^{\cos (\pi -\theta )}\right|$

Answer 1

The correct option is C $\frac{2}{3}\ln \left|{\left(\frac{\cos \theta +\cos \frac{\pi }{6}}{\cos \theta -\cos \frac{\pi }{6}}\right)}^{\tan \frac{\pi }{6}}{e}^{\sec (\pi -\theta )}\right|$

$I=\int \frac{2\sin \theta \cos \theta d\theta }{{\cos }^2\theta (4{\cos }^3\theta -3\cos \theta )}$

Put $x=\cos \theta \Rightarrow dx=-\sin \theta d\theta$
$I=-2\int \frac{dx}{x^2(4x^2-3)}=-\frac{2}{3}\int \frac{4x^2-(4x^2-3)}{x^2(4x^2-3)}=-\frac{2}{3}\left\{\int \frac{dx}{x^2-{\left(\frac{\sqrt{3}}{2}\right)}^2}-\int \frac{dx}{x^2}\right\}=-\frac{2}{3}\{\frac{1}{\sqrt{3}}\ln \left|\frac{2x-\sqrt{3}}{2x+\sqrt{3}}\right|+\frac{1}{x}\}+C=\frac{2}{3}\{\ln \left|{\left(\frac{\cos \theta +\cos \frac{\pi }{6}}{\cos \theta -\cos \frac{\pi }{6}}\right)}^{\tan \frac{\pi }{6}}\right|+\ln e^{(-x^{-1})}\}+CI=\frac{2}{3}\ln \left|{\left(\frac{\cos \theta +\cos \frac{\pi }{6}}{\cos \theta -\cos \frac{\pi }{6}}\right)}^{\tan \frac{\pi }{6}}{e}^{\sec (\pi -\theta )}\right|+C$