Question

$\int \frac{d\theta }{(sin\theta -2\cos \theta )(2\sin \theta +\cos \theta )}=$

• A. $\frac{1}{5}\log |\frac{\tan \theta -2}{2\tan \theta +1}|+c$
• B. $\frac{1}{2}\log |\frac{\tan \theta -2}{2\tan \theta +1}|+c$
• C. $\frac{1}{5}\log |\frac{2\tan \theta +1}{\tan \theta -2}|+c$
• D. $\frac{1}{2}\log |\frac{2\tan \theta -1}{\tan \theta +2}|+c$

Answer 1

The correct option is A $\frac{1}{5}\log |\frac{\tan \theta -2}{2\tan \theta +1}|+c$

Given, $\int \frac{d\theta }{(sin\theta -2\cos \theta )(2\sin \theta +\cos \theta )}$
Dividing numerator and denominator by ${\cos }^2\theta$ we have;
$\Rightarrow$ $\int \frac{{\sec }^2\theta d\theta }{(\tan \theta -2)(2\tan \theta +1)}$
Breaking it in partial fraction form we have:
$\Rightarrow$ $\int \left[\frac{1}{5}\left(\frac{se{c}^2\theta }{\tan \theta -2}\right)-\frac{2}{5}\left(\frac{se{c}^2\theta }{2\tan \theta +1}\right)\right]d\theta$
$\Rightarrow$ $\frac{1}{5}log(\tan \theta -2)-\frac{2}{5\times 2}log(2\tan \theta +1)+c$
$\Rightarrow$ $\frac{1}{5}log\frac{(\tan \theta -2)}{(2\tan \theta +1)}+c$
Hence, option 'A' is correct.