Question

If $\int \frac{d\theta }{\left({\cos }^2\theta \right(\tan 2\theta +\sec 2\theta )}=\lambda \tan \theta +2{\log }_e|f\left(\theta \right)|+C$ where $C$ is constant of integration, then the ordered pair $(\lambda ,f(\theta \left)\right)$ is equal to:

• A. $(-1,1-\tan \theta )$
• B. $(-1,1+\tan \theta )$
• C. $(1,1+\tan \theta )$
• D. $(-1,1-\tan \theta )$

Answer 1

The correct option is B $(-1,1+\tan \theta )$

Let $I=\int \frac{d\theta }{{\cos }^2\theta (\sec 2\theta +\tan 2\theta )}$

$I=\int \frac{{\sec }^2\theta d\theta }{\left(\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }\right)+\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)}$

$I=\int \frac{(1-{\tan }^2\theta )\left({\sec }^2\theta \right)d\theta }{(1+\tan \theta {)}^2}$
Let $\tan \theta =k\Rightarrow {\sec }^2\theta d\theta =dk$
$I=\int \frac{(1-k^2)}{(1+k{)}^2}dk=\int \frac{(1-k)}{(1+k)}dk$
$I=(\frac{2}{1+k}-1)dk$
$I=2\ln |1+k|-k+c$
$I=2\ln |1+\tan \theta |-\tan \theta +c$
Given $I=\lambda \tan \theta +2\log f\left(\theta \right)+c$
$\therefore \lambda =-1,f\left(\theta \right)=1+\tan \theta$