Question

$\int e^{\tan \theta }(\sec \theta -\sin \theta )d\theta$ equals

• A. $-e^{\tan \theta }\sin \theta +c$
• B. $e^{\tan \theta }\sin \theta +c$
• C. $e^{\tan \theta }\sec \theta +c$
• D. $e^{\tan \theta }\cos \theta +c$

Answer 1

The correct option is D $e^{\tan \theta }\cos \theta +c$

Let $I=\int e^{\tan \theta }(\sec \theta -\sec \theta )d\theta$
Substitute $\tan \theta =t\Rightarrow {\sec }^2\theta d\theta =dt$
$\therefore I=\int e^t(\cos \theta -\tan \theta \sec \theta )dt$
$=\int e^t(\frac{1}{\sqrt{1+t^2}}-t\sqrt{1+t^2})dt$
As $\frac{d}{dx}\left(\frac{1}{\sqrt{1+t^2}}\right)=t\sqrt{1+t^2}$
$\therefore I=e^t\left(\frac{1}{\sqrt{1+t^2}}\right)=e^{\tan \theta }\cos \theta +c$