Q. $\int_{}^{}{e^{x}(\log{\sin{x + \cot x}})dx}$ is equal to
A) $e^{x}\cot{x + c}$
B) $e^{x}\log{\sin{x + c}}$
C) $e^{x}\log{\sin{x + \tan{x + c}}}$
D) $e^{x} + \sin{x + c}$
Explanation
(b)
We know, $\int_{}^{}{e^{x}\lbrack f(x) + f^{'}(x)\rbrack dx} = e^{x}f(x) + c$
$\therefore$ If $f(x) = \log{\sin x}$
$\Rightarrow \ f^{'}(x) = \frac{1}{\sin x}.\cos x = \cot x$
$\therefore\ \int_{}^{}e^{x}\left( \log{\sin x} + \cot x \right)dx = e^{x}\log{\sin x + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.