$\int\left\{\sin \left(\log _e x\right)+\cos \left(\log _e x\right)\right\} d x$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$I =\int\left\{\sin \left(\log _e x\right)+\cos \left(\log _e x\right)\right\} d x$
$\Rightarrow I =\int e^t(\sin t+\cos t) d t$, where $t=\log _e x$
$\Rightarrow I=e^t \sin t+C~~~~~\left[∵ \int e^x\left\{f(x)+f'(x)\right\} d x=e^x f(x)+C\right] $
$\Rightarrow I=x \sin \left(\log _e x\right)+C$