Q. $\int_{}^{}{\lbrack\sin{(\log{x) + \cos{(\log{x)\rbrack dx}}}}}$ is equal to
A) $x\cos{(\log{x) + c}\ }$
B) $\cos{(\log{x) + c}}$
C) $x\sin{(\log{x) + c}}$
D) $\sin{(\log{x) + c}}$
Explanation
(c)
$\int_{}^{}{\lbrack\sin{(\log{x) + \cos{(\log{x)\rbrack dx}}}}}$
$= \int_{}^{}{\frac{d}{dx}\lbrack x\sin{(\log{x)}}\}}dx$
$= x\sin{(\log{x) + c}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.