( _e x) ,dx is equal to - Vidyadip

MCQ

$\int {\cos ({{\log }_e}x)\,dx} $ is equal to

✓
$\frac{1}{2}x\{ \cos ({\log _e}x) + \sin ({\log _e}x)\} $
B
$x\{ \cos ({\log _e}x) + \sin ({\log _e}x)\} $
C
$\frac{1}{2}x\{ \cos ({\log _e}x) - \sin ({\log _e}x)\} $
D
$x\{ \cos ({\log _e}x) - \sin ({\log _e}x)\} $

✓

Answer

Correct option: A.

$\frac{1}{2}x\{ \cos ({\log _e}x) + \sin ({\log _e}x)\} $

a
(a) Let $I = \int {\cos ({{\log }_e}x)\,dx} $$ = \int {\cos ({{\log }_e}x)\,.\,1\,dx} $
$I = \cos ({\log _e}x).\,x - \int {\frac{{ - \sin ({{\log }_e}x)}}{x}} .\,x\,\,dx$
$ = x\cos ({\log _e}x) + \int {\sin ({{\log }_e}x)} \,\,dx$
$ = x\cos \,({\log _e}x) + \int {\sin \,({{\log }_e}x)} \,\,1\,\,dx$
$ = x\cos ({\log _e}x) + \sin ({\log _e}x).\,x - \int {\frac{{\cos ({{\log }_e}x)}}{x}x\,dx} $
$ = x\,\cos ({\log _e}x) + x\sin ({\log _e}x) - I$
$ \Rightarrow 2I = x\,[\cos \,({\log _e}x) + \sin \,({\log _e}x)]$
$ \Rightarrow I = \frac{x}{2}\,[\cos \,\,({\log _e}x) + \sin \,({\log _e}x)]$.

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