_1^2 ( x) x ,dx = - Vidyadip

MCQ

$\int_1^2 {\frac{{\cos (\log x)}}{x}} \,dx = $

A
$\sin \,(\log 3)$
✓
$\sin \,(\log 2)$
C
$\cos \,(\log 3)$
D
None of these

✓

Answer

Correct option: B.

$\sin \,(\log 2)$

b
(b) Put $t = \log x \Rightarrow dt = \frac{1}{x}dx$.

As $x = 2 \Rightarrow t = \log 2$

and $x = 1 \Rightarrow t = 0$, we have

$\int_1^2 {\frac{{\cos (\log x)}}{x}} dx = - \int_0^{\log 2} {\cos t\,dt} = [\sin t]_0^{\log 2}$$ = \sin (\log 2)$.

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