MCQ
$\int_1^2 {\frac{{\cos (\log x)}}{x}} \,dx = $
- A$\sin \,(\log 3)$
- ✓$\sin \,(\log 2)$
- C$\cos \,(\log 3)$
- Dએકપણ નહીં.
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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