MCQ
$\int_1^2 \frac{\cos (\log x)}{x} d x=$
- A$\sin (\log 3)$
- ✓$\sin (\log 2)$
- C$\cos (\log 3)$
- D$\cos (\log 2)$
✓
Answer
Correct option: B.
$\sin (\log 2)$
(B)
Put $\log x= t \Rightarrow \frac{1}{x} d x= dt$
When $x=1, t =0$ and when $x=2, t =\log 2$
$\therefore \quad \int_1^2 \frac{\cos (\log x)}{x} d x=\int_0^{\log 2} \cos t d t$
$=[\sin t]_0^{\log 2}=\sin (\log 2)$
Put $\log x= t \Rightarrow \frac{1}{x} d x= dt$
When $x=1, t =0$ and when $x=2, t =\log 2$
$\therefore \quad \int_1^2 \frac{\cos (\log x)}{x} d x=\int_0^{\log 2} \cos t d t$
$=[\sin t]_0^{\log 2}=\sin (\log 2)$
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