Question
Evaluate: $\int_1^3 \frac{\cos (\log x)}{x} d x$
✓
Answer
$\text { Let } I =\int_1^3 \frac{\cos (\log x)}{x} d x$
Put $\log x=t$
$\therefore \frac{1}{x} d x= dt$
When $x =1, t =\log 1=0$ and when $x =3, t =\log 3$
$ \therefore I=\int_0^{\log 3} \cos t d t$
$=[\sin t]_0^{\log 3}$
$=\sin (\log 3)-\sin 0$
$=\sin (\log 3) $
Put $\log x=t$
$\therefore \frac{1}{x} d x= dt$
When $x =1, t =\log 1=0$ and when $x =3, t =\log 3$
$ \therefore I=\int_0^{\log 3} \cos t d t$
$=[\sin t]_0^{\log 3}$
$=\sin (\log 3)-\sin 0$
$=\sin (\log 3) $
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