$\int [sin (log x) + cos (log x)] d x =$

e^{x} sin (log x) + c

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e^{x} cos (log x) + c

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x sin (log x) + c

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x cos (log x) + c

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Solution

The correct option is C $x sin (log x) + c$
$\int [s i n (l o g x) + c o s (l o g x)] d x$
$= \int s i n (l o g x) d x + \int c o s (l o g x) d x$

Integrating by parts,
$= x s i n (l o g x) - \int c o s (l o g x) d x + \int c o s (l o g x) d x$
$= x s i n (l o g x) + c$

Suggest Corrections

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