How do you integrate $int xcosx$ by integration by parts method?

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Answer 1 · 2016-07-31T02:09:15

$int xcosx dx = xsinx + cosx +C$

Integration by parts tell us : $int f(x)g'(x) dx = f(x)g(x) - int f'(x)g(x) dx$

In this example
$f(x) = x$ and $g'(x) = cosx -> g(x) = sinx$

Therefore, using integration by parts:
$int xcosx dx = xsinx - int 1 * sinx dx$

$= xsinx + cosx + C$

How do you integrate int xcosxint xcosx by integration by parts method?