How do you integrate $\int x cos 4 x$ $int xcos4x$ by integration by parts method?

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Answer 1 · 2016-10-22T16:19:52

$\int x sin 4 x d x = \frac{x sin 4 x}{4} + \frac{cos 4 x}{16} + C$ $intxsin4xdx=(xsin4x)/4+(cos4x)/16+C$

Use the formula

$intu'v=uv -intuv'$

So here $u'=cos4x$ so $u=(sin4x)/4$

and $v=x$ so $v'=1$

Putting these in the formula

$intxsin4xdx=(xsin4x)/4-int(sin4xdx)/4$

$=(xsin4x)/4+(cos4x)/16+C$

How do you integrate int xcos4x∫xcos4xint xcos4x by integration by parts method?