How do you integrate #int xcos4x# by integration by parts method?

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Answer 1 · 2016-10-22T16:19:52

#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#