How do you integrate $int xsin(10x)$ by integration by parts method?

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Answer 1 · 2017-01-29T08:35:37

$x/10cos10x+1/100sin10x+C$

It is important that the integration by parts formula be memorised

$intuv'dx=uv-intvu'dx$

when using the IBP the choice of $u " & " v'$ is crucial.

in this case

$u=x=>u'=1$

$v'=sin10x=>v=-1/10cos10x$

$:.I=intuv'dx=uv-intvu'dx$

becomes

$I=-x/10cos10x-(int1/10cos10xdx)$

$I=-x/10cos10x+1/100sin10x+C$

How do you integrate int xsin(10x)int xsin(10x) by integration by parts method?