Question #1c46e

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Answer 1 · 2016-02-07T15:55:42

Trisha - integrate by parts twice ...

$u=sin(lnx)$

$du=[cos(lnx)]/x dx$

$dv=x dx$

$v=x^2/2$

$int xsin(lnx) dx=(x^2/2)(sin(lnx))-int(xcos(lnx))/2 dx$

Do it again on the integral ...

$u=cos(lnx)/2$

$du=-[sin(lnx)]/(2x) dx$

$dv=x dx$

$v=x^2/2$

Using the integration by parts formula one more time ...

$int xsin(lnx) dx=(x^2/2)(sin(lnx))-x^2cos(lnx)/4-1/4(intxsin(lnx))$

Combine integral terms ...

$(5/4)int xsin(lnx) dx=(x^2/2)(sin(lnx))-x^2cos(lnx)/4$

Simplify ...

$int xsin(lnx) dx=-x^2/5[cos(lnx)-2sin(lnx)]+C$

hope that helped