How do you integrate $int sin(lnx)$ using integration by parts?

Question

1727 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-03T22:07:46

$1/2([xsin(ln(x))]-[xcos(ln(x))])$

$intsin(ln(x))dx$

Let's $u = ln(x)$

$du = dx/x$

$dx=xdu$

$x=e^u$

then

$inte^usin(u)du$

By part : $dv = e^u ; v = e^u ; w = sin(u) ; dw = cos(u)$

$[v*w]-intdw*v$

$inte^usin(u)du=[e^usin(u)]-inte^ucos(u)du$

By part again

$dv=e^u ; v = e^u ; w = cos(u) ; dw = -sin(u)$

$inte^usin(u)du=[e^usin(u)]-([e^ucos(u)]+inte^usin(u)du)$

$inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]-inte^usin(u)du$

$2inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]$

$inte^usin(u)du=1/2([e^usin(u)]-[e^ucos(u)])$

Substitute back for $u = ln(x)$

$intsin(ln(x))dx=1/2([xsin(ln(x))]-[xcos(ln(x))])$

How do you integrate int sin(lnx)int sin(lnx) using integration by parts?