How do you integrate $\int sin (ln x) d x$ $int sin(lnx) dx$ ?

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Answer 1 · 2018-08-07T13:53:31

$\int sin (ln x) d x = \frac{x}{2} [sin (ln x) - cos (ln x)] + C$ $intsin(lnx)dx=x/2[sin(lnx)-cos(lnx)]+C$

Here,

$I=intsin(lnx)dx.................(A)$

Subst. $color(red)(lnx=u=>x=e^u=>dx=e^udu$

$:.I=intsinu*e^udu$

$I=sinuinte^udu-int(d/(du)(sinu)inte^udu)du$

$:.I=sinuxxe^u-intcosue^udu$

$I=e^usinu-{cosuxxe^u-int(-sinue^u)du}$

$:.I=e^usinu-e^ucosu-inte^usinudu+c$

$I=e^usinu-e^ucosu-I+c.to$ from $(A)$

$:.I+I=e^usinu-e^ucosu+c$

$:.2I=e^usinu-e^ucosu+c$

$:.I=1/2[e^usinu-e^ucosu]+c/2$

Subst.back $color(red)(lnx=u=>x=e^u$

$intsin(lnx)dx=1/2[x*sin(lnx)-xcos(lnx)]+C$

Hence ,

$intsin(lnx)dx=x/2[sin(lnx)-cos(lnx)]+C$

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