How do you integrate #int sin(lnx) dx#?

Question

Cameron Alexander · Accepted Answer

#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#

Applying Integration in Pieces:

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

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

Utilizing Integration by Parts Once More:

#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#

Charles Anctil · Answer

undefined To integrate $ \int \sin(\ln x) \, dx $, we use substitution. Let $ u = \ln x $. Then, $ du = \frac{1}{x} \, dx $. Substituting these into the integral:

$ \int \sin(\ln x) \, dx = \int \sin(u) \, du $

Now, integrate $ \sin(u) $ with respect to $ u $:

$ \int \sin(u) \, du = -\cos(u) + C $

Finally, substitute back for $ u $:

$ = -\cos(\ln x) + C $

So, $ \int \sin(\ln x) \, dx = -\cos(\ln x) + C $, where $ C $ is the constant of integration.

Emily Ammerman · Answer

undefined To integrate $ \int \sin(\ln(x)) \, dx $, you can use a substitution method. Let $ u = \ln(x) $. Then $ du = \frac{1}{x} \, dx $. Substituting $ u $ and $ du $ into the integral, it becomes $ \int \sin(u) \, du $. This integral can be easily evaluated to give $ -\cos(u) + C $, where $ C $ is the constant of integration. Finally, substitute $ u = \ln(x) $ back into the expression to get the final result: $ -\cos(\ln(x)) + C $.