How do you integrate #sin( ln x )#?

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Answer 1 · 2016-07-21T11:24:36

#I = x/2 ( sin(ln x) - cos(ln x) )+C#

#I = int \ sin (ln x) \ dx#

this is in the IBP section meaning you don't really have much choice how to take this, so ...

#I =int \ d/dx(x) * sin (ln x) \ dx#

which by IBP

#= x sin(ln x) - int \ x *d/dx( sin (ln x)) \ dx#

#= x sin(ln x) - int \ x cos (ln x)* 1/x \ dx#

#= x sin(ln x) - int \ cos (ln x) \ dx#

and now another round of IBP

#= x sin(ln x) - int \ d/dx(x) * cos (ln x) \ dx#

#= x sin(ln x) - x cos(ln x) + int \ x d/dx( cos (ln x)) \ dx#

#= x sin(ln x) - x cos(ln x) + int \ x (-sin (ln x) 1/x) \ dx#

#= x sin(ln x) - x cos(ln x) - int \ sin (ln x) \ dx#

#= x sin(ln x) - x cos(ln x) - I#

#2I = x sin(ln x) - x cos(ln x) +C# making sure to add in the integration constant

#implies I = x/2 ( sin(ln x) - cos(ln x) )+C#

Answer 2 · 2016-07-21T11:48:08

#int sin(log_e x) = x/2(sin(log_ex)-cos(log_ex))#

#e^{i phi} = cos phi+ i sin phi# (Moivre's identity)

so

#e^{i log_e x} = cos(log_e x)+i sin(log_e x)#

but

#e^{i log_e x} = (e^{log_e x})^i = x^i#

and

#int x^i dx = 1/2(1-i)x^{1+i} = x/2(1-i)x^i#

Substituting for #x^i# and taking the imaginary component we get

#int sin(log_e x) = x/2(sin(log_ex)-cos(log_ex))#