How do you integrate #int cos(lne(x)) # using integration by parts?

1 Answer
May 19, 2018

#intcos(ln(x))dx=x(cos(ln(x))-sin(ln(x)))/2+C#, #C in RR#

Explanation:

#intcos(ln(x))dx#
let : #X=ln(x)#
#x=e^X#
#dx=e^XdX#
so :
#intcos(ln(x))dx=intcos(X)e^XdX#
Using integration by parts :
#intuv'dX=uv-intu'vdX#
there :
#u=e^X# #v'=cos(X)#

#u'=e^X# #v=sin(X)#

So : #intcos(X)e^XdX=(-sin(X)e^X)-int(sin(X)e^X)dx#

#=-sin(X)e^X-intsin(X)e^XdX#

Using integration by parts again :

#intsin(X)e^XdX=-cos(X)e^X+int(cos(X)e^XdX#

So :

#intcos(X)e^XdX=-sin(X)e^X+cos(X)e^X-int(cos(X)e^XdX#

#2intcos(X)e^XdX=e^(X)(cos(X)-sin(X))#

#intcos(X)e^XdX=(e^(X)(cos(X)-sin(X)))/2+C#, #C in RR#

#intcos(ln(x))dx=cancel(e^(ln(x)))^(=x)(cos(ln(x))-sin(ln(x)))/2+C#, #C in RR#

#intcos(ln(x))dx=x*(cos(ln(x))-sin(ln(x)))/2+C#, #C in RR#

\0/ here's our answer !