How do I evaluate #int3 sin(ln(x))dx#?

1 Answer
Massimiliano
Jan 27, 2015

The answer is #I=(3/4)x[sin(ln(x))-cos(ln(x))]+c#

This integral has to be done two times by parts because it is a cyclic integral. A cyclic integral is an integral of a cyclic function, like #sinf(x)# or #cosf(x)#.

The theorem of integration by parts says:

#intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx#

In this case, it seems that there is only one function, but it is not true because we can assume that #f(x)=sin(ln(x))# and #g'(x)dx=1dx#.

For the first step it is necessary to find #g(x)=intg'(x)dx=int1dx=x#, and it is necessary to find #f'(x)=cos(ln(x))1/x#.

So:

#I=int 3sin(ln(x))dx=3intsin(ln(x))dx=3[xsin(ln(x))-intcos(ln(x))(1/x)xdx]=3[xsin(ln(x))-intcos(ln(x))dx]=#

Now it is necessary to make another time the integration by parts:

#3[xsin(ln(x))-(xcos(ln(x))-int-sin(ln(x))(1/x)xdx]=3[xsin(ln(x))-xcos(ln(x))-intsin(ln(x))dx]#

So:

#I=3xsin(ln(x))-3xcos(ln(x))-3intsin(ln(x))dx=3xsin(ln(x))-3xcos(ln(x))-I#

because the last integral is identical to the first.

So:

#2I=3xsin(ln(x))-3xcos(ln(x))rArr#
#I=(3/2)x[sin(ln(x))-cos(ln(x))]+c#

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