Expert: Ahmed Salami - 12/5/2005

Question
Hi, do you know how to integrate sin(ln(x))dx between the limits 1 and e using integration by parts? Thanks for your time, Adam.

Answer
Hi Adam,
I hope you're already familiar with the topic because my explanation would require that.
Integrate as u = sin(lnx) and dv = 1
we get du = cos(lnx). 1/x  and v = x
Therefore,
Integral sin(lnx) dx = xsin(lnx) - integral cos(lnx)dx
integral cos(lnx) is similar to our initial problem and we proceed in the previous manner to get
integral cos(lnx) = xcos(lnx) + integral sin(lnx) dx
Now,
Integral sin(lnx) dx
= xsin(lnx) - xcos(lnx)- Integral sin(lnx) dx
We now have something similar on both sides. Now, let's call integral sin(lnx)dx, say I. Then we have
I = xsin(lnx) - xcos(lnx)- I
2I = xsin(lnx) - xcos(lnx)
I = [xsin(lnx) - xcos(lnx)] / 2
 = (x/2)[sin(lnx) - cos(lnx)]
inserting our limits gives us
I = (e/2)[sin 1 - cos 1] - (1/2)[sin 0 - cos 0]
noting that ln e = 1 and ln 1 = 0
Simplify the rest and you'll be fine.
I hope you get it, you can always get back to me.
Regards.