**Problem**: (2003 Mu Alpha Theta National Convention) Evaluate $ \displaystyle \int \cos{(\ln{x})} dx $.

**Solution**: Consider the substitution $ u = \ln{x} $. We than have $ du = \frac{1}{x}dx $ so $ dx = xdu = e^udu $. It remains to evaluate the integral

$ \displaystyle \int e^u(\cos{u}) du $.

We can evaluate this by parts (set $ v = \cos{u} $, $ dw = e^udu $); we find that

$ \displaystyle \int e^u(\cos{u}) du = e^u\cos{u}+\int e^u(\sin{u})du $.

Now evaluating $ \displaystyle \int e^u(\sin{u})du $ by parts as well ($ v = \sin{x} $, $ dw = e^udu $), we have

$ \displaystyle \int e^u(\sin{u})du = e^u\sin{u}-\int e^u(\cos{u})du $,

so

$ \displaystyle \int e^u(\cos{u})du = e^u\cos{u}+e^u\sin{u}-\int e^u(\cos{u})du $

and

$ \displaystyle \int e^u(\cos{u})du = \frac{e^u\cos{u}+e^u\sin{u}}{2} $.

Substituting $ u = \ln{x} $ back, we have

$ \displaystyle \int \cos{(\ln{x})}dx = \frac{x\cos{(\ln{x})}+x\sin{(\ln{x})}}{2} $. QED.

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Comment: At first glance, the integral doesn't suggest any direct way to proceed. After the substitution, we have a more familiar integral that we know how to evaluate by parts.

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Practice Problem: (2004 Mu Alpha Theta National Convention) Evaluate $ \displaystyle \int_0^1 \int_y^1 e^{-x^2}dx dy $.

intgrate with respect to y first ( don't forget to change the bounds), then the problem will become trivial.

ReplyDeleteYou actually don't even need to do a substitution: you can plunge right into by parts. Try u = cos ( ln x) and dv = dx, and you'll end up with:

ReplyDeleteS cos (ln x) dx = - x cos (ln x) + S sin (ln x) dx

and then you do by parts on the remaining integral as before and get your solution.

The second integral is equivalent to

S (0,1) S (0,x) e^(-x^2) dy dx

and when you perform the inner integration (which is trivial), you see that the second integration is straightforward via substitution.