## Friday, March 24, 2006

### Integrate It. Topic: Calculus.

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.

--------------------

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.

--------------------

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

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

2. You 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:

S 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.