This is not a typo and it’s hard, so great question!!
First off, most functions that you write down randomly don’t have an anti-derivative that can be expressed as a finite combination of elementary functions. For example, Wolfram Alpha says
\int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\text{erf}(x).
Which is great and all but you’re probably wonder what that \text{erf} thing is. Well, I just typed erf into Google and the first hit is the Wikipedia page for the Error function, which defines \text{erf} to be
\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} \, dt.
That is, \text{erf} is literally defined to be the anti-derivative of the function that we started with!
Now, whatever you typed in into whatever software you typed it into expresses your integral in terms of a special function called the Gamma function; that’s \Gamma - the upside down L that you see. It too is a special function. It might relate your integral to some other things that \Gamma is related to but, really, it’s probably pretty worthless for our purposes.
But, keep in mind that your problem is a definite integral - not an indefinite integral. Finding antiderivatives to apply the fundamental theorem of calculus is one way to evaluate definite integrals but it’s not the only way. For many integrals, it’s not even the best way.
Here are a couple of approaches to this problem:
Approach 1
First (and as discussed in class), let’s use symmetry. The integrand f(x) = x\cos(x^{12}) is an odd function. That is, its graph is symmetric about the the origin and, like any odd function, the value of a definite integral over an interval that’s symmetric about the orgin is zero. You can see why by looking at a graph of the function, which looks something like so:
Thus, the green area on top cancels with the red area on the bottom and we get
\int_{-\pi}^{\pi} x\,\cos(x^{12}) \, dx = 0.
Approach 2
Well, the problem did recommend u-substitution so let’s see what we can do with it. I guess that we’ve got an x^{12} nested inside a cosine function, which suggests that we let u=x^{12}. If we do so, then we could express x^{11}dx as \frac{1}{12}du. We’ve got only a x^1 outside but we could write
x\,dx = \frac{1}{x^{10}} \, x^{11} dx = \frac{1}{12} \frac{1}{u^{10/12}} du.
Thus, we get
\begin{aligned}
\int_{-\pi}^{\pi} x \cos(x^{12}) \, dx &= \int_{-\pi}^{\pi} \frac{1}{x^{10}} \cos(x^{12}) \, x^{11} dx \\
&= \frac{1}{12}\int_{\pi^{12}}^{\pi^{12}} \frac{1}{u^{10/12}} \cos(u) \, du = 0.
\end{aligned}
Note that the value is again zero, since we’re integrating from \pi^{12} to \pi^{12}; any time the bounds of integration are the same, you’re guaranteed that the value of the integral is zero.
Finally, are a couple of key issues that I’ve really tried to drive home these first couple of weeks:
- Quite generally, you need to be able to think geometrically and picture what you’re trying to do. That’s what math is really all about.
- More specifically (for this problem), changing bounds of integration is worthwhile and even essential in some problems.
