When evaluating $\int x \sin x \, dx$ explain how letting $ u =\sin x $ and $dv = x dx$ makes the solution more difficult to find. Integration by parts. The solution doesn't seem any more difficult for me. I'm just curious on how someone would answer this.
As per Ross Millikan's request, I'll provide an answer:

From integration by parts, we have $\displaystyle\int u \, \mathrm dv = uv-\displaystyle\int v \, \mathrm du$. The entire point of this is to go from a more complicated integral to an easier one. If $\displaystyle\int v \, \mathrm du$ is simpler than $\displaystyle\int u \, \mathrm dv$, you've made the "right" choice for $u$ and $\mathrm dv$.

In this example, you have $\displaystyle\int x\sin(x) \, \mathrm dx$. If you choose $u = x$ and $\mathrm dv = \sin(x) \, \mathrm dx$, you'll get $\mathrm du = \mathrm dx$ and $v = -\cos(x)$. This gives:

$$\int x\sin(x) \, \mathrm dx = -x\cos(x)-\color{blue}{\int-\cos(x) \, \mathrm dx}$$

Notice how $\color{blue}{\displaystyle\int v \, \mathrm du}$ is simple to integrate: it's just $-\sin(x)+C$. Notice that this is because choosing $x$ as $u$ (which is differentiated) will result in the exponent going down, which makes things simpler.

However, what if you pick $u = \sin(x)$ and $\mathrm dv = x \, \mathrm dx$? Then, you'll have $\mathrm du = \cos(x) \, \mathrm dx$ and $v = \dfrac{1}{2}x^2$. This gives:

$$\int x\sin(x) \, \mathrm dx = \frac{1}{2}x^2\sin(x)-\color{blue}{\int\frac{1}{2}x^2\cos(x) \, \mathrm dx}$$

How are you going to integrate $\color{blue}{\displaystyle\int v \, \mathrm du}$ now? By choosing $x \, \mathrm dx$ as $\mathrm dv$ (which is integrated), the exponent goes up. This means you've only made the problem harder. Going back in reverse would just waste two steps, so you might as well have just make the "right" choice to start with.

The same idea applies to other integration by parts problems as well. As it's been pointed out in the comments, the LIATE rule is usually used as the rule of thumb, but once you get used to these problems, you don't really have to rely on mnemonics anymore.
$$ \int x \sin x \, dx = -\int x \,  d\cos x =- x\cos x + \int \cos x \,  dx = \sin x - x\cos x + \textrm{const.}$$

The reason why the other way is more "difficult," is that it does not lead to the solution, but rather, a more compicated expression, as KM101 pointed out in the comments.