Now that the weekend is over, here is my solution:
Using \(\displaystyle \sin x \leq 1\), we get
\(\displaystyle \int_0^{\pi/2} x^r \sin x \, dx \leq \frac{1}{r+1} ( \pi/2 )^{r+1}.\)
Since \(\displaystyle \sin\) is concave down, the secant line is a lower bound, i.e. \(\displaystyle \frac{2}{\pi}x \leq \sin x\) on the interval \(\displaystyle x \in [0,\pi/2]\), which gives us
\(\displaystyle \frac{1}{r+2} ( \pi/2 )^{r+1} \leq \int_0^{\pi/2} x^r \sin x \, dx.\)
That tells us the numerator, asymptotically, which is very promising.
If we try something similar with the denominator, we quickly come to the conclusion that 1 is not a very good upper bound for \(\displaystyle \cos\), so we might try
\(\displaystyle 1 - \frac{2}{\pi}x \leq \cos x \leq 1 - \left( \frac{2}{\pi}x \right)^2,\)
which tells us that
\(\displaystyle \left( \frac{1}{r+1} - \frac{1}{r+2} \right) ( \pi/2 )^{r+1} \leq \int_0^{\pi/2} x^r \cos x \, dx \leq \left( \frac{1}{r+1} - \frac{1}{r+3} \right) ( \pi/2 )^{r+1}.\)
Unfortunately, this isn't good enough. It tells us that we need \(\displaystyle c=-1\), but about \(\displaystyle L\) it only says that \(\displaystyle 1 \leq L \leq 2\). So we think about that integral and ask where does the weight of the integral come from? When \(\displaystyle 0 < x < 1\), the integrand drops to 0. When \(\displaystyle x = \pi/2\), the integrand is always zero, but in \(\displaystyle 1 \leq x < \pi/2\), the cosine is positive and the \(\displaystyle x^r\) term grows to infinity. So all of the weight is coming from just left of the right edge. So we need our bounds on cosine to be accurate at \(\displaystyle x = \pi/2\). So we use
\(\displaystyle \cos x = \sin (\pi/2 - x) \leq \pi/2 - x\)
and
\(\displaystyle (\pi/2 - x) - (\pi/2 - x)^2 \leq \sin (\pi/2 - x) = \cos x.\)
Using these bounds gives us \(\displaystyle c=-1\) and \(\displaystyle L = \pi/2.\)
