Sandwich it, May be

Relevant wiki: Jensen's Inequality

Assume $f(r) = \displaystyle\int_{0}^{\frac{\pi}{2}} x^{r} \sin x \text{ d}x$

Using integration by parts, $\begin{aligned} \displaystyle\int_{0}^{\frac{\pi}{2}} x^{r} \cos x \text{ d}x &= {\left.\frac{x^{r+1} \cos x}{r+1}\right\vert_{0}^{\frac{\pi}{2}}} + \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{x^{r+1} \sin x}{r+1} \text{ d}x\\\displaystyle\int_{0}^{\frac{\pi}{2}} x^{r} \cos x \text{ d}x &=\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{x^{r+1} \sin x}{r+1}\text{ d}x = \frac{f(r+1)}{r+1} \end{aligned}$

Thus we have,

$\text{L}=\lim_{r \rightarrow \infty} \frac{r^c \displaystyle\int_{0}^{\frac{\pi}{2}} x^r \sin (x) \space \text{d}x}{\displaystyle\int_{0}^{\frac{\pi}{2}} x^r \cos (x) \space \text{d}x} = \lim_{r \rightarrow \infty} \large{\frac{r^c (r+1) f(r)}{f(r+1)}}$

Now,

$\begin{aligned} f(r)&<\displaystyle\int_{0}^{\frac{\pi}{2}} x^r \text{ d}x \\ f(r) &< \frac{\left(\frac{\pi}{2}\right)^{r+1}}{r+1} \\ \color{#3D99F6}{\text{also},} \\ f(r) &> \displaystyle\int_{0}^{\frac{\pi}{2}} \frac{2}{\pi}x^{r+1} \text{ d}x \quad \quad \quad \small\color{#3D99F6}{\because \sin x > \frac{2x}{\pi}, \quad \forall x\in\left(0,\frac{\pi}{2}\right)} \\f(r) &>\frac{\left(\frac{\pi}{2}\right)^{r+1}}{r+2} \\\\ \Rightarrow \frac{r}{r+2} < r \left(\frac{2}{\pi}\right)^{r+1} f(r) <\frac{r}{r+1} \end{aligned}$

Consider,

$\lim_{r \rightarrow \infty} \dfrac{f(r)}{f(r+1)} = \lim_{r \rightarrow \infty} \dfrac{r\space \left(\dfrac{2}{\pi}\right)^{r+1} f(r)}{(r+1)\left(\dfrac{2}{\pi}\right)^{r+2} f(r+1)}\cdot \dfrac{2(r+1)}{\pi r} = \dfrac{2}{\pi}$

$\begin{aligned} \text{L} &= \lim_{r \rightarrow \infty} r^{c} \space (r+1) \cdot \dfrac{f(r)}{f(r+1)} \\ &= \lim_{r \rightarrow \infty} \quad r^{c} \space (r+1) \cdot \dfrac{2}{\pi}\end{aligned}$

Since $\text{L}>0 \Rightarrow c$ needs to be $-1$ .

Thus, $\text{L}= \dfrac{2}{\pi}$

$\boxed{\large\pi \text{L} -c=\pi \cdot \dfrac{2}{\pi} -(-1)=3}$