<s>[INST] Proof that $ \frac{3\pi}{8}< \int_{0}^{\pi/2} \cos{\sin{x}} dx < \frac{49\pi}{128}$ Proof that
$$
\frac{3\pi}{8} <
\int_{0}^{\pi/2} \cos\left(\sin\left(x\right)\right)\,\mathrm{d}x <
\frac{49\pi}{128}
$$
Can somebody give me some instruction how to deal with inequality like that? My current idea is:
I see $\frac{3\pi}{8}$ on the left. So I think that I can prove that
$$ \frac{3}{4}<\cos{\sin{x}} $$
And after take integral:
$$ \frac{3x}{4} \rightarrow \frac{3}{4} \cdot \frac{\pi}{2} = \frac{3\pi}{8} $$
But it is not true because
$$ \cos{\sin{x}} \geqslant \cos{1} \approx 0.5403 < 3/4$$
What have I do in such situation?
 [/INST] Hint for one part, using this inequality and this one
$$\cos{x} \geq 1 - \frac{x^2}{2}$$
we have
$$\int\limits_{0}^{\frac{\pi}{2}}\cos{\sin{x}} dx > \int\limits_{0}^{\frac{\pi}{2}} \left(1-\frac{\sin^2{x}}{2}\right)dx=\frac{3 \pi}{8}$$
</s>