<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] Follow-up to @rtybase:
$$
\cos x\le 1-\frac{x^2}2+\frac{x^4}{24}\implies
\int_0^{\pi/2}\cos\sin xdx<\int_0^{\pi/2}\left(1-\frac{\sin^2 x}2+\frac{\sin^4 x}{24}\right)dx=\frac{49\pi}{128}.
$$
</s>