Proof that $ \frac{3\pi}{8}< \int_{0}^{\pi/2} \cos{\sin{x}} dx < \frac{49\pi}{128}$

Question

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?

Answer 1

Hint for one part, using this inequality and this one $$\cos{x} \geq 1 - \frac{x^2}{2}$$ we have $$\int\limits_{0}^{\frac{\pi}{2}}\cos{\sin{x}} dx > \int\limits_{0}^{\frac{\pi}{2}} \left(1-\frac{\sin^2{x}}{2}\right)dx=\frac{3 \pi}{8}$$

Answer 2

Follow-up to @rtybase:

$$ \cos x\le 1-\frac{x^2}2+\frac{x^4}{24}\implies \int_0^{\pi/2}\cos\sin xdx<\int_0^{\pi/2}\left(1-\frac{\sin^2 x}2+\frac{\sin^4 x}{24}\right)dx=\frac{49\pi}{128}. $$