Definite Integral of $\sin(\sin x)$

Question

I recently came across this problem:

Prove that $\int_{0}^{\frac{\pi}{2}} \sin(\sin y) dy \leq 1$.

However, from my understanding, the expression $\sin(\sin y)$ does not have an indefinite integral (How to find the indefinite integral of sin(sin(x))dx?). So, how will we go about solving the above problem using just the Riemann sums?

Note that you're not being asked to evaluate the integral, only to bound it — MPW, Apr 25 '20 at 12:18
That is true. So, the next step should be to compare it with a function that is known? — throwawayAccountForRandomStuff, Apr 25 '20 at 12:21
Wolfy says that the integral is $ (π H_0(1))/2≈0.893244$ so it is fairly close to 1. — marty cohen, Apr 25 '20 at 13:07

score 2 · Accepted Answer · answered Apr 25 '20 at 12:11

2

Define $f(x) = \sin(\sin x)$. We know that $0 \le \sin x \le x$. As $\sin x$ is increasing on $[0, \pi/2]$, $0 \le f(x) \le \sin x$ on this interval.

Taking the integral, you're done.

answered Apr 25 '20 at 12:11

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Oh, that's pretty straightforward from here. Thank you. But do you think there might be a way of going about it without integrating $sinx$? – throwawayAccountForRandomStuff Apr 26 '20 at 09:06
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You can check my answer here without integrating $sinx$ :) – Tortar Apr 30 '20 at 08:41
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Thank you so much! That's a pretty smart way of going about it. – throwawayAccountForRandomStuff Apr 30 '20 at 15:41

Definite Integral of $\sin(\sin x)$

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