Since it looks like you've found a solution ... assuming ##a\leq b## gives $$\left|\int_a^b\sin x\ \mathrm{d}x\right|\leq\int_a^b|\sin x|\ \mathrm{d}x\leq\int_a^b1\ \mathrm{d}x$$.
If ##a>b##, you just need to flip the limits on the last two integrals. The desired inequality isn't too incredibly difficult to get from there.
I like the MVT proof better, though. It's, like, one step.
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