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If $$g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t)dt$$ then
$$g'(\frac{\pi}{2})=-2\pi$$
$$g'(-\frac{\pi}{2})=2\pi$$
$$g'(\frac{\pi}{2})=2\pi$$
$$g'(-\frac{\pi}{2})=-2\pi$$
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