derivate $\displaystyle \int_{sin(x)}^{0} \frac{dt}{\sqrt{1-t^2}}$, $\displaystyle |x|< \frac{\pi}{2}$
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Originally Posted by peteryellow derivate $\displaystyle \int_{sin(x)}^{0} \frac{dt}{\sqrt{1-t^2}}$, $\displaystyle |x|< \frac{\pi}{2}$ $\displaystyle \int_{sin(x)}^{0} \frac{dt}{\sqrt{1-t^2}} = -\int_0^{sin(x)} \frac{dt}{\sqrt{1-t^2}}$ $\displaystyle \frac{d}{dx}\left[-\int_0^{sin(x)} \frac{dt}{\sqrt{1-t^2}} \right]= -\frac{1}{1-\sin^2{x}} \cdot \cos{x} = -\sec{x}$
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