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All I know is that it uses the fundamental theorem of calculus.

$$\large\frac{d}{dx}\int_{x^2}^{\sin x} e^{xt^2}dt = e^{x\;\sin^2 x}\cos x - e^{x^5}2x+\int_{x^2}^{\sin x} t^2e^{xt^2}dt$$

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Zev Chonoles, thanks for the edit. I have to learn how to write those formulas. –  Dokkat Jan 16 '12 at 9:51

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This is more an application of differentiation under the integral sign, which is a generalization of the fundamental theorem (and can be proved using it). $$ \frac{d}{dx}\,\int_{a(x)}^{b(x)}f(x,t)\,dt = f(x,b(x))\,b'(x) - f(x,a(x))\,a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}\, f(x,t)\; dt $$

In this case, $a(x) = x^2$, $b(x) = \sin(x)$, and $f(x, t) = e^{x t^2}$. $$ a'(x) = 2x $$ $$ b'(x) = \cos(x) $$ $$ \frac{\partial}{\partial x} f(x, t) = t^2 e^{x t^2} $$ So $$ \large\frac{d}{dx}\int_{x^2}^{\sin x} e^{xt^2}dt = e^{x\;\sin^2 x}\cos x - 2xe^{x^5}+\int_{x^2}^{\sin x} t^2e^{xt^2}dt $$

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