# What was done in this equation involving the fundamental theorem of calculus?

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

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