# How do you use the second fundamental theorem of Calculus to find the derivative of given int ((sin^3)(t))dt from [0, e^x]?

Jul 9, 2016

${e}^{x} {\sin}^{3} {e}^{x}$

#### Explanation:

you want

$\frac{d}{\mathrm{dx}} {\int}_{0}^{{e}^{x}} {\sin}^{3} \left(t\right) \setminus \mathrm{dt}$

by the FTC we know that

$\frac{d}{\mathrm{du}} {\int}_{a}^{u} f \left(t\right) \setminus \mathrm{dt} = f \left(u\right)$

and if $u = u \left(x\right)$ we also know that

$\frac{d}{\mathrm{dx}} {\int}_{a}^{u} f \left(t\right) \setminus \mathrm{dt} = f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

we apply that here with $f \left(t\right) = {\sin}^{3} t$, $a = 0$ and $u = {e}^{x}$ to get

${\sin}^{3} {e}^{x} \cdot {e}^{x}$