# How do you calculate the derivative of int3(sin(t))^4 dt from [e^x,1]?

Jun 12, 2015

Use the Fundamental Theorem of Calculus, Part 1 (after rewriting) and use the chain rule.

#### Explanation:

g(x) = ${\int}_{{e}^{x}}^{1} 3 {\sin}^{4} t \mathrm{dt}$

To use the fundamental theorem in one form, we must have the constant as the lower limit of integration, so we rewrite:

g(x) = $- {\int}_{1}^{{e}^{x}} 3 {\sin}^{4} t \mathrm{dt}$

With $u = {e}^{x}$, we have: $g \left(x\right) = h \left(u\right) = - {\int}_{1}^{u} 3 {\sin}^{4} t \mathrm{dt}$

Use the chain rule to find $g ' \left(x\right) = h ' \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}}$.

By FTC 1, $h ' \left(u\right) = 3 {\sin}^{4} u$

We also have $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {x}^{e}$.

So,

$g ' \left(x\right) = 3 {\sin}^{4} u \frac{\mathrm{du}}{\mathrm{dx}} = 3 {\sin}^{4} \left({e}^{x}\right) \cdot {e}^{x}$