How do you use the Fundamental Theorem of Calculus to find the derivative of int 6*(sin(t))^2 dt from 1 to e^x?

Jul 15, 2016

$= 6 {e}^{x} {\left(\sin \left({e}^{x}\right)\right)}^{2}$

Explanation:

we have $\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{{e}^{x}} \setminus 6 \cdot {\left(\sin \left(t\right)\right)}^{2} \setminus \mathrm{dt}$

STEP 1: FTC pt2 states that

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

STEP 2: from the chain rule $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

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

here that means that

$\frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{{e}^{x}} \setminus 6 \cdot {\left(\sin \left(t\right)\right)}^{2} \setminus \mathrm{dt} = 6 \cdot {\left(\sin \left({e}^{x}\right)\right)}^{2} \frac{d}{\mathrm{dx}} {e}^{x}$

$= 6 {e}^{x} {\left(\sin \left({e}^{x}\right)\right)}^{2}$