# How do you find the derivative of y= int 6*(sin(t))^2 dt from a=e^x to b=1?

Jan 10, 2016

Use second fundamental theorem of calculus to find the answer. The working is given below.

#### Explanation:

$y = {\int}_{{e}^{x}}^{1} 6 \cdot \left({\sin}^{2} \left(t\right)\right) \mathrm{dt}$
$y = - {\int}_{1}^{{e}^{x}} 6 \cdot \left({\sin}^{2} \left(t\right)\right) \mathrm{dt}$

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

Using second fundamental theorem of calculus we replace $t$ by ${e}^{x}$ and multiply the result by derivative of ${e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(6 {\sin}^{2} \left({e}^{x}\right)\right) \left({e}^{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 6 {e}^{x} {\sin}^{2} \left({e}^{x}\right)$ Answer