# How do you use part 1 of Fundamental Theorem of Calculus to find the derivative of the function y=int (6+v^2)^10 dv from sinx to cosx?

##### 1 Answer
Apr 20, 2015

$\left[- {\left(6 + {\cos}^{2} x\right)}^{10} \sin x - {\left(6 + {\sin}^{2} x\right)}^{10} \cos x\right]$

If it is F(v) = ${\int}_{\sin} {x}^{\cos} x \left(6 + {v}^{2}\right)$dv, it can be written as a sum of two integrals= ${\int}_{\sin} {x}^{0} \left(6 + {v}^{2}\right) \mathrm{dv}$ +${\int}_{0}^{\cos} x \left(6 + {v}^{2}\right) \mathrm{dv}$

=${\int}_{0}^{\cos} x \left(6 + {v}^{2}\right) \mathrm{dv} - {\int}_{0}^{\sin} x \left(6 + {v}^{2}\right) \mathrm{dv}$

F'(v) =$\frac{d}{\mathrm{dx}} {\int}_{0}^{\cos} x \left(6 + {v}^{2}\right) \mathrm{dv} - \frac{d}{\mathrm{dx}} {\int}_{o}^{\sin} x \left(6 + {v}^{2}\right) \mathrm{dv}$

Now applying the Fundamental Theorem of Calculus, the derivative would be $\left[- {\left(6 + {\cos}^{2} x\right)}^{10} \sin x - {\left(6 + {\sin}^{2} x\right)}^{10} \cos x\right]$