How do you use the Fundamental Theorem of Calculus to find the derivative of int sqrt(1+ sec(t)) dt from x to pi?

Jul 2, 2016

$= - \sqrt{1 + \sec \left(x\right)}$

Explanation:

you're interested in the second part of the Fundamental Theorem which states that

if $F \left(x\right) = {\int}_{a}^{x} \setminus f \left(t\right) \setminus \mathrm{dt}$

then $\frac{\mathrm{dF}}{\mathrm{dx}} = f \left(x\right)$

here you want

$\frac{d}{\mathrm{dx}} {\int}_{x}^{\pi} \setminus \sqrt{1 + \sec \left(t\right)} \setminus \mathrm{dt}$

to get that into the form required by the FTC pt 2, and using just the summation signs to save time, we observe that

${\int}_{x}^{\pi} = - {\int}_{\textcolor{red}{\pi}}^{\textcolor{red}{x}}$

so if we are to use the FTC part 2, we re-write it as

$\frac{d}{\mathrm{dx}} \left({\int}_{x}^{\pi} \setminus \sqrt{1 + \sec \left(t\right)} \setminus \mathrm{dt}\right)$

$\frac{d}{\mathrm{dx}} \left(- {\int}_{\textcolor{red}{\pi}}^{\textcolor{red}{x}} \setminus \sqrt{1 + \sec \left(t\right)} \setminus \mathrm{dt}\right)$

$- \frac{d}{\mathrm{dx}} \left({\int}_{\pi}^{x} \setminus \sqrt{1 + \sec \left(t\right)} \setminus \mathrm{dt}\right)$

$= - \sqrt{1 + \sec \left(x\right)}$