# How do you use the second fundamental theorem of Calculus to find the derivative of given int (1)/(2+sin(t)) from [0, lnx]?

Nov 7, 2016

$\frac{d}{\mathrm{dx}} {\int}_{0}^{\ln x} \frac{1}{2 + \sin t} \mathrm{dt} = \frac{1}{x \left(2 + \sin \left(\ln x\right)\right)}$

#### Explanation:

The second fundamental theorem of calculus states that if F is any anti-derivative of f, then

${\int}_{a}^{b} f \left(t\right) \mathrm{dt} = F \left(b\right) - F \left(a\right)$, and $F ' \left(x\right) = f \left(x\right)$

Note that as this is a definite integral, the RHS is a number.

If $f \left(x\right) = \frac{1}{2 + \sin t}$ and $F ' \left(x\right) = f \left(x\right)$ Then

${\int}_{0}^{\ln x} \frac{1}{2 + \sin t} \mathrm{dt} = F \left(\ln x\right) - F \left(0\right)$

If we differentiate wrt $x$ we get

$\frac{d}{\mathrm{dx}} {\int}_{0}^{\ln x} \frac{1}{2 + \sin t} \mathrm{dt} = \frac{d}{\mathrm{dx}} \left(F \left(\ln x\right) - F \left(0\right)\right)$
$\therefore \frac{d}{\mathrm{dx}} {\int}_{0}^{\ln x} \frac{1}{2 + \sin t} \mathrm{dt} = \frac{d}{\mathrm{dx}} F \left(\ln x\right)$ (as $F \left(0\right)$ is a constant)
$\therefore \frac{d}{\mathrm{dx}} {\int}_{0}^{\ln x} \frac{1}{2 + \sin t} \mathrm{dt} = \frac{d}{\mathrm{dx}} \left(\ln x\right) F ' \left(\ln x\right)$ (using the chain rule)
$\therefore \frac{d}{\mathrm{dx}} {\int}_{0}^{\ln x} \frac{1}{2 + \sin t} \mathrm{dt} = \frac{1}{x} \cdot \frac{1}{2 + \sin \left(\ln x\right)}$
$\therefore \frac{d}{\mathrm{dx}} {\int}_{0}^{\ln x} \frac{1}{2 + \sin t} \mathrm{dt} = \frac{1}{x \left(2 + \sin \left(\ln x\right)\right)}$