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

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

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

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To find the derivative of the integral (\int_{0}^{\ln(x)}\frac{1}{2+\sin(t)} , dt) with respect to (x) using the Second Fundamental Theorem of Calculus, you first evaluate the integrand at the upper limit of integration, (\ln(x)), and then multiply by the derivative of the upper limit, which is (1/x). Then subtract from that the value of the integrand evaluated at the lower limit of integration, (0). This gives you the derivative of the integral with respect to (x). So the derivative is:

[\frac{d}{dx} \left( \int_{0}^{\ln(x)}\frac{1}{2+\sin(t)} , dt \right) = \frac{1}{2+\sin(\ln(x))} \cdot \frac{1}{x} - \frac{1}{2+\sin(0)} \cdot 0]

Simplify as needed.

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To find the derivative of the integral ( \int_{0}^{\ln(x)} \frac{1}{2+\sin(t)} , dt ) using the second fundamental theorem of calculus, you first need to express the integral in terms of a function of ( x ) using the Chain Rule. Then, you can differentiate the integral with respect to ( x ) to find its derivative.

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