Question

How do you find the derivative of `F(x) = int sqrt(1+sec(3t)) dt`?

Answer 1

Without knowing the limits of integration all we can say if the following.

Explanation:

For `F(x) = int_(f(x))^g(x) sqrt(1+sec(3t)) dt`

The Fundamental Theorem of Calculus gives us

`F'(x) = sqrt(1+sec(3g(x))) g'(x) - sqrt(1+sec(3f(x))) f'(x)`

So

Example 1
For `F(x) = int_(-pi/8)^x sqrt(1+sec(3t)) dt`

The Fundamental Theorem of Calculus gives us

`F'(x) = sqrt(1+sec(3x)`

Example 2

For `F(x) = int_0^x^2 sqrt(1+sec(3t)) dt`

The Fundamental Theorem of Calculus gives us

`F'(x) = 2xsqrt(1+sec(3x^2))`

Example 3

For `F(x) = int_x^x^2 sqrt(1+sec(3t)) dt`

The Fundamental Theorem of Calculus gives us

`F'(x) = 2xsqrt(1+sec(3x^2)) - sqrt(1+sec(3x))`