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

1 Answer
Oct 24, 2015

Answer:

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))#