Calculate the Derivative.

Question

d/dt 83∫t (sec(29x+3)) dx=

 

On what interval is the derivative defined?

Answer 1

Take the integral of sec(29x + 4) from x=t to x=83.

 

∫sec(29x + 4) dx

 

Let          u = 29x + 4

             du = 29 dx

           (1/29)du = dx

 

(1/29)∫sec(u) du =

 

(1/29)ln(sec(u) + tan(u))

 

(1/29)ln(sec(29x + 4) + tan(29x + 4))

 

 

Now plug in the bounds to get the integral in terms of t.  Then, evaluate the derivative with respect to t.

 

The derivative is defined under these conditions:

 

1) denominator terms cannot be zero

2) -1 < sinx < 1

3) -1 < cosx < 1

4) -1 < tanx < 1