How do you find an equation for the function #f'(x)=sec^2(2x)# whose graph passes through the point (pi/2,2)?

1 Answer
Oct 30, 2016

I take this to be the challenge:

Given #f'(x) = sec^2(2x)# and #f(pi/2) = 2#, we are to find an explicit expression for #f(x)#.

From #f'(x) = sec^2(2x)#, we can find

#f(x) = int sec^2(2x) dx#

Integrate (antidifferentiate) by substitution to get

#f(x) = 1/2 tan(2x) +C#

Use the point we were given to find #C#.

#f(pi/2) = 1/2 tan(2(pi/2)) +C = 2# so

#1/2tan(pi)+C = 2#

Since #tan(pi)=0#, we get #C = 2#.

We finish with

#f(x) = 1/2 tan(2x) +2#