How do you find an equation for the function #f'(x)=pisecpixtanpix# whose graph passes through the point (1/3,1)?

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Answer 1 · 2017-03-28T11:14:39

# f(x) = sec (pix) -1#

We have;

# f'(x) = pi \ sec pix \ tan pix #

Using #d/dx \ secx=secxtanx# we can integrate to get:

# f(x) = pi \ (sec pix) / pi + C#
# " " = sec pix + C#

We are given #f(1/3) = 1#

# :. sec (pi/3) + C = 1 #
# :. 2 + C = 1 #
# :. C = -1 #

Hence the equation of the function is:

# f(x) = sec (pix) -1#