What is #f(x) = int -cos^2x dx# if #f(pi/3) = 0 #?

1 Answer
Roy E.
Jan 31, 2017

#f(x)=pi/6+sqrt(3)/8-1/2x-1/4sin2x#

Explanation:

#int-cos^2x dx#
#=-1/2int(1+cos2x)dx#
#=-1/2x-1/4sin2x+c#
But
#f(pi/3)=0#
Therefore
#0=-1/2(pi/3)-1/4sin(2pi/3)+c#
#0=-pi/6-1/8sqrt(3)+c#
So
#c=pi/6+sqrt(3)/8#

Notes:
#cos2x=cos^2x-sin^2x#
#=cos^2x-(1-cos^2x)=2cos^2x-1#
#sin((2pi)/3)=sqrt(3)/2#

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