What is f(x) = int -cos^2x dxf(x)=∫−cos2xdx if f(pi/3) = 0 f(π3)=0?

1 Answer
Roy E.
Jan 31, 2017

f(x)=pi/6+sqrt(3)/8-1/2x-1/4sin2xf(x)=π6+√38−12x−14sin2x

Explanation:

int-cos^2x dx∫−cos2xdx
=-1/2int(1+cos2x)dx=−12∫(1+cos2x)dx
=-1/2x-1/4sin2x+c=−12x−14sin2x+c
But
f(pi/3)=0f(π3)=0
Therefore
0=-1/2(pi/3)-1/4sin(2pi/3)+c0=−12(π3)−14sin(2π3)+c
0=-pi/6-1/8sqrt(3)+c0=−π6−18√3+c
So
c=pi/6+sqrt(3)/8c=π6+√38

Notes:
cos2x=cos^2x-sin^2xcos2x=cos2x−sin2x
=cos^2x-(1-cos^2x)=2cos^2x-1=cos2x−(1−cos2x)=2cos2x−1
sin((2pi)/3)=sqrt(3)/2sin(2π3)=√32

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