Question

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

Answer 1

`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`