What is f(x) = int -cos^2x dx if f(pi/3) = 0 ? Calculus Techniques of Integration Evaluating the Constant of Integration 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 Answer link Related questions How do you find the constant of integration for intf'(x)dx if f(2)=1? What is a line integral? What is f(x) = int x^3-x if f(2)=4 ? What is f(x) = int x^2+x-3 if f(2)=3 ? What is f(x) = int xe^x if f(2)=3 ? What is f(x) = int x - 3 if f(2)=3 ? What is f(x) = int x^2 - 3x if f(2)=1 ? What is f(x) = int 1/x if f(2)=1 ? What is f(x) = int 1/(x+3) if f(2)=1 ? What is f(x) = int 1/(x^2+3) if f(2)=1 ? See all questions in Evaluating the Constant of Integration Impact of this question 2235 views around the world You can reuse this answer Creative Commons License