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 2170 views around the world You can reuse this answer Creative Commons License