What is #f(x) = int -cos^2x dx# if #f(pi/3) = -6 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Eddie Sep 1, 2016 #f(x)= - 1/2 ( x + 1/2sin 2 x) -6+sqrt(3)/8+pi/6# Explanation: #f(x) = int -cos^2x dx# # =- int cos^2x dx# use identity #cos 2A = cos^2 A - sin^2 A = 2 cos^2 A - 1# #implies - 1/2 int 1 + cos 2 x dx# #f(x)= - 1/2 ( x + 1/2sin 2 x) + C# #f(pi/3) = -6# #implies -6 = - 1/2 ( pi/3 + 1/2sin (2pi )/3) + C# #C = -6+sqrt(3)/8+pi/6 # #f(x)= - 1/2 ( x + 1/2sin 2 x) -6+sqrt(3)/8+pi/6# 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 1243 views around the world You can reuse this answer Creative Commons License