What is #f(x) = int x-sin2x-6cosx dx# if #f(pi/2)=3 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer sjc Dec 22, 2016 #f(x)=x^2/2+1/2cos2x-6sinx+7/2-pi^2/8# Explanation: #f(x)=int(x-sin2x-6cosx)dx##" "f(pi/2)=3# integrate term by term #f(x)=x^2/2+1/2cos2x-6sinx+C# #f(pi/2)=1/2xx(pi/2)^2+1/2cos(2xxpi/2)-6sin(2xxpi/2)+C=3# #pi^2/8+1/2cancel(cospi)^-1- 6cancel(sinpi)^0+C=3# #pi^2/8-1/2+C=3# #C=3+1/2-pi^2/8=7/2-pi^2/8# #f(x)=x^2/2+1/2cos2x-6sinx+7/2-pi^2/8# 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 1468 views around the world You can reuse this answer Creative Commons License