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 1531 views around the world You can reuse this answer Creative Commons License