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 1326 views around the world You can reuse this answer Creative Commons License