What is f(x) = int x^2sinx^3 + 6cosx dx if f(pi/12)=-8 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Ananda Dasgupta Feb 26, 2018 f(x) = -1/3 cos((pi/12)^3)+6 sin (pi/12)-9.22 Explanation: f(x) = int x^2 sin(x^3) dx+ int 6 cos x dx = 1/3 int sin(x^3) times 3x^2 dx +6 sin x = 1/3 int sin(x^3) d(x^3) +6 sin x = -1/3 cos(x^3)+6 sin x +C We can find C from the given condition f(pi/12)=-8 -8 = f(pi/12) = -1/3 cos((pi/12)^3)+6 sin (pi/12) +C so C = 1/3 cos((pi/12)^3)-6 sin (pi/12)-8 ~~-9.22 and thus f(x) = -1/3 cos((pi/12)^3)+6 sin (pi/12)-9.22 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 1389 views around the world You can reuse this answer Creative Commons License