What is f(x) = int sec^2x- cos^2x dx if f((5pi)/4) = 0 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Monzur R. Feb 26, 2017 f(x)=1/8(2sin2x-4+5pi-10)+tanx Explanation: f(x)=intsec^2x-cos^2x dx, f(5/4pi)=0 f(x)=intsec^2x-cos^2x dx=intsec^2x dx-intcos^2x dx= tanx+1/4(sin2x-2x)+"C" f(5/4pi)=0=tan(5/4pi)+1/4(sin(5/2pi)-5/2pi)+"C"= 1+1/4-5/8pi+"C"=0 "C"=(5pi-10)/8 f(x)=tanx+1/4(sin2x-2x)+(5pi-10)/8= 1/8(2sin2x-4+5pi-10)+tanx 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 2000 views around the world You can reuse this answer Creative Commons License