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 1903 views around the world You can reuse this answer Creative Commons License