What is #f(x) = int xsin2x-6cotx dx# if #f(pi/4)=1 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Cem Sentin Feb 9, 2018 #f(x)=-x/2*cos2x+1/4*sin2x-6ln(sinx)+1+(pisqrt2)/16-3Ln2# Explanation: Due to #int xsin2x*dx=x*(-1/2*cos2x)-int (-1/2)*cos2x*dx# =#-x/2*cos2x+1/2intcos2x*dx# =#-x/2*cos2x+1/4sin2x+C# #f(x)=int (xsin2x-6cotx)*dx=-x/2*cos2x+1/4*sin2x-6ln(sinx)+C# After imposing #f(pi/4)=1# condition, #-pi/8*sin(pi/4)-6Ln(sin(pi/4))+C=1# #-(pisqrt2)/16-6Ln(sqrt2/2)+C=1# #-(pisqrt2)/16+3Ln2+C=1# #C=1+(pisqrt2)/16-3Ln2# Thus, #f(x)=-x/2*cos2x+1/4*sin2x-6ln(sinx)+1+(pisqrt2)/16-3Ln2# 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 1380 views around the world You can reuse this answer Creative Commons License