How do you integrate #int x^2 cos3 x dx # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer Lovecraft Jan 9, 2016 #I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c# Explanation: #I = intx^2cos(3x)dx# Say #u = x^2# so #du = 2x# and #dv = cos(3x)# so #v = sin(3x)/3# #I = (x^2sin(3x))/3 - 2/3intxsin(3x)dx# Say #u = x# so #du = 1# and #dv = sin(3x)# so #v = -cos(3x)/3# #I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +1/3intcos(3x)dx)# #I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +sin(3x)/9) + c# #I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 2853 views around the world You can reuse this answer Creative Commons License