How do you integrate #int x^2 sin x dx # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer Lovecraft Jan 7, 2016 #I = -x^2cos(x) + 2xsin(x) + 2cos(x) + c# Explanation: Say #u = x^2# so #du = 2x#, #dv = sin(x)# so #v = -cos(x)# #I = -x^2cos(x) +2intxcos(x)dx# Say #u = x# so #du = 1#, #dv = cos(x)# so #v = sin(x)# #I = -x^2cos(x) + 2xsin(x) - 2intsin(x)dx# #I = -x^2cos(x) + 2xsin(x) + 2cos(x) + 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 3408 views around the world You can reuse this answer Creative Commons License