How do you integrate #int x^3 sin x^2 dx # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer Cesareo R. Sep 24, 2016 # 1/2(sin(x^2)-x^2cos(x^2))+C# Explanation: #d/dx(x^2cos(x^2))=2xcos(x^2)-2x^3sin(x^2)# but #2xcos(x^2)= d/dx(sin(x^2))# so #d/dx(x^2cos(x^2)) = d/dx(sin(x^2))-2x^3sin(x^2)# then #int x^3sin(x^2)dx = 1/2(sin(x^2)-x^2cos(x^2))+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 4046 views around the world You can reuse this answer Creative Commons License