How do you find the integral of #x^3 * sin( x^2 ) dx#? Calculus Techniques of Integration Integration by Parts 1 Answer Sasha P. Oct 11, 2015 #I=1/2(-x^2cosx^2+sinx^2)+C# Explanation: #x^2=t => 2xdx=dt, x^3dx=1/2x^2 2xdx = 1/2tdt# #int x^3 sinx^2 dx = 1/2 int tsintdt = I# #u=t => du=dt# #dv=sintdt => v=int sintdt= -cost# #I=1/2[-tcost + int costdt] = 1/2(-tcost+sint)+C# #I=1/2(-x^2cosx^2+sinx^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 30334 views around the world You can reuse this answer Creative Commons License