How do you find the integral of #x^3 cos(x^2) dx#? Calculus Techniques of Integration Integration by Parts 1 Answer Konstantinos Michailidis Sep 15, 2015 #intx^3cos(x^2)dx=1/2x^2sin(x^2)+1/2cos(x^2)# Explanation: We set #t=x^2# hence #dt=2xdx# hence we have that #int x^3cosx^2dx=int 1/2*x^2 cos(x^2) 2xdx=int1/2*tcostdt=1/2inttcostdt=1/2intt(sint)'dt=1/2[tsint]-1/2intt'sintdt=1/2tsint-1/2intsintdt=1/2tsint+1/2cost# Hence #intx^3cos(x^2)dx=1/2x^2sin(x^2)+1/2cos(x^2)# 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 2178 views around the world You can reuse this answer Creative Commons License