How do you integrate #int t^2 * cos(1-t^3) dt#? Calculus Techniques of Integration Integration by Parts 1 Answer maganbhai P. Jun 13, 2018 #I=-1/3*sin(1-t^3)+c# Explanation: Here, #I=intt^2*cos(1-t^3)dt=intcos(1-t^3)*t^2dt# Let, #1-t^3=u=>-3t^2dt=du=>t^2dt=-1/3du# So, #I=intcosu(-1/3)du# #=>I=-1/3int cosudu# #=>I=-1/3 (sinu)+c# Subst. back, #u=1-t^3# ,we get #I=-1/3*sin(1-t^3)+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 3495 views around the world You can reuse this answer Creative Commons License