How do you integrate #int tsin(2t)# by integration by parts method? Calculus Techniques of Integration Integration by Parts 1 Answer t0hierry Aug 14, 2016 #I = -t cos(2t)/2 + 1/4 sin(2t)# Explanation: Set #sin(2t)dt = dv# then #v = -cos(2t)/2# #u = t# #du = dt# Using #int udv = [uv] - int v du# now you get #I = -t cos(2t)/2 + 1/2 int cos(2t) dt# #I = -t cos(2t)/2 + 1/4 sin(2t)# 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 1730 views around the world You can reuse this answer Creative Commons License