How do you integrate #int xsin2x# by integration by parts method? Calculus Techniques of Integration Integration by Parts 1 Answer Andrea S. Jan 16, 2017 #int xsin2xdx = -1/2xcos2x +1/4sin2x# Explanation: As: #d(cos2x) =-2sin2x dx# we can integrate by parts in this way: #int xsin2xdx = -1/2 int x d(cos2x) = -1/2 xcos2x +1/2 int cos2x dx= -1/2xcos2x +1/4sin2x# 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 15233 views around the world You can reuse this answer Creative Commons License