How do you integrate int xsin(2x) by integration by parts method? Calculus Techniques of Integration Integration by Parts 1 Answer Euan S. Aug 8, 2016 =1/4sin(2x) - x/2cos(2x) + C Explanation: For u(x), v(x) int uv'dx = uv ' - int u'vdx u(x) = x implies u'(x) = 1 v'(x) = sin(2x) implies v(x) = -1/2cos(2x) intxsin(2x)dx = -x/2cos(2x) +1/2intcos(2x)dx = -x/2cos(2x)+1/4sin(2x)+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 2921 views around the world You can reuse this answer Creative Commons License