Book Calculus by Stewart Edition 6 chapter 81 problem 8 one

Solution

Yes, that is right. v = sin(mx) / m I\'ll go through the first integration by parts, and let you do the second one. Let u = x^2, and dv = cos(mx) dx, then du = 2x dx and v = (1/m) sin(mx) integral (udv) = uv - integral (vdu) = x^2 (1/m) sin(mx) - (2/m) integral (x sin(mx) dx) Then do another integration by parts. Let u = x, and dv = sin(mx), then du = dx and v = (-1/m) cos(mx)