I will use this symbol ⌠ to indicate the integral operation. Because you do not have a dx after the C I am unsure whether it is part of the integral. If it is part of the integral, then you can break it up into two integrals as follows: ⌠[{x}cos(x)sin(2x) + C]dx = ⌠{x}cos(x)sin(2x)dx + ⌠Cdx ⌠Cdx is a trivial integral and I will leave that to you. If C is not part of the integral, then it can just be added to the answer afterward. Therefore, let us turn our attention to the integral: ⌠[{x}cos(x)sin(2x)dx Use the trig identity sin(2x) = 2sin(x)cos(x) This makes the integral: 2⌠[{x}cos^2(x)sin(x)dx To integrate this, use integration by parts http://en.wikipedia.org/wiki/Integration_by_parts : let u = x let dv = cos^2(x)sin(x)dx so now we must compute v and du du = dx v = ⌠cos^2(x)sin(x)dx This adds yet another integral within the integration by parts but this integral can be solved by using a u substitution but I will not use the letter u because it would be confusing; I will use y, instead. let y = cos(x) then dy = -sin(x)dx Therefore: v = -⌠y^2dy v = -(y^3)/3 v = -{cos^3(x)}/3 Now that we have equations for u,v, du, and dv we can go back to the integration by parts formula: ⌠udv = uv - ⌠vdu 2⌠[{x}cos^2(x)sin(x)dx = 2[x{-cos^3(x)}/3] - 2⌠-{cos^3(x)}/3 dx Let's bring the minus signs outside of the brackets and the integral: 2⌠[{x}cos^2(x)sin(x)dx = -2/x{cos^3(x)} + 2/3⌠{cos^3(x)}dx We must now left with the integral: ⌠{cos^3(x)}dx This is a table look up. ⌠{cos^3(x)}dx = sin(x) - 1/3{sin^3(x)} 2⌠[{x}cos^2(x)sin(x)dx = -2/3x{cos^3(x)} + 2/3[sin(x) - 1/3{sin^3(x)}] 2⌠[{x}cos^2(x)sin(x)dx = -2/3[x{cos^3(x)}] + 2/3{sin(x)} - 2/9{sin^3(x)} + K where K is a constant of integration. You can add the C or the integral of C back in depending on what you meant.

8-21-2009 Where is your question mark?