∫5x2cos(2x)dx

To solve the integral ∫5x²cos(2x)dx, we can use the method of integration by parts. The formula for integration by parts is ∫udv = uv - ∫vdu.

Step 1: Choose u and dv Let's choose u = x² and dv = 5cos(2x)dx.

Step 2: Compute du and v Then, du = 2xdx and v = (5/2)sin(2x).

Step 3: Apply the integration by parts formula ∫5x²cos(2x)dx = uv - ∫vdu = x² * (5/2)sin(2x) - ∫(5/2)sin(2x) * 2xdx = (5/2)x²sin(2x) - ∫5xsin(2x)dx

Step 4: Solve the remaining integral The remaining integral ∫5xsin(2x)dx can be solved again by integration by parts. Let's choose u = x and dv = 5sin(2x)dx. Then, du = dx and v = -(5/2)cos(2x).

Apply the integration by parts formula again: ∫5xsin(2x)dx = uv - ∫vdu = x * -(5/2)cos(2x) - ∫-(5/2)cos(2x)dx = -(5/2)xcos(2x) - (5/2)∫cos(2x)dx = -(5/2)xcos(2x) - (5/4)sin(2x) + C (where C is the constant of integration)

Step 5: Substitute back into the original equation Substitute this result back into the original equation from Step 3: ∫5x²cos(2x)dx = (5/2)x²sin(2x) - [-(5/2)xcos(2x) - (5/4)sin(2x) + C] = (5/2)x²sin(2x) + (5/2)xcos(2x) + (5/4)sin(2x) + C

So, the integral ∫5x²cos(2x)dx = (5/2)x²sin(2x) + (5/2)x*cos(2x) + (5/4)sin(2x) + C.