Evaluate §xsin^2xdx? ?

To evaluate the integral of x*sin^2(x) dx, we can use integration by parts. Let u = x and dv = sin^2(x) dx. Then, du = dx and v = (1/2)(x - sin(2x)/2).

Using the formula for integration by parts, we have:

∫x*sin^2(x) dx = uv - ∫v du
= x(1/2)(x - sin(2x)/2) - ∫(1/2)(x - sin(2x)/2) dx
= (1/2)x^2 - x*sin(2x)/4 - (1/4)∫x dx + (1/4)∫sin(2x) dx
= (1/2)x^2 - x*sin(2x)/4 - (1/8)x + (1/8)(-1/2)cos(2x)
= (1/2)x^2 - x*sin(2x)/4 - (1/8)x - (1/16)cos(2x) + C

Therefore, the integral of x*sin^2(x) dx is (1/2)x^2 - x*sin(2x)/4 - (1/8)x - (1/16)cos(2x) + C, where C is the constant of integration.