solve: integral of (x^2)(sin0.5x)dx?

Question

please help me solve:

integral of (x^2)(sin0.5x)dx

preferably using integration by parts or tabular integration

ty

Answer 1

Integration by parts:
∫f(x)g'(x) dx = [f(x)g(x)] - ∫f'(x)g(x) dx

We want f(x) to equal x^2, so that in the integration by parts, we're left with 2x sin .5x. Integrating by parts again gets rid of the x, leaving something we can integrate normally.

∫(x^2)(sin0.5x)dx =
f(x) = x^2
g'(x) = sin .5x
g'(x) = -1/.5 cos .5x = -2 cos .5x
= [f(x)g(x)] - ∫f'(x)g(x) dx
= [-2 x^2 cos .5x + C] - ∫(2x) -2 cos .5x dx
= [-2 x^2 cos .5x + C] + 4∫x cos .5x dx

Do integration by parts for ∫x cos .5x dx
∫x cos .5x dx
f(x) = x
g'(x) = cos .5x
g(x) = 1/.5 sin .5x = 2 sin .5x
= [f(x)g(x)] - ∫f'(x)g(x) dx
= [(2x)(2 sin .5x + C] - ∫1 * 2 sin .5x dx
= [4 x sin .5x + C] - 2 ∫ sin .5x dx

Now, we have:
= [-2 x^2 cos .5x + C] + 4([4 x sin .5x + C] - 2 ∫ sin .5x dx)
= [-2 x^2 cos .5x + C] + [16 x sin .5x + C] - 8 ∫ sin .5x dx

Finally, let's integrate sin .5x dx:
∫ sin .5x dx = -1/.5 cos .5x dx = -2 cos .5x dx

Which gives us:
= [-2 x^2 cos .5x + C] + [16 x sin .5x + C] - 8 [-2 cos .5x dx]
= [-2 x^2 cos .5x + C] + [16 x sin .5x + C] + [16 cos .5x dx]
= [(16 - 2 x^2) cos .5x + C] + [16 x sin .5x + C] (solution!)