Integral Question?

Question

Integral of x^2 * cos(x)

Answer 1

Put u = x^2 and dv = cos(x) dx., so that v = sin(x) and du = 2x dx. Then, by parts,

Int u dv = Int x^2 cos(x) dx = uv - Int v du = x^2 sin(x) - Int 2x sin(x) dx = x^2 sin(x) - 2 Int x sin(x) dx

Now, again by parts, put u = x, du = dx, dv = sin(x) dx, v = - cos(x). Therefore,

Int x sin(x) dx = - x cos(x) - Int (- cos(x) dx = - xcos(x) + Int cos(x) dx = - x cos(x) + sin(x) + C. So, our integral is

Int x^2 cos(x) dx = x^2 sin(x) - 2 (- x cos(x) + sin(x) + C) = x^2 + 2x cos(x) -2 sin(x) + K, where K = 2C is the integration constant.

Answer 2

2 x Cos[x] + (-2 + x^2) Sin[x]

So you have to do integration by parts twice.

Int[udv] = uv - Int[vdu]

Keep making the x^2 and then the x the "u" in the integration by parts formula.

Answer 3

2 x Cos[x] + (-2 + x^2) Sin[x]

Answer 4

use "the integrator"

http://integrals.wolfram.com/index.jsp