integral problem?

Question

integrate

x^2(sinx)dx

Answer 1

Integral (x^2 sin(x) dx )

To solve this, we have to use integration by parts (twice).

Let u = x^2. dv = sin(x) dx
du = 2x dx. v = -cos(x)

Our formula for parts goes

uv - Integral (v du), so we get

-x^2 cos(x) - Integral ( -cos(x) 2x dx )

which we can simplify to

-x^2 cos(x) + Integral (2x cos(x) dx )

And we can even pull the constant 2 out of the integral, to give us

-x^2 cos(x) + 2 * Integral (x cos(x) dx)

We have to use integration by parts again.

Let u = x. dv = cos(x) dx
du = dx. v = sin(x).

-x^2 cos(x) + 2 [x sin(x) - Integral (sin(x) dx)] + C

Which we simplify into

-x^2 cos(x) + 2x sin(x) - 2 * Integral (sin(x) dx) + C

And now, this is an easy integral.

-x^2 cos(x) + 2x sin(x) - 2 [-cos(x)] + C
-x^2 cos(x) + 2x sin(x) + 2cos(x) + C

Answer 2

This is obviously an integral by parts because separated they are easy to integrate and x^2 will eventually become a constant when split a couple of times.

The formula is integral udv = uv - integral(vdu) so u=x^2, dv=sin(x)dx and du=2xdx and v=-cos(x) so

S(x^2sin(x)dx) = -x^2cos(x) + 2*S(xcos(x)dx).
Then more integration by parts with u=x, dv=cos(x)dx and du=1 and v=sin(x):
S(x^2sin(x)dx) = -x^2cos(x) +2xsin(x) - 2S(sin(x)dx) =
-x^2cos(x) + 2xsin(x) + 2cos(x) + C.

Wait until the end for the constants because you can always add arbitrary constants into a single arbitrary constant.