Hello, dadon!

How would you solve the following integral: (by using integration by parts)

. . -∫4x²·cos x·dx

First, take out the "-4" and leave it out front: .-4∫x²·cos x·dx

There is a "tabular" method, but I'll do it the long way . . .

We must do "by parts"twice.

Let u = x² . . . dv = cos x·dx

. . du = 2x·dx . .v = sin x

And we have: .x²·sin x - 2∫x·sin x·dx

We do "by parts" on this new integral.

Let u = x . . . dv = sin x·dx

. . du = dx . . .v = -cos x

And we have: .x²·sin x -2[-x·cos x + ∫cos x·dx]

. . . . . . . .= . x²·sin x + 2x·cos x - 2∫cos x·dx

. . . . . . . .= . x²·sin x + 2x·cos x - 2·sin x + C

Don't forget to replace the "-4".