Hey All,
How would you solve the following integral: (by using integration by parts)
-∫4x^2cosxdx
∫udv = uv -∫vdu
Thank you
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".