Question

Use the tabular method to find the integral.

5x^3 sin(x)dx

Answer 1

u = x^3 or 5x^3
du = 3x^2 or 15x^2

I believe I can pull the 5 out tho?

dv = sin(x)dx
v = -cos(x)

Answer 2

\[5x^3* -\cos(x) - \int\limits -\cos(x)15x^2dx\]

Answer 3

u = 15x^2
du = 30xdx

dv = -cos(x)dx
v = -sin(x)

Answer 4

\[5x^3* -\cos(x) + \sin(x)30xdx + \int\limits \sin(x) 30xdx\]

Answer 5

love how openstudy just crashed had rto re-write all of that :P

Answer 6

now 1 more time

u = 30x
du = 30dx

dv = sin(x)dx
v = -cos(x)

Answer 7

Continuously tabular until the third time:
= 5 ( - x³cosx + 3x² sinx + 6x cosx - 6sinx ) + C

Answer 8

\[5x^3 *-\cos(x) + \sin(x) 30x + 30x * -\cos(x) + 30 \int\limits \cos(x)dx\]

Answer 9

???

Answer 10

\[5x^3 *-\cos(x) + 15x^2\sin(x) + 30x * -\cos(x) + 30 \sin(x) +c\]

Answer 11

You should learn Tabular method!

Answer 12

did I not do it correctly?

Answer 13

seems to be the same answer as yours besides some signage issues.

Answer 14

Your final one is correct, but your method isn't Tabulation which plainly simplify!

Answer 15

That's why you got wrong signs!

Answer 16

yeah I see it now, creating a table....

I realized when I was doing it that if I said it to be +sin(x) it wou;dn't have cancelled, but I left it as -sin(x) so it did... weird.

Answer 17

Ic.. hmm oh wells :P.

Answer 18

\[-5x^3\cos(x) + 15x^2\sin(x) +30xcos(x) - 30\sin(x) +c\]

Answer 19

I just multiply in, see if it matches with yours:
= -5 x³cosx + 15x² sinx + 30 x cosx - 30 sinx ) + C

Answer 20

mhm