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C4 integration (sign change?)

erm, when u integrate 4x sinx dx

i get -4x cosx - 4 sinx +c

however, the actual answer is:

-4x cosx + 4 sinx + c

Erm... :o:
Reply 1
Avatar for pl224
pl224
Using integration by parts, the setup for your integral should be:

u = 4x, du = 4dx
dv = sinx dx, v = -cosx

The integral is uv - ∫vdu (basically the product rule for differentiation), so substituting in, you get:

-4xcosx - ∫-4cosx dx

-4xcosx + 4∫cosx dx

-4xcosx + 4sinx + C

Most likely you accidentally omitted one of the negative signs or said the integral of cosx was -sinx.
Reply 2
Avatar for SOTEK
SOTEK
With regards to sign changes:

(sinx) ---> (cosx) ---> (-sinx) ---> (-cosx) ---> (sinx)

And so it goes on. (For differentiation anyway, follow in reverse for integration.)
Reply 3
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
pl224
Using integration by parts, the setup for your integral should be:

u = 4x, du = 4dx
dv = sinx dx, v = -cosx

The integral is uv - ∫vdu (basically the product rule for differentiation), so substituting in, you get:

-4xcosx - ∫-4cosx dx

-4xcosx + 4∫cosx dx

-4xcosx + 4sinx + C

Most likely you accidentally omitted one of the negative signs or said the integral of cosx was -sinx.

awesome explanation..cheeers

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