Integrals(antiderivative) plz help!!!?

Question

what is the derivative of x2cosx and its anti-derivative..plz help!!!!!!
thnx 4 youre support

Answer 1

derivative of x2cosx :



y = x^2cosx



y' = 2xcosx - x^2.sinx >=================< ANSWER





anti-derivative of x^2cosx + ?

Let,

I = ∫x^2cosx.dx = x^2.sinx - ∫2xsinx.dx



I = x^2.sinx -2xcosx + ∫2cosx.dx



I = x^2.sinx -2xcosx +2sinx +c >==============< ANSWER

Answer 2

The derivative is the easy part. Use the product rule. d/dx of u(v) is uv' +vu'. So x^2(-sin x) +cos x (2x) =-x^2 sin x +2x cos x. The anti-derivative requires integration by parts. There's a formula for that too. If you consider part of your function to be u and part to be dv (v', that is) then the answer is uv-

integral of v du. Let x^2 be the u, and cos be dv. V would be sin x. Du would be 2x. So x^2 sinx -

integral of 2x sinx. Do parts on that one again. U is 2x du is 2. Dv is sin x, v is -cos x. So uv=--2xcos x. Remember there was a minus sign there, so this becomes positive.and you take the integral of 2 cos x, which is 2 sin x and it would be negative. So the answer is x^2 sin x +2x cos x =-2 sin x. Try differentiating this and you'll get back to the original.

Answer 3

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Answer 4

so you find derivative

(x^2cosx)'=2xcosx-x^2sinx

so your after ∫x^2cosxdx?

no you wrote the question wrong, because then it deals with double integrals, it was maybe ∫x^2sinxdx

so

(x^2cosx)'=2xcosx-x^2sinx

x^2sinx=2xcosx-(x^2cosx)'

∫x^2sinxdx=∫2xcosxdx-x^2cosx

to find the ∫2xcosxdx

find the derivative of 2xsinx

so

(2xsinx)'=2sinx+2xcosx

2xcosx=(2xsinx)'-2sinx

∫2xcosxdx=-∫2sindx+2xsinx

∫2xcosxdx=2cosx+2xsinx



so

∫x^2sinxdx=∫2xcosxdx-x^2cosx

∫x^2sinxdx=2cosx+2xsinx-x^2cosx

=2xsinx-(x^2-2)cosx



otehrwise if the question was ∫x^2cosxdx that take long