- #1

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int x^3 cosx dx = (Ax^3 + Bx^2+Cx+D)cos x + K

or

int x^3 coxs dx = (Ax^3+Bx)sinx + (Cx^2+D)cosx +K

but I'm having trouble... any help would be appreciated!

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- Thread starter anderma8
- Start date

- #1

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int x^3 cosx dx = (Ax^3 + Bx^2+Cx+D)cos x + K

or

int x^3 coxs dx = (Ax^3+Bx)sinx + (Cx^2+D)cosx +K

but I'm having trouble... any help would be appreciated!

- #2

AKG

Science Advisor

Homework Helper

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Why do you think you don't need a term involving sin here?I'm trying to integtrate x^3 sin x without using integration by parts. I have set up the equation to either:

int x^3 cosx dx = (Ax^3 + Bx^2+Cx+D)cos x + K

Why do you think the polynomial in front of sin doesn't need a square or constant term? Why do you think the polynomial in front of cos doesn't need a cube or linear term?int x^3 coxs dx = (Ax^3+Bx)sinx + (Cx^2+D)cosx +K

- #3

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Thanks for the reply...

essientially, we need to set this up and are having problems. We have an example of int x^4ex + dx. We get (Ax^4 + Bx^3 + Cx^2 + Ex +F) e^x + k. But we are stuck for finding the starting equation for int x^3 cosx.

- #4

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Differentiate and solve for the coefficients.

- #5

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- #6

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Save yourself the trouble and do integration by parts.

- #7

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we would but we can't use that as per the project instructions X-) Any suggestions?

- #8

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How in the world do you get that?

Do you know how to calculate

[tex]\frac{d}{dx}\left(\int x^3\,\sin x\, dx\right)[/tex]

Hint: Its fundamental.

- #9

VietDao29

Homework Helper

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You should note that once you differentiate x^{n}cos(x), you'll get:

(x^{n}**cos**(x))' = nx^{n - 1}**cos**(x) - *x*^{n}**sin**(x), i.e, you have both cos, and sin function on the right, and the highest power of the polynomial is still n.

If you differentiate x^{n}sin(x), you'll get almost the same.

(x^{n}**sin**(x))' = nx^{n - 1}**sin**(x) + *x*^{n}**cos**(x)

Note that, if you differentiate x^{n}cos(x), the highest power polynomial will go with sin(x), and vice versa. Note the italic, and underlined part.

So, if you want to find a function, such that its derivative is x^{3}cos(x) (note that the power of the polynomial term is 3, and is highest), the function should look something like:

f(x) = (Ax^{3} + Bx^{2} + Cx + D) sin(x) + (Ex^{2} + Fx + G) cos(x)

Now, if you try to integrate the function x^{n}cos(x) by Parts, like this:

[tex]\int x ^ n \fbox{\cos x} dx = x ^ {n} \fbox{\sin x} - n \int x ^ {n - 1} \sin (x) dx[/tex]

[tex]= x ^ {n} \fbox{\sin x} + n x ^ {n - 1} \cos(x) - n (n - 1) \int x ^ {n - 2} \cos (x) dx[/tex]

[tex]= x ^ {n} \fbox{\sin x} + n x ^ {n - 1} \cos(x) - n (n - 1) x ^ {n - 2} \sin (x) + n (n - 1) (n - 2)\int x ^ {n - 3} \sin (x) dx = ...[/tex]

So, you can see that the result will look something like x^{n} **sin**(x) + n x^{n - 1} **cos**(x) - n (n - 1) x^{n - 2} **sin**(x) + ... i.e, one sin, and then one cos, and then back to one sin, and...

So, you can conclude that the constants B, D, and F in the function f(x) should all be 0, leaving you with:

f(x) = (Ax^{3} + Cx) sin(x) + (Ex^{2} + G) cos(x)

Can you get it? :)

(x

If you differentiate x

(x

Note that, if you differentiate x

So, if you want to find a function, such that its derivative is x

f(x) = (Ax

Now, if you try to integrate the function x

[tex]\int x ^ n \fbox{\cos x} dx = x ^ {n} \fbox{\sin x} - n \int x ^ {n - 1} \sin (x) dx[/tex]

[tex]= x ^ {n} \fbox{\sin x} + n x ^ {n - 1} \cos(x) - n (n - 1) \int x ^ {n - 2} \cos (x) dx[/tex]

[tex]= x ^ {n} \fbox{\sin x} + n x ^ {n - 1} \cos(x) - n (n - 1) x ^ {n - 2} \sin (x) + n (n - 1) (n - 2)\int x ^ {n - 3} \sin (x) dx = ...[/tex]

So, you can see that the result will look something like x

So, you can conclude that the constants B, D, and F in the function f(x) should all be 0, leaving you with:

f(x) = (Ax

Can you get it? :)

Last edited:

- #10

AKG

Science Advisor

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(Ax

Even though there are a lot of extra terms, differentiating polynomials and trigonometric functions is easy, and determining these undetermined coefficients ends up being very quick and easy.

- #11

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The interesting part is that we too noticed that the highest exponent went with the opposite trig funciton being integrated. Again, thanks for your reply!

- #12

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Thanks!

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