- #1

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**Please Help!! Integration!**

can anyone help me solve the following integration? thanks a lot.

[tex]\int x^5 cos(x^3) dx[/tex]

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- Thread starter mckallin
- Start date

- #1

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can anyone help me solve the following integration? thanks a lot.

[tex]\int x^5 cos(x^3) dx[/tex]

- #2

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try to solve it by parts

- #3

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If I make [tex] u=cos(x^3), dv=x^5 dx [/tex], the grade of x, which is in the [tex] cos(x^3) [/tex], won't be reduce.

If I make [tex] u=x^5, dv=cos(x^3) dx [/tex], I can't solve the [tex] \int cos(x^3) dx[/tex].

Could you give me some more advice? thanks

- #4

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[tex] \int cos(x^3) dx[/tex]

be solvable? Can you choose slightly different u and dv to accomplish that?

- #5

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- #6

HallsofIvy

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(I notice now that that is essentially what Rainbow Child said.)

- #7

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How it can help?

How this can help?Make the substitution u= x^{3},thenintegrate by parts!

- #8

Hootenanny

Staff Emeritus

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I don't think there's any harm in showing the substitution,How this can help?

[tex]u = x^3 \Rightarrow \frac{du}{dx} = 3x^2 \Rightarrow dx = \frac{du}{3x^2}[/tex]

Hence when we make the substitution the integral becomes,

[tex]\int \frac{x^5}{3x^2}\cos(u)du = \frac{1}{3}\int u\cos(u)du[/tex]

Which is a simple integral to solve.

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