- #1

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Integrate (x^2 / sinx) / (1 + x^6):

How to do this? I just need some clues...

How to do this? I just need some clues...

- Thread starter frozen7
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- #1

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Integrate (x^2 / sinx) / (1 + x^6):

How to do this? I just need some clues...

How to do this? I just need some clues...

- #2

siddharth

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[tex] \int \frac{x^2 dx}{(\sin x)(1+x^6)} [/tex]?

Also, are you sure that it is not a definite integral?

- #3

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Sorry...should be (x^2 )( sinx) / (1 + x^6):

- #4

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and from x = - pi / 2 to x = pi / 2

- #5

siddharth

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Ah! That makes the problem solvable.frozen7 said:and from x = - pi / 2 to x = pi / 2

Here is a hint:Find if your function is even or odd

- #6

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Erm...may I know what is odd and even function?

- #7

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A function, f(x), is even iffrozen7 said:Erm...may I know what is odd and even function?

f(-x) = f(x)

and odd if

f(-x) = -f(x)

the integral of an odd from function from -a to a is zero. and the integral of an even function from -a to a is equal to two times the integral of that function from 0 to a.

- #8

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hint2: sin(x) = - sin(-x)

- #9

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Let x=([email protected])^3

- #10

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What`s the point to let x=([email protected])^3 ?

- #11

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Now that i think about it, such substitution might not work...

Do you have a limit for your integral? Are you testing for convergence?

Do you have a limit for your integral? Are you testing for convergence?

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- #12

arildno

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- #13

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- #14

arildno

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The answer to that is SUFFICIENT to answer what your definite integral is going to be.

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- #16

arildno

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It IS a proper way!!frozen7 said:

- #17

VietDao29

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There's a little proof:frozen7 said:

You can easily see this from a graph.

I'll do

[tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = \int \limits_{-\alpha} ^ 0 f(x) dx + \int \limits_{0} ^ \alpha f(x) dx[/tex]

Let's first integrate this:

[tex]\int \limits_{-\alpha} ^ 0 f(x) dx[/tex]. Substitude x = -u, so: dx = -du. x = 0 => u = 0 [itex]x = -\alpha \Rightarrow u = \alpha[/itex]

[tex]\int \limits_{-\alpha} ^ 0 f(x) dx = - \int \limits_{\alpha} ^ 0 f(-u) du = \int \limits_{0} ^ \alpha f(-u) du[/tex]. Using the fact that f(-u) = f(u), you'll have:

[tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx[/tex], u is the dummy variable.

So:

[tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = \int \limits_{-\alpha} ^ 0 f(x) dx + \int \limits_{0} ^ \alpha f(x) dx = \int \limits_{0} ^ \alpha f(x) dx + \int \limits_{0} ^ \alpha f(x) dx = 2 \ \int \limits_{0} ^ \alpha f(x) dx[/tex]. (Q.E.D)

Can you do the same to #1, and the other part of #2?

- #18

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Thanks. The explanation is very clear. :)

- #19

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frozen7 said:

This surely is convergent. Let u= x^3

- #20

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how come [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx[/tex]

Shouldn`t it be [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx[/tex]

Shouldn`t it be [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx[/tex]

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- #21

VietDao29

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Okay, youfrozen7 said:how come [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx[/tex]

Shouldn`t it be [tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx[/tex]

[tex]\int \limits_{\alpha} ^ {\beta} f(x) dx = \int \limits_{\alpha} ^ {\beta} f(t) dt = \int \limits_{\alpha} ^ {\beta} f(z) dz = \int \limits_{\alpha} ^ {\beta} f(y) dy = ...[/tex]

Since you have:

[tex]\int \limits_{\alpha} ^ {\beta} f(x) dx = F(\beta) - F(\alpha)[/tex]

It shows that the result

You can also think this way, in the

-------------------

I'll give you an illustration:

Let f(x) = 4x

f(t) = 4t

Now let's find the area under these graphs from 0 to 2:

[tex]\int \limits_{0} ^ {2} f(x) dx = \left. x ^ 4 \right|_{0} ^ {2} = 2 ^ 4 = 16[/tex].

[tex]\int \limits_{0} ^ {2} f(t) dt = \left. t ^ 4 \right|_{0} ^ {2} = 2 ^ 4 = 16 = \int \limits_{0} ^ {2} f(x) dx[/tex].

Do you get it now?

-------------------

By the way, are you sure that:

[tex]\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx[/tex]?

Note that: x = -u, so

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- #22

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Thanks a lot. :)

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