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

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

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

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siddharth
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Do you mean
$$\int \frac{x^2 dx}{(\sin x)(1+x^6)}$$?

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

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

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

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

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

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

frozen7 said:
Erm...may I know what is odd and even function?
A function, f(x), is even if

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.

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

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

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

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arildno
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Use the even or odd suggestion. VISUALIZE what it means that a function is odd, and how that will affect the value of a definite integral over that function between limits -a and a.

By plotting the graph of sinx , the area within $$\frac{-\Pi}{2}$$ and $$\frac{\Pi}{2}$$ is zero and also know that the function is an odd function. So, what I get is only the value of sinx , how about $$\frac{x^2}{1+x^6}$$?

arildno
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Is the function $f(x)=\frac{x^{2}\sin(x)}{1+x^{6}}$ even or odd?

Ya. Thanks. Finally i catch the ball. But how should I answer this question in a proper way? Is there any other answering method for this question?

arildno
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frozen7 said:
Ya. Thanks. Finally i catch the ball. But how should I answer this question in a proper way? Is there any other answering method for this question?
It IS a proper way!!

VietDao29
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frozen7 said:
Ya. Thanks. Finally i catch the ball. But how should I answer this question in a proper way? Is there any other answering method for this question?
There's a little proof:
#1. If f(x) is an odd function then: $$\int \limits_{-\alpha} ^ \alpha f(x) dx = 0$$.
#2. If f(x) is an even function then: $$\int \limits_{-\alpha} ^ \alpha f(x) dx = 2 \ \int \limits_{0} ^ \alpha f(x) dx = 2 \ \int \limits_{-\alpha} ^ 0 f(x) dx$$.
You can easily see this from a graph.
I'll do #2, leaving #1 for you.
$$\int \limits_{-\alpha} ^ \alpha f(x) dx = \int \limits_{-\alpha} ^ 0 f(x) dx + \int \limits_{0} ^ \alpha f(x) dx$$
Let's first integrate this:
$$\int \limits_{-\alpha} ^ 0 f(x) dx$$. Substitude x = -u, so: dx = -du. x = 0 => u = 0 $x = -\alpha \Rightarrow u = \alpha$
$$\int \limits_{-\alpha} ^ 0 f(x) dx = - \int \limits_{\alpha} ^ 0 f(-u) du = \int \limits_{0} ^ \alpha f(-u) du$$. Using the fact that f(-u) = f(u), you'll have:
$$\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx$$, u is the dummy variable.
So:
$$\int \limits_{-\alpha} ^ \alpha f(x) dx = \int \limits_{-\alpha} ^ 0 f(x) dx + \int \limits_{0} ^ \alpha f(x) dx = \int \limits_{0} ^ \alpha f(x) dx + \int \limits_{0} ^ \alpha f(x) dx = 2 \ \int \limits_{0} ^ \alpha f(x) dx$$. (Q.E.D)
Can you do the same to #1, and the other part of #2?

Thanks. The explanation is very clear. :)

frozen7 said:
By plotting the graph of sinx , the area within $$\frac{-\Pi}{2}$$ and $$\frac{\Pi}{2}$$ is zero and also know that the function is an odd function. So, what I get is only the value of sinx , how about $$\frac{x^2}{1+x^6}$$?

This surely is convergent. Let u= x^3

how come $$\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx$$

Shouldnt it be $$\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx$$

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VietDao29
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frozen7 said:
how come $$\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ \alpha f(x) dx$$
Shouldnt it be $$\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx$$
Okay, you should look back at your text-book. There should be a part that states:
$$\int \limits_{\alpha} ^ {\beta} f(x) dx = \int \limits_{\alpha} ^ {\beta} f(t) dt = \int \limits_{\alpha} ^ {\beta} f(z) dz = \int \limits_{\alpha} ^ {\beta} f(y) dy = ...$$
Since you have:
$$\int \limits_{\alpha} ^ {\beta} f(x) dx = F(\beta) - F(\alpha)$$
It shows that the result only depends on F, $\alpha , \mbox{ and } \beta$.
You can also think this way, in the two-dimensional coordinate system, replace the axis Ox by Ot, draw the graph of f(t). It's exactly the same graph as the graph f(x) (i.e when the horizontal axis is Ox). Now change the horizontal axis to Oz, draw the graph of f(z), it's exactly the same as the graph of f(x), and f(t). Hence, the 3 areas under the curves f(x), f(t), f(z) from $x = \alpha , \ t = \alpha , \mbox{ or } z = \alpha$ to $x = \beta , \ t = \beta , \mbox{ or } z = \beta$ are exactly the same.
-------------------
I'll give you an illustration:
Let f(x) = 4x3, that means:
f(t) = 4t3, right?
Now let's find the area under these graphs from 0 to 2:
$$\int \limits_{0} ^ {2} f(x) dx = \left. x ^ 4 \right|_{0} ^ {2} = 2 ^ 4 = 16$$.
$$\int \limits_{0} ^ {2} f(t) dt = \left. t ^ 4 \right|_{0} ^ {2} = 2 ^ 4 = 16 = \int \limits_{0} ^ {2} f(x) dx$$.
Do you get it now?
-------------------
By the way, are you sure that:
$$\int \limits_{0} ^ \alpha f(-u) du = \int \limits_{0} ^ \alpha f(u) du = \int \limits_{0} ^ {-\alpha} f(x) dx$$?
Note that: x = -u, so dx = -du

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Thanks a lot. :)