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

1. Dec 27, 2005

### frozen7

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

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

2. Dec 27, 2005

### siddharth

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

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

3. Dec 27, 2005

### frozen7

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

4. Dec 27, 2005

### frozen7

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

5. Dec 27, 2005

### siddharth

Ah! That makes the problem solvable.

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

6. Dec 27, 2005

### frozen7

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

7. Dec 27, 2005

### d_leet

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.

8. Dec 27, 2005

### maverick6664

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

9. Dec 28, 2005

### kant

Let x=(tan@)^3

10. Dec 28, 2005

### frozen7

Whats the point to let x=(tan@)^3 ?

11. Dec 28, 2005

### kant

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

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

Last edited: Dec 28, 2005
12. Dec 28, 2005

### arildno

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.

13. Dec 28, 2005

### frozen7

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 $$\frac{x^2}{1+x^6}$$?

14. Dec 28, 2005

### arildno

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

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

15. Dec 28, 2005

### frozen7

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?

16. Dec 28, 2005

### arildno

It IS a proper way!!

17. Dec 28, 2005

### VietDao29

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?

18. Dec 28, 2005

### frozen7

Thanks. The explanation is very clear. :)

19. Dec 28, 2005

### kant

This surely is convergent. Let u= x^3

20. Dec 28, 2005

### frozen7

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

Last edited: Dec 28, 2005