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

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

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

 

Ah! That makes the problem solvable.

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

 

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)

 

Let x=(tan@)^3

 

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


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

 

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.

 

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

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

 

It IS a proper way!!

 

There's a little proof:
#1. If f(x) is an odd function then: [tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = 0[/tex].
#2. If f(x) is an even function then: [tex]\int \limits_{-\alpha} ^ \alpha f(x) dx = 2 \ \int \limits_{0} ^ \alpha f(x) dx = 2 \ \int \limits_{-\alpha} ^ 0 f(x) dx[/tex].
You can easily see this from a graph.
I'll do #2, leaving #1 for you.
[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?

 

This surely is convergent. Let u= x^3