1. Mar 5, 2014

### lap

Integrate ( (sqrt (x^2 - 9))/x )( exp x^2 )( cos 7x )( sin(x^4 + 5x^2 + 100) ) dx
with upper limit = 3 and lower limit = -3

I have tried to use integration by part and set u = ( (sqrt (x^2 - 9))/x )( exp x^2 ) and
dv = ( cos 7x )( sin(x^4 + 5x^2 + 100) ) dx

Last edited: Mar 5, 2014
2. Mar 5, 2014

What have you tried so far?

3. Mar 5, 2014

### Ray Vickson

I very much doubt there is any closed-form formula for the antiderivative, so you probably need to contemplate numerical integration for the general case of $\int_a^b f(x) \, dx$. However, before doing that, sit down and think carefully about your specific problem.

4. Mar 5, 2014

### lap

How to integrate ( (sqrt (x^2 - 9))/x )( exp x^2 )( cos 7x )( sin(x^4 + 5x^2 + 100) ) dx ?

5. Mar 5, 2014

### Ray Vickson

I have already told you it cannot be done with formulas---even very long ones having billions of complicated terms and taking millions of pages to write out. However, that was not your original question: you wanted $\int_{-3}^3 f(x) \, dx$. As I suggested, think hard about the problem first.

6. Mar 6, 2014

7. Mar 6, 2014

### dirk_mec1

The answer is correct but can you prove it?

8. Mar 6, 2014

### lap

I know the answer is 0 because the positive area canceled the negative area but I don't know how to prove it

9. Mar 6, 2014

### Dick

If f(x) is that big expression you are integrating, can you prove that f(-x)=(-f(x))? Then show $\int_{-a}^0 f(x) dx = -\int_{0}^a f(x) dx$.

Last edited: Mar 6, 2014
10. Mar 6, 2014

### lap

I proved that f(-x)=(-f(x)) and solved it. Thank you very much !

11. Mar 6, 2014

### HallsofIvy

In order to use symmetry here you must also show that this is not an improper integral. Your integrand is a fraction with sin(x^4+ 5x^2+ 100) in the denominator. Can you show that this never 0 for x between -3 and 3?

12. Mar 6, 2014

### Saitama

There is an x in the denominator instead of sine. Moreover, the function doesn't seem to defined within the given limits.