# Thread: integral problem

1. ## integral problem

the integral of (X^3sin(6x)/sin(5x)) from -4 to 4

using the riemann sum as well?

post the answer if you know it.

I have been working on this for almost 4 hours now and can not figure it out.

2. ## Re: integral problem

Originally Posted by richman18
$\int_{-4}^{4} \frac{x^3 \sin (6x)}{\sin (5x)}dx$

using the riemann sum as well?

post the answer if you know it.

I have been working on this for almost 4 hours now and can not figure it out.
Maybe you would care to reply with what work you have done and/or deadends that you have encountered in your four hours of work?

(edit)...
Having now looked at the actual integral.
The integrand is odd, integrated over an interval centered at zero. You might not be expected to find an antiderivative, but rather an analytic argument for why this integral equals...

3. ## Re: integral problem

My initial impression is, "This is an eyeball problem. The value of the integral is zero (0), by inspection. The argument is an odd function and the limits are symmetric about the origin."

A little more thought suggests I should determine that it actually exists. That sin(5x) in the denominator needs some thought. What do you think about it?

$x = 0, \frac{\pi}{5} + \frac{k\cdot\pi}{5}\;for\;k = \pm 1, 2, 3, 4, 5$

4. ## Re: integral problem

one more observation ... the integrand has a discontinuity at x = 0

5. ## Re: integral problem

Originally Posted by skeeter
one more observation ... the integrand has a discontinuity at x = 0
And at a few other places ...

6. ## Re: integral problem

Thanks for the tips. I will post my work that I have done so far in a little bit. Once again thanks for the help.

7. ## Re: integral problem

Originally Posted by skeeter
one more observation ... the integrand has a discontinuity at x = 0
In x=0 the function has a removable singularity [You can easily verify that using l'Hopital rule...] and therefore that has no effects on integrability. Unfortunately in $[-4,4]$ there are several [non-removable...] singularities so that, according to the 'standard' definition, the integral doesn't exist. If we use however the definition of 'Cauchy Principal Value Integral'...

Cauchy Principal Value -- from Wolfram MathWorld

... and consider that f(x) is an odd function, the we can say that is...

$\text{P V} \int_{- a}^{a} f(x)\ dx =0$

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