# Integration Problem - given a function find a value

1. Mar 27, 2009

### carlodelmundo

Integration Problem -- given a function... find a value

1. The problem statement, all variables and given/known data

If f(3) = 7 and f'(x) = sin (1 + x^2) / (x^3 - 2x), then f(5) = ...

2. Relevant equations

We can use a calculator (TI-84+) to solve this problem. I'm having a brain fart ... we're given initial conditions, and we're given f'(x) a function that cannot be integrated with conventional methods.

3. The attempt at a solution

I'm thinking this has something to do with the manipulation of the integrand (and its respective limits).

Usually, when I do these problems I integrate... find "C" and once "C" is a value, I plug in the value we're trying to find. But this function isn't easily integrable. What rules am I missing? I need hints not answers please, thank you!

2. Mar 27, 2009

### tiny-tim

Hi carlodelmundo!

(try using the X2 tag just above the Reply box )

Have you tried the obvious substitution … u = 1 + x2 ?

3. Mar 27, 2009

### carlodelmundo

Re: Integration Problem -- given a function... find a value

Hello Tim!

If we let u = 1 +x2 .. then du = 2x dx.

This does us no good since the bottom term has degree 3. I know the bottom term can be factored but how can that help us?

By the way... the equation is written as f'(x) = [ sin (1+x2) ] / (x3 - 2x). Thought I might clarify.

4. Mar 27, 2009

### tiny-tim

You don't seem to recognise the difference between a problem and an opportunity

multiply top and bottom by x, and then the bottom will have degree 4

5. Mar 27, 2009

### carlodelmundo

Re: Integration Problem -- given a function... find a value

That is a very clever trick. Here is my work thus far:

$$\int$$$$\frac{xsin (1 + x^2)}{x^4 - 2x^2}$$

$$\int$$$$\frac{xsin (1 + x^2)}{x^2(x^2-2)}$$

I let u = 1 + x2

u - 1 = x2

du = 2x dx

The resulting integrand is:

$$\frac{1}{2}$$$$\int$$$$\frac{sin (u) du}{(u-1)(u-3)}$$

From here I am completely lost. I'm thinking we need to use partial fractions to get rid of the ugly denominator? And sorry for the long response, I tried scanning only to use LaTex

6. Mar 28, 2009

### tiny-tim

Hi carlodelmundo!

just got up :zzz: …
Yes, partial fractions give you two integrals of the form ∫sinx/x …

the integral of that is the "sine integral function", Si(x) … see http://en.wikipedia.org/wiki/Sine_integral

I don't think there's any other way of doing it, except expanding sinx/x in powers of x, integrating each term, and then calculating to the required order of accuracy.
General tip: when you're desperate

go back to the obvious!