Integration Problem - given a function find a value

[carlodelmundo]

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

Homework Statement

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

Homework 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.

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!

 

[tiny-tim]

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

Hi carlodelmundo!

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

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

 

[carlodelmundo]

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.

 

[tiny-tim]

This does us no good since the bottom term has degree 3.

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 ​

 

[carlodelmundo]

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

 

[tiny-tim]

Hi carlodelmundo!

just got up :zzz: …

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?

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.

That is a very clever trick.

General tip: when you're desperate …

go back to the obvious! ​