Is this improper?

1. May 18, 2008

JinM

I had this integral on my final exam and I was wondering at what point in the interval is the improper integral discontinuous?

$$\displaystyle\int^{\pi/2}_{0} \frac{cos(x)dx}{sin^{2}(x) - 3sin(x) -4}$$

I know this is solved by partial fraction decomposition, but I don't see how the integral is improper?

Last edited: May 18, 2008
2. May 18, 2008

Dick

It's improper if the denominator vanishes, isn't it? But I don't think it does, so I don't think it is improper. Why do you think it is?

3. May 18, 2008

JinM

I don't! But that was what the final asked of me -- evaluate the improper integral!

4. May 18, 2008

Dick

Hard to say why they said that then. If you got it right, I wouldn't worry about it.

5. May 18, 2008

rocomath

Factor the denominator.

Last edited: May 18, 2008
6. May 18, 2008

rootX

Why I couldn't solve it using partials?

1/[(sinx-4)*(sinx+1)] = A/(sin x - 4) + B/(sin x + 1 )

so,
A(sinx+1) + B(sinx-4) = 1

Am I doing some stupid math?

7. May 18, 2008

Dick

The OP already said it was known it could be solved by partials. Want's to know why it's 'improper'.

8. May 18, 2008

rootX

I just wanted solve it for my own fun ;)
I couldn't but matlab did:

>> syms x;
>> f = cos(x)/((sin(x))^2-3*(sin(x))-4);
>> int(f,x,0,pi/2)

ans =

-3/5*log(2)+1/5*log(3) = -0.0851937465(approx using google calc)

It's weird that I got "-0.1962" using other software: "Graph"...

P.S. I don't think it was already known though

9. May 18, 2008

rocomath

It's improper bc of pi/2.

10. May 18, 2008

Dick

The OP said "I know this is solved by partial fraction decomposition". It's not weird that you got "-0.1962", because that's the correct answer. What is strange is that "google calc" evaluated -3/5*log(2)+1/5*log(3) as -0.0851937465. That's wrong.

11. May 18, 2008

Dick

Why??? What's going on with this thread?? The denominator is (-6) at x=pi/2!!!

Last edited: May 18, 2008
12. May 18, 2008

Dick

Ok. It's not that weird. Why did you use log10?

13. May 18, 2008

rootX

oo thnx!
matlab uses ln for log .. so that was the problem

14. May 18, 2008

rocomath

Oh sorry, I read my factorization incorrectly. I have like sinx+1 but read it as sinx-1, LOL.

15. May 18, 2008

Dick

Ok, now what's rocomath's problem?

16. May 18, 2008

Dick

So is everybody happy then?