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Is this improper?

  • Thread starter JinM
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  • #1
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I had this integral on my final exam and I was wondering at what point in the interval is the improper integral discontinuous?

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

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

  • #2
Dick
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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
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I don't! But that was what the final asked of me -- evaluate the improper integral!
 
  • #4
Dick
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I don't! But that was what the final asked of me -- evaluate the improper integral!
Hard to say why they said that then. If you got it right, I wouldn't worry about it.
 
  • #5
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Factor the denominator.
 
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  • #6
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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
Dick
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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?
The OP already said it was known it could be solved by partials. Want's to know why it's 'improper'.
 
  • #8
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The OP already said it was known it could be solved by partials. Want's to know why it's 'improper'.
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
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It's improper bc of pi/2.
 
  • #10
Dick
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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
Dick
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It's improper bc of pi/2.
Why??? What's going on with this thread?? The denominator is (-6) at x=pi/2!!!
 
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  • #12
Dick
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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
Ok. It's not that weird. Why did you use log10?
 
  • #13
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oo thnx!
matlab uses ln for log .. so that was the problem
 
  • #14
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Why??? What's going on with this thread?? The denominator is (-6) at x=pi/2!!!
Oh sorry, I read my factorization incorrectly. I have like sinx+1 but read it as sinx-1, LOL.
 
  • #15
Dick
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oo thnx!
matlab uses ln for log .. so that was the problem
Ok, now what's rocomath's problem?
 
  • #16
Dick
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Oh sorry, I read my factorization incorrectly. I have like sinx+1 but read it as sinx-1, LOL.
So is everybody happy then?
 

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