Solving Improper Integrals: Is discontinuity at interval 0 to π/2?

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Homework Help Overview

The discussion revolves around the evaluation of an improper integral involving trigonometric functions, specifically the integral \(\int^{\pi/2}_{0} \frac{\cos(x)dx}{\sin^{2}(x) - 3\sin(x) - 4}\). Participants are examining the conditions under which the integral is considered improper and the implications of discontinuities within the specified interval.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the integral and question the conditions that make it improper, particularly focusing on the behavior of the denominator. There are attempts to factor the denominator and explore the implications of discontinuities at specific points in the interval.

Discussion Status

The discussion is ongoing, with various interpretations of the integral's properties being explored. Some participants have provided computational results using software, while others are questioning the correctness of these results and the reasoning behind the classification of the integral as improper.

Contextual Notes

There is confusion regarding the evaluation of the integral and the behavior of the denominator at \(x = \pi/2\). Participants are also addressing potential discrepancies in numerical results obtained from different software tools.

JinM
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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?

\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?
 
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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?
 
I don't! But that was what the final asked of me -- evaluate the improper integral!
 
JinM said:
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.
 
Factor the denominator.
 
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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?
 
rootX said:
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'.
 
Dick said:
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
 
It's improper bc of pi/2.
 
  • #10
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
rocomath said:
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
rootX said:
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
oo thnx!
matlab uses ln for log .. so that was the problem
 
  • #14
Dick said:
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
rootX said:
oo thnx!
matlab uses ln for log .. so that was the problem

Ok, now what's rocomath's problem?
 
  • #16
rocomath said:
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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