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

  • Thread starter JinM
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In summary, the conversation revolved around the integration of the function cos(x)/(sin(x)^2-3*sin(x)-4) over the interval [0, pi/2]. The original poster was wondering if the integral was improper, but it was determined that it was not. The conversation then shifted to the use of partial fraction decomposition to solve the integral and the importance of using the correct logarithmic function in calculations. In the end, the conversation concluded with a humorous mix-up in the factorization of the denominator.
  • #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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  • #2
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
I don't! But that was what the final asked of me -- evaluate the improper integral!
 
  • #4
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.
 
  • #5
Factor the denominator.
 
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  • #6
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
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'.
 
  • #8
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
 
  • #9
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?
 

1. What is an improper integral?

An improper integral is an integral where either the upper or lower limit goes to infinity, or where the integrand has a discontinuity within the interval of integration.

2. What is a discontinuity at interval 0 to π/2?

A discontinuity at interval 0 to π/2 means that the function being integrated has a break or jump at some point within the interval, which can cause issues when trying to evaluate the integral.

3. Why is it important to solve improper integrals?

Improper integrals are important because they allow us to evaluate integrals for functions that would otherwise be impossible to integrate using traditional methods. They also have many practical applications, such as calculating areas under curves and finding the average value of a function.

4. How do you solve an improper integral with a discontinuity at interval 0 to π/2?

The first step is to identify the type of discontinuity present (removable, infinite, or jump). Then, you can use techniques such as splitting the integral into two separate integrals, using the limit comparison test, or using a change of variables to solve the integral.

5. Are there any common mistakes to avoid when solving improper integrals with a discontinuity at interval 0 to π/2?

Yes, some common mistakes to avoid include forgetting to identify and address the discontinuity, using incorrect limits of integration, and not considering the behavior of the function near the point of discontinuity. It is important to carefully analyze the problem and use the appropriate techniques to solve the integral correctly.

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