# Is this improper?

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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Dick
Homework Helper
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!

Dick
Homework Helper
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?

Dick
Homework Helper
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'.

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.

Dick
Homework Helper
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.

Dick
Homework Helper
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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Dick
Homework Helper
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?

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

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.

Dick
Homework Helper
oo thnx!
matlab uses ln for log .. so that was the problem
Ok, now what's rocomath's problem?

Dick