Homework Help: Is this improper?

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?

 

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

 

Factor the denominator.

 

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'.

 

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.

 

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.

 

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

 

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

 

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

 

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

 

Ok, now what's rocomath's problem?

 

So is everybody happy then?