Finding the general integral

  • Thread starter BlackMamba
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  • #1
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Hello,

I have a problem that I am having difficulties with. I'm told to find the general integral.

So here is the problem:
[itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx} {1 - sin^2 x} dx[/itex]

Here is my partial solution:

[itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx} {cos^2 x} dx[/itex]

[itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {1} {cosx} * \frac {sinx} {cosx} dx[/itex]

[itex]\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} secxtanx dx[/itex]

[itex]= [tanx][/itex]

But here is where I am stuck, because when you evaluate [itex]tanx[/itex] from [itex]\frac {\pi} {2}[/itex] to [itex]\frac {-\pi} {2}[/itex] you get an error because no value exists for [itex]tan \frac {\pi} {2}[/itex]


Any help would be greatly appreciated. :smile:
 

Answers and Replies

  • #2
746
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wouldn't it be zero? you're integrating from pi/2 to -pi/2 & tan is an odd function.
 
  • #3
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Well that is what I was unsure of. I just double checked the math on my calculator, and I realized I made an error inputting the values. So that would explain the error.

Thanks for the input. :smile:
 
  • #4
1,074
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The problem might be that you have the wrong antiderivative of sec(x)tan(x). What is the derivative of sec(x)?
 
  • #5
746
13
OOPS i guess it has been a while since i last did calculus... sorry about that :eek: :redface:
 
  • #6
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d_leet said:
The problem might be that you have the wrong antiderivative of sec(x)tan(x). What is the derivative of sec(x)?

But surely sec (pi/2)=1/0 as well??:yuck:
 
  • #7
HallsofIvy
Science Advisor
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I'm a little puzzled by your saying you are asked to find the "general integral" and then asking for a specific value for a definite integral.

Of course, since the denominator, 1- sin2x is 0 at both [itex]\frac{pi}{2}[/itex] and [itex]-\frac{\pi}{2}[/itex], this is an "improper integral". It's value is defined as
[tex]\lim_{\alpha\rightarrow -\frac{\pi}{2}} \lim_{\beta\rightarrow \frac{\pi}{2}}\int_\beta^\alpha \frac{sin x}{1- sin^2 x}dx[/tex]
So what you want to do is find the "general anti-derivative" and then take limits. Of course, that is precisely what you did- you can't evaluate tan(x) at the endpoints precisely because those limits do not exist.

The "Cauchy Principal Value" for this integral is defined as
[tex]\lim_{\alpha\rightarrow \frac{pi}{2}}\int_\alpha^{-\alpha}\frac{sin x}{1- sin^2x} dx[/tex]
and since the integrand is odd, that is 0.
 
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