Integration problem ?

  • #1

Homework Statement


[tex]\int^{0}_{-\pi}\sqrt{1-cos^{2} x}[/tex]


Homework Equations





The Attempt at a Solution


I substitute into (sin x)^2 and get an answer of -2 but the answer should be 2 . how to i do this question.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
cristo
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How do you get an answer of -2? Perhaps you should show your work.
 
  • #3
How do you get an answer of -2? Perhaps you should show your work.
just substitute (sin x)^2 and become sqr((sin x)^2) and then become sin x and substitute the limit from -pi to 0 get -2
 
  • #4
cristo
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just substitute (sin x)^2 and become sqr((sin x)^2) and then become sin x and substitute the limit from -pi to 0 get -2

You need to integrate sin(x) before you plug in the limits.
 
  • #5
You need to integrate sin(x) before you plug in the limits.

I have integrate it into -cos x and plugin the limit , i got -2 .But the answer is 2 . How ?
 
  • #6
cristo
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So you have [tex]\left[-\cos(x)\right]^0_{-\pi}=cos(0)-cos(-\pi)[/tex]. Can you evaluate that?
 
  • #7
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maybe you should be interpreting [itex]\sqrt{\sin^2(x)}[/itex] as [itex]|\sin(x)|[/itex]
 
  • #8
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So you have [tex]\left[-\cos(x)\right]^0_{-\pi}=cos(0)-cos(-\pi)[/tex]. Can you evaluate that?[/QUOTE

[tex]\left[-\cos(x)\right]^0_{-\pi}=-cos(0)+cos(-\pi)[/tex]
shouldn't is be this way cristo.
 
Last edited:
  • #9
maybe you should be interpreting [itex]\sqrt{\sin^2(x)}[/itex] as [itex]|\sin(x)|[/itex]
Then , how to integrate [itex]|\sin(x)|[/itex]
 
  • #10
HallsofIvy
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For x between [itex]-\pi[/itex] and 0, sin(x)< 0. In that range, |sin(x)| is just -sin(x). Integrating that will obviously give you the negative of your previous answer.
 
  • #11
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but ((sin x)^2 )^0.5 gives us two answers

sinx and -sinx

???
 
  • #12
1,631
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but ((sin x)^2 )^0.5 gives us two answers

sinx and -sinx

???


yeah, generally it does, but look here we are only integrating in [-pi.0], and obviously sinx, where x is from [-pi,0] is always negative, so

I sin(x) I = -sinx, whenever x is from the interval [-pi. 0]
now as halls said, integrating this you will get the desired answer.
 
  • #13
yeah, generally it does, but look here we are only integrating in [-pi.0], and obviously sinx, where x is from [-pi,0] is always negative, so

I sin(x) I = -sinx, whenever x is from the interval [-pi. 0]
now as halls said, integrating this you will get the desired answer.

But i use mathematica to integrate , it shows -Cot x ((Sin x)^2)^(1/2)
How to integrate to get this form ?
 
  • #14
HallsofIvy
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Why would you want to? This is a definite integral. The result is a number. The integrand reduces to |sin(x)| which, for [itex]-\pi\le x\le 0[/itex] is -sin(x). That's easy to integrate.

I've never used mathematica and what you give makes me glad I haven't! It's clearly using some general algorithm and then not recognizing that, since [itex](sin^2(x))^(1/2)[/itex] is |sin(x)|, [itex]-cot(x)(sin^2(x))^(1/2)= -(cos(x)/sin(x))(-sin(x))= cos(x)[/itex].
 

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