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QRAwarrior

  • 2 years ago

Suppose you have, in the denominator of your integral, the expression of the form x^2 - a^2. You want to use trigonometric substitution with this. How would the resulting triangle look like? ATTEMPT:

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  1. QRAwarrior
    • 2 years ago
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    |dw:1337434470844:dw|

  2. asnaseer
    • 2 years ago
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    I'm sure I follow you, but some of the rules for trig substitution in integrals are described here: http://en.wikipedia.org/wiki/Trigonometric_substitution maybe that will help you.

  3. asnaseer
    • 2 years ago
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    *I'm NOT sure...

  4. QRAwarrior
    • 2 years ago
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    Got any ideas @shivam_bhalla?

  5. shivam_bhalla
    • 2 years ago
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    Your trigonometric substitution for such a case will be \[x= asec (\theta)\] So \[\sec(\theta)=x/a\] |dw:1337436343334:dw|

  6. QRAwarrior
    • 2 years ago
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    Yes!

  7. shivam_bhalla
    • 2 years ago
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    yes @QRAwarrior , you have done it correctly :)

  8. QRAwarrior
    • 2 years ago
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    Because I thought that there would be a side with "x^2 - a^2"

  9. QRAwarrior
    • 2 years ago
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    Anyways, thanks a lot.

  10. shivam_bhalla
    • 2 years ago
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    yw :)

  11. apoorvk
    • 2 years ago
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    okay, think about this --> if I have something like \[\frac{1}{x^2 - a^2}\] I would be very happy, if I can reduce (x^2 - a^2) to a monomial. Right? Because that become very easy to integrate after I convert it into a form in the numerator. Now, if I remember the basics: \[tan^2m +1 = sec^2m\] and \[sin^2m + cos^2m = 1\] Now, I see, if I can get some think like a^2 *(sec^2(m) instead of the x^2, I can manage to get a a^2 tan^x in the denominator, and some ".....dm" in the numerator, which would be pretty easy to integrate then. So, I use the substitution: x=asecm and proceed. I get: dx= a secm tanm dm i plug this in the numerator in the place of dx. and then evaluate the integral pretty easily.

  12. apoorvk
    • 2 years ago
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    I just have to think of some ways to remove the radical sign if present, or make the denominator simpler.

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