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Math Help - An integration of sorts.

  1. #1
    Super Member Showcase_22's Avatar
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    An integration of sorts.



    That's the question I can't do. It would be so much easier if it was a cos squared!

    I think I have to use a substitution but i'm not quite sure what it should be.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post


    That's the question I can't do. It would be so much easier if it was a cos squared!

    I think I have to use a substitution but i'm not quite sure what it should be.
    Multiply by \frac{\sqrt{1+\cos{\theta}}}{\sqrt{1+\cos{\theta}}  }

    Can you do it now?
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  3. #3
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    Or, you can rewrite \sqrt{1-\cos{\theta}} as \sqrt{2}\sin{\left(\frac{x}{2}\right)}. I'll leave it to you to figure out why they're equivalent.
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  4. #4
    Super Member Showcase_22's Avatar
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    I can do it, and I have solved it.

    I tried deriving the second identity myself but I couldn't figure that part out.

    can you show me how it's done?
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  5. #5
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    \sqrt{1-\cos{\theta}}

    = \displaystyle\sqrt{\underbrace{1-\cos^2{\frac{\theta}{2}}}_{\sin^2{\frac{\theta}{2}  }}+\sin^2{\frac{\theta}{2}}}

    = \sqrt{2\sin^2{\frac{\theta}{2}}}

    = \sqrt{2}\sin{\frac{\theta}{2}}
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  6. #6
    Super Member Showcase_22's Avatar
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    oh!! I don't know what I was doing. It's just a simple double angle formulae.

    That's rather useful.
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