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Integration - proof

  1. Apr 27, 2005 #1
    Prove that, for even powers of sine,

    [tex]\int _0 ^{\pi /2} \sin ^{2n} x \: dx = \frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n} \frac{\pi}{2}[/tex]

    Here's what I've got:

    The reduction formula

    [tex]\int \sin ^n x \: dx = -\frac{1}{n}\cos x \sin ^{n-1} x + \frac{n-1}{n}\int \sin ^{n-2} x \: dx \qquad n \geq 2 \mbox{ is an integer}[/tex]

    allows us to obtain

    [tex]\int _0 ^{\pi /2} \sin ^{2n} x \: dx = \left. -\frac{1}{2n} \cos x \sin ^{2n-1} x \right] _0 ^{\pi /2} + \frac{2n-1}{2n} \int _0 ^{\pi /2} \sin ^{2n-2} x \: dx[/tex]

    [tex]\int _0 ^{\pi /2} \sin ^{2n} x \: dx = \frac{2n-1}{2n} \int _0 ^{\pi /2} \sin ^{2n-2} x \: dx[/tex]

    [tex]\int _0 ^{\pi /2} \sin ^{2n-2} x \: dx = \frac{2n-3}{2n-2} \int _0 ^{\pi /2} \sin ^{2n-4} x \: dx [/tex]

    [tex]\int _0 ^{\pi /2} \sin ^{2n-4} x \: dx = \frac{2n-5}{2n-4} \int _0 ^{\pi /2} \sin ^{2n-6} x \: dx [/tex]

    Hence, we can deduce that

    [tex]\int _0 ^{\pi /2} \sin ^{2n} x \: dx = \frac{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdot \cdots \cdot 5\cdot 3\cdot 1}{2n \cdot (2n-2)\cdot (2n-4)\cdot \cdots 6 \cdot 4\cdot 2}[/tex]

    As you can see, I've missed the [tex]\frac{\pi}{2}[/tex] from the statement above. I double-checked my solution, but I can't tell where the mistake is.

    Any help is highly appreciated.
     
  2. jcsd
  3. Apr 27, 2005 #2

    AKG

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    [tex]\int _0 ^{\pi /2} \sin ^{2n-4} x \: dx = \frac{2n-5}{2n-4} \int _0 ^{\pi /2} \sin ^{2n-6} x \: dx [/tex]

    .
    .
    .

    [tex]\int _0 ^{\pi /2} \sin ^{2n-(2n-2)} x \: dx = \frac{2n-(2n-2)-1}{2n-(2n-2)} \int _0 ^{\pi /2} \sin ^{2n-2n} x \: dx [/tex]

    [tex]\int _0 ^{\pi /2} \sin ^2 x \: dx = \frac{1}{2} \int _0 ^{\pi /2} \sin ^0 x \: dx = \frac{1}{2} \int _0 ^{\pi /2} 1 \: dx = \frac{1}{2}\frac{\pi}{2}[/tex]

    You forgot to actually evaluate that last integral
     
  4. Apr 27, 2005 #3
    Thanks for your input. It makes sense now.
     
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