Integration: Proof of Sine Powers Int. Formula

In summary, using the reduction formula and repeatedly applying it, we can prove that for even powers of sine, the integral from 0 to pi/2 is equal to the product of odd numbers over the product of even numbers, multiplied by pi/2.
  • #1
DivGradCurl
372
0
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.
 
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  • #2
[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]

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.
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[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
 
  • #3
Thanks for your input. It makes sense now.
 

Related to Integration: Proof of Sine Powers Int. Formula

What is the integration formula for sine powers?

The integration formula for sine powers is ∫ sinn x dx = -1/n cos x sinn-1 x + (n-1)/n ∫ sinn-2 x dx, where n is any real number except for -1.

How do you prove the integration formula for sine powers?

The integration formula for sine powers can be proved using the substitution method and the trigonometric identity sin2 x = (1-cos 2x)/2.

What is the significance of the integration formula for sine powers?

The integration formula for sine powers is useful in solving integrals involving powers of sine, which often arise in physics and engineering problems.

What is the domain of the integration formula for sine powers?

The integration formula for sine powers is valid for all real numbers except for -1, as this would result in a division by 0 error.

Can the integration formula for sine powers be extended to other trigonometric functions?

Yes, the integration formula for sine powers can be extended to other trigonometric functions using similar substitution and trigonometric identity methods, such as the integration formula for cosine powers.

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