Integration: Proof of Sine Powers Int. Formula

[DivGradCurl]

Prove that, for even powers of sine,

\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}

Here's what I've got:

The reduction formula

\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}

allows us to obtain

\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

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

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

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

Hence, we can deduce that

\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}

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

Any help is highly appreciated.

 

[AKG]

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

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\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

\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}

You forgot to actually evaluate that last integral

 

[DivGradCurl]

Thanks for your input. It makes sense now.