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Let $$I_n = \int^{\pi/2}_0 (\sin x)^n dx.$$ The reduction formula is $$I_n = \frac{n-1}{n} I_{n-2}$$ which I derived myself.

$$I_{2n} = \frac{1\cdot 3\cdot 5 \cdot \dots \cdot (2n-1)\pi}{2\cdot 4 \cdot 6 \cdot \dots \cdot 2n 2}$$

which I also derived myself, and

$$I_{2n+1} = \frac{2\cdot 4\cdot 6\cdot \dots \cdot 2n}{3\cdot 5 \cdot 7\cdot \dots \cdot(2n+1)}$$

I am supposed to show that $I_{2n+2} \leq I_{2n+1} \leq I_{2n}$

Now, it is quite easy to show that $I_{2n+2} \leq I_{2n}$, but I have found it literally impossible to prove the inside term. Can you show me how to do it?

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I_2n got messed up in editing. The denominator is supposed to be multiplied by another 2. I don't know how to fix it. –  Mark T Mar 26 '13 at 0:41
    
did it say you have to show it from your expression of $I_n$? If not, then you can just observe the integrand is decreasing in your integral representation? –  Lost1 Mar 26 '13 at 0:44
    
How do you see that? –  Mark T Mar 26 '13 at 0:47
    
$\sin^n(x)\geq \sin^{n+1}(x)$ since $0\leq \sin(x)\leq 1$ on $[0,\pi/2]$ –  Lost1 Mar 26 '13 at 0:49
1  
Yeah, that's what I immediately thought. I suppose I should just say that –  Mark T Mar 26 '13 at 0:52

3 Answers 3

You show this is true for $n = 0$. Then you can do it by induction?

Suppose it hold for some $n$.

we have $I_{2n}\leq I_{2n+1}$, so

$I_{2n}\frac{2n+1}{2n+2}\leq I_{2n+1} \frac{2n+2}{2n+3}$ because we multiplied something bigger on the righthand than left and they are both positive.

Similiar we can do it for $I_{2n+1}$ and $I_{2n+2}$.

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We can argue as follows. Look at $x\in[0,\pi/2]$. Then $0\leq\sin x\leq 1$. Let $r=\sin x$. This means that $$r^{m}\leq r^{n} $$ whenever $m\geq n$. Do you see why? Then note that $2n+2\geq 2n+1\geq 2n$. Finally, integration will preserve this relation.

I suppose you're looking forward to proving Wallis' formula?

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As $n$ increases, the integral decreases since the integrand's range is in $[0,1]$; it lies in $(0,1)$ except at a (very finite) number of points.

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