Show $n I_n I_{n-1} = \frac{\pi}{2}$ where $I_n = \displaystyle\int_0^{\pi/2} \cos^n x dx$. Consider the sequence $(I_n)_{n \in \mathbb{N}}$:
$$ I_0 = \frac{\pi}{2} $$
$$ I_n = \int_0^{\pi/2} \cos ^n x dx$$
For this sequence I have to prove that the following is true:
$$n I_n I_{n-1} = \frac{\pi}{2}$$
for $n \in \mathbb{N}^*$. This is what I've done so far:
$$I_n = \int_0^{\pi/2} \cos^n x dx = \int_0^{\pi/2} \cos x \cos ^{n-1} x dx$$
$$= \int_0^{\pi/2} (\sin x)' \cos ^{n-1} x dx = \sin x \cos ^{n-1} x \Bigg|_0^{\pi/2} + (n-1)\int_0^{\pi/2} \cos ^{n-2}x \sin^2x dx$$
$$ = 0 - 0 + (n-1)\int_0^{\pi/2} \cos ^{n-2} x (1 - \cos ^2 x) dx $$
$$ = (n-1) \int_0^{\pi/2} (\cos ^ {n-2} x - \cos^n x) dx $$
$$= (n-1)(I_{n-2} - I_{n})$$
So that means we have:
$$ I_n = nI_{n-2}  - nI_n - I_{n-2} + I_n $$
$$nI_n = (n-1) I_{n-2}$$
And if we multiply with $I_{n-1}$ we get:
$$n I_n I_{n-1} = (n-1) I_{n-2} I_{n-1}$$
But that's as far as I got. I don't see how I could show that the above is equal to $\frac{\pi}{2}$.
 A: I'll continue from where you got to : 
$$nI_{n}I_{n-1} = (n-1)I_{n-2}I_{n-1} $$
Now,
$$nI_{n}I_{n-1} = (n-1)I_{n-2}I_{n-1}$$
$$(n-1)I_{n-1}I_{n-2} = (n-2)I_{n-3}I_{n-2}$$
$$(n-2)I_{n-2}I_{n-3} = (n-3)I_{n-4}I_{n-3}$$
$$.....$$
$$(2)I_2I_1 =I_0I_1$$
Multiply all these equations: You can see most of the terms get cancelled.....you end up with:
$$nI_nI_{n-1} = I_0I_1 $$
$$=\int_0^{\frac{\pi}{2}} \cos^0x\,dx\int_0^{\frac{\pi}{2}} \cos x\,dx$$
$$nI_nI_{n-1} = \left(\frac{\pi}{2}\right) (1)$$

Alternatively,
$$nI_nI_{n-1}=n\left(\int_0^\frac{\pi}{2} \cos^nx\,dx \right)\left(\int_0^\frac{\pi}{2} \cos^{n-1}x\,dx \right) =A$$
Use the substitution :
$$\cos^2x=t$$
$$\cos x = t^{\frac{1}{2}}$$
$$-\sin x \,dx = \frac{1}{2} t^{-\frac{1}{2}}\,dt$$
$$\,dx=-\frac{1}{2} t^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}}\,dt $$
$$$$
$$\therefore A = n \left( \int_0^1 \frac{1}{2} (t)^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}} (t)^\frac{n}{2} \,dt  \right) \left( \int_0^1 \frac{1}{2} (t)^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}} (t)^\frac{n-1}{2} \,dt  \right)$$
$$A=\frac{n}{4}\left( \int_0^1(t)^{\frac{n+1}{2} -1} (1-t)^{\frac{1}{2}-1}\,dt \right) \left( \int_0^1(t)^{\frac{n}{2} -1} (1-t)^{\frac{1}{2}-1}\,dt \right)$$
$$=\frac{n}{4}\beta\left( \frac{n+1}{2},\frac{1}{2} \right).\beta\left( \frac{n}{2},\frac{1}{2} \right)$$
Here, $\beta(x,y)$ is the Beta Function.
Now, we can use the relation between the beta function and the Gamma function ($\Gamma (n)$) : $$\beta(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ 
$$$$
$$A = \frac{n}{4} \frac{\Gamma\left( \frac{n+1}{2} \right) \Gamma\left( \frac{1}{2} \right)  }{\Gamma\left( \frac{n}{2} +1 \right)}.\frac{\Gamma\left( \frac{n}{2}  \right) \Gamma\left( \frac{1}{2} \right)}{\Gamma\left( \frac{n+1}{2} \right)} $$
The $\Gamma\left( \frac{n+1}{2} \right)$ gets cancelled out from num and denom.
The $\Gamma\left( \frac{1}{2} \right)$ terms have a known value of $\sqrt\pi$
The rest : $\frac{\Gamma\left( \frac{n}{2}  \right)}{\Gamma\left( \frac{n}{2} +1 \right)}$ can be reduced by the property of the Gamma function: $=\frac{\left( \frac{n}{2} -1 \right)!}{\left( \frac{n}{2} \right)!} = \frac{2}{n}$
So, we are left with :
$$A=\frac{n}{4} \left(\Gamma\left(\frac{1}{2} \right)\right)^2 \frac{2}{n}$$
$${A}=\frac{(\sqrt\pi)^2}{2} = \frac{\pi}{2}$$
