Integral inequalities and recurrence Let
$$U_{n}=\int_{0}^{1}x^{n}\sqrt {1-x^{2}}$$
Prove that:     
$$U_{n+2}=\dfrac {n+2} {n+4}\times U_{n}$$
I've done some alike this ,But I couldn't solve this . 
I also have this one :   Let  
$$I_{n}=\int _{0}^{\pi/4}\tan ^{n}\left( x\right) dx $$
  I've shown that $$I_{n+2}+I_{n}=\dfrac {1} {n+1}$$ and $$I_{n} \geq I_{n+1}$$
Now I'm stuck at 
$$\dfrac {1} {2\left( n+1\right) }\leq I_{n}\leq \dfrac {1} {n+1} $$
Help!
 A: Through the substitution $x=\sin\theta$ we have:
$$ U_n=\int_{0}^{\pi/2}\sin^n(\theta)\cos^2(\theta)\,d\theta \tag{1}$$
and by integration by parts, since $\int\sin^n(\theta)\cos(\theta)\,d\theta=C+\frac{\sin^{n+1}(\theta)}{n+1}$, we have:
$$ U_n = \frac{1}{n+1}\int_{0}^{\pi/2}\sin^{n+2}(\theta)\,d\theta \tag{2}$$
from which:
$$\begin{eqnarray*} (n+1)U_n-U_{n+2} &=& \int_{0}^{\pi/2}\sin^{n+2}(\theta)(1-\cos^2\theta)\,d\theta\\ &=& \int_{0}^{\pi/2}\sin^{n+4}(\theta)\,d\theta\\ &=& (n+3) U_{n+2} \tag{3}\end{eqnarray*}$$
and $$ U_{n+2} = \frac{n\color{red}{+1}}{n+4} U_n.\tag{4} $$
So, no wonder you could not prove the claim, since your claim was incorrect.

About the second part, if we set
$$ I_n = \int_{0}^{\pi/4}\tan(x)^n\,dx = \int_{0}^{1}\frac{u^n}{u^2+1}\,du \tag{5} $$
we obviously have $I_n\leq \int_{0}^{1}u^n\,du = \frac{1}{n+1}$ and $I_n\geq \int_{0}^{1}\frac{u^n}{2}\,du = \frac{1}{2(n+1)}$.
A: The first one looks wrong to me, since, for example $U_0=\pi/4$ while $U_2=\pi/16=U_0/4$. If the formula would be correct, you would have $U_2=U_0/2$.
For the second, with $u=\tan x$:
$$
I_{n+2}+I_n=\int_0^{\pi/4}\tan^n x(1+\tan^2x)\,dx=\int_0^1 u^n\,du=\frac{1}{1+n}.
$$
