Evaluate $A_r=\int_{{0}}^{\frac{\pi}{2}} \sin^{r}x \ \ dx$ Let $$A_r=\int_{{0}}^{\frac{\pi}{2}} \sin^{r}x \ \ dx$$
What is the relation beween $A_r$ and $A_{r-2}$ ? Then, find the value of $A_r$ for all $r$.
 A: Assume $r\ge 2$.  To integrate by parts, let $u=\sin^{r-1} x$, $dv=\sin x\,dx$. Then $du=(r-1)\cos x\,\sin^{r-2}x\,dx$ and we can take $v=-\cos x$. So we find that
$$A_r=\left.(-\cos x)\sin^{r-1} x\right|_0^{\pi/2}+\int_0^{\pi/2}(r-1)\cos^2 x \sin^{r-2} x\,dx.$$
Since $r \ge 2$, the first term on the right-hand side is $0$ (the cosine term dies at $\pi/2$, the sine term dies at $0$).  Rewrite $\cos^2 x$ as $1-\sin^2 x$. Then we obtain
$$A_r=\int_0^{\pi/2}(r-1)(1-\sin^2 x)\sin^{r-2} x\,dx=(r-1)A_{r-2}-(r-1)A_r.$$
From this we obtain
$$rA_r=(r-1)A_{r-2},\quad\text{or equivalently}\quad A_r=\frac{r-1}{r}A_{r-2}.\tag{$1$}$$
Now for an integer $r$ there are $2$ quite different cases, $r$ odd and $r$ even. Treat them separately. 
After a while, you will be able to find a general formula. But note that our recurrence $(1)$ gives $A_r$ in terms of the "simpler" $A_{r-2}$. Then we can use the recurrence to express $A_{r-2}$ in terms of the simpler $A_{r-4}$, and so on.  I suggest you first handle the cases $r=1$, and $r=0$. For example, it is easy to verify that $A_1=1$. 
Now let's deal with odd $r$.  It is best to deal first with a concrete case. Let's for example find $A_7$. By the recurrence $(1)$, we have $A_7=\frac{6}{7}A_5$. But again by recurrence $(1)$, $A_5=\frac{4}{5}A_3$. But $A_3=\frac{2}{3}A_1$. So $A_7=\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}$.
Now you should be able to see what happens in the general case $r$ odd.
For $r$ even, go through a similar procedure for say $r=6$ or $r=8$ to see what's going on. Then you should be able to handle the general case.
