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 $r$.

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First write: $A_r = \int_0^{\pi /2} (1 - \cos^2 x) \cdot \sin^{r-2} x \, dx = A_{r-2} - \int_0^{\pi/2} \cos^2 x \, \sin^{r-2} x \, dx$ and for integration by parts take: $u = \cos x, \; v' = \cos x \, \sin^{r-2} x$. – qoqosz Jun 10 '12 at 16:09
Answer to your question is here. – Amihai Zivan Jun 10 '12 at 17:12

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.

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10x for help, I will try to solve the rest. – nour Jun 10 '12 at 19:19
For $r$ is odd, I got this general equation by reccurrence $A_r=\frac{r-1}{r}.\frac{r-3}{r-2}.\frac{r-5}{r-4}. \dots .\frac{r-7}{r-6}$ but for example if I plug in $r$=3 in this general equation I will get zero as an answer...similarly for other values of $r$...I don't know where is my mistake...any help? thanks – nour Jun 10 '12 at 20:59
You just have to know when to stop. It is the product $\prod_{i=0}^{(r-1)/2} \frac{r-2i-1}{r-2i}$. You can also write it as a product in the other direction, starting at the right end. – André Nicolas Jun 10 '12 at 21:10