Chapter 5.6, Problem 48E

a.

To determine

To show: that $I_{2n+2}\le I_{2n+1}\le I_{2n}$ .

Explanation of Solution

Given information:

The integral is $I_n={\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx}$ .

Formula used:

  ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx}=\frac{n-1}{n}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}xdx}$

Proof: to proof, use the reduce formula of sine and then simplify,

  $\begin{array}{l}{I}_{2n+2}=\frac{2n+2-1}{2n+2}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n+2-2}xdx}\text{ [substitute]}\\ I_{2n+2}=\frac{2n+1}{2n+2}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx}\text{ [simplify]}\\ I_{2n+1}=\frac{2n+1-1}{2n+1}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n+1-2}xdx}\text{ [substitute]}\\ I_{2n+1}=\frac{n}{n+1}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n-1}xdx}\text{ [simplify]}\\ I_{2n}=\frac{2n-1}{2n}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n-2}xdx}\text{ [substitute]}\end{array}$

Thus, $I_{2n+2}\le I_{2n+1}\le I_{2n}$ .

b.

To determine

To show: that $\frac{I_{2n+2}}{I_{2n}}=\frac{2n+1}{2n+2}$ .

Explanation of Solution

Given information:

The integral is $I_n={\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx}$ .

Proof: to proof, use the formulas then find ratio of them,

  $I_{2n}=\frac{1·3·5···\left(2n-1\right)}{2·4·6···2n}\frac{\pi }{2}$

  $I_{2n+2}=\frac{1·3·5···\left(2n+1\right)}{2·4·6···2n+2}\frac{\pi }{2}$

Now, divide the both,

  $\frac{I_{2n+2}}{I_{2n}}=\frac{2n+1}{2n+2}$

c.

To determine

To show: that $\frac{2n+1}{2n+2}\le \frac{I_{2n+2}}{I_{2n}}\le 1$ .

Explanation of Solution

Given information:

The integral is $I_n={\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx}$ .

Proof: from the part (a) and part(b) it can be observed that,

  $\frac{2n+1}{2n+2}\le \frac{I_{2n+2}}{I_{2n}}\le 1$