Chapter 5.6, Problem 37E

a.

To determine

Toshow: the reduction formula.

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}x}=\frac{n-1}{n}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}xdx}$ .

Proof: to proof, first use the reduction formula and then substitute,

Since, the reduction formula,

  $\begin{array}{l}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx}={\left[-\frac{1}{n}{\sin }^{n-1}x\cos x\right]}_0^{\frac{\pi }{2}}+{\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\left(n-1\right)}{n}{\sin }^{n-2}xdx}\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx}=0+{\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\left(n-1\right)}{n}{\sin }^{n-2}xdx}\text{ [substitute and simplify]}\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx}=\frac{\left(n-1\right)}{n}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}xdx}\end{array}$

b.

To determine

To find: the value of integral.

Answer to Problem 37E

The values of integral are $\frac{2}{3}\text{ and }\frac{8}{15}$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx}\text{ and }{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{5}xdx}$ .

Calculation: to find the value of integral, first substitute in the formula and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{3}x}=\frac{3-1}{3}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{3-2}xdx}\text{ [substitute }n=3]\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{3}x}=\frac{2}{3}{\left[-\cos x\right]}_0^{\frac{\pi }{2}}\text{ [integral]}\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{3}x}=\frac{2}{3}\left(0+1\right)\text{ [substitute and simplify]}\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{3}x}=\frac{2}{3}\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{5}x}=\frac{5-1}{5}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{5-2}xdx}\text{ [substitute }n=3]\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{5}x}=\frac{4}{5}\left(\frac{2}{3}\right)\text{ [simplify]}\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{5}x}=\frac{8}{15}\end{array}$

Thus, the integral values are $\frac{2}{3}\text{ and }\frac{8}{15}$ .

c.

To determine

To show: the reduction formula for odd powers of sine.

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n+1}dx}=\frac{2·4·6···2n}{3·5·7···\left(2n+1\right)}$ .

Proof: to proof, use the part (a) and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}x}=\frac{n-1}{n}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}xdx}\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n+1}x}=\frac{2n+1-1}{2n+1}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n+1-2}xdx}\text{ [substitute }n=2n+1]\\ {\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{2n+1}dx}=\frac{2·4·6···2n}{3·5·7···\left(2n+1\right)}\text{ [by expanding and simplify]}\end{array}$