Question

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^(n-2)dx where n>=2 is an integer.

(b)Use part (a) to evaluate integral from 0 to pi/2 of sin(x)^3dx and integral from 0 to pi/2 of sin(x)^5dx.

(c) Use part (a) to show that, for odd powers of sine, integral from 0 to pi/2 sin(x)^(2n 1)dx is (2*4*6*...*2n)/[3*5*7...*(2n 1)].

Answer 1

The reduction formula shows how to express integrals of sine functions raised to the nth power in terms of integrals with lesser powers. Specific integrals of sin(x)^3 and sin(x)^5 can be evaluated using this method, and a general form for odd powers of sine can be determined.

Reduction Formula for Sine Integrals

The reduction formula is used to determine the integral of sine functions raised to a power, specifically when dealing with the nth power of the sine function. The reduction formula for the integral of sin(x)^n from 0 to pi/2 is shown by integrating by parts and using the recursive relationship, which is:

integral of sin(x)^n dx = (n-1)/n * integral of sin(x)^(n-2) dx, where n ≥ 2 and n is an integer.

Evaluation of Specific Sine Integrals

Using the reduction formula:

1. The integral from 0 to pi/2 of sin(x)^3 dx evaluates to 2/3 * integral from 0 to pi/2 of sin(x) dx.
2. The integral from 0 to pi/2 of sin(x)^5 dx evaluates using the same approach and further reductions.

General Form for Odd Powers of Sine

For odd powers of sine, the integral from 0 to pi/2 sin(x)^(2n+1)dx follows a pattern and is equal to (2*4*6*...*2n)/[3*5*7...*(2n+1)], which can be derived from the initial reduction formula iteratively applied.

Answer 2

Hello,

a)

`I= \int\limits^{ (\pi)/(2) }_0 {sin^n(x)} \, dx = \int\limits^{ (\pi)/(2) }_0 {sin(x)*sin^(n-1)(x)} \, dx \\ = [-cos(x)*sin^(n-1)(x)]_0^ (\pi)/(2)+(n-1)*\int\limits^{ (\pi)/(2) }_0 {cos(x)*sin^(n-2)(x)*cos(x)} \, dx \\ =0 + (n-1)*\int\limits^{ (\pi)/(2) }_0 {cos^2(x)*sin^(n-2)(x)} \, dx \\ = (n-1)*\int\limits^{ (\pi)/(2) }_0 {(1-sin^2(x))*sin^(n-2)(x)} \, dx \\ = (n-1)*\int\limits^{ (\pi)/(2) }_0 {sin^(n-2)(x)} \, dx - (n-1)*\int\limits^{ (\pi)/(2) }_0 {sin^n(x) \, dx\\`

`I(1+n-1)= (n-1)*\int\limits^{ (\pi)/(2) }_0 {sin^(n-2)(x)} \, dx \\ I= (n-1)/(n) *\int\limits^{ (\pi)/(2) }_0 {sin^(n-2)(x)} \, dx \\`

b)

`\int\limits^{ (\pi)/(2) }_0 {sin^(3)(x)} \, dx \\ = (2)/(3) \int\limits^{ (\pi)/(2) }_0 {sin(x)} \, dx \\ = (2)/(3)\ [-cos(x)]_0^{(\pi)/(2)}=(2)/(3) \\`


`\int\limits^{ (\pi)/(2) }_0 {sin^(5)(x)} \, dx \\ = (4)/(5)*(2)/(3) \int\limits^{ (\pi)/(2) }_0 {sin(x)} \, dx = (8)/(15)\\`

c)


`I_n= (n-1)/(n) * I_(n-2) \\ I_(2n+1)= (2n+1-1)/(2n+1) * I_(2n+1-2) \\ = (2n)/(2n+1) * I_(2n-1) \\ = ((2n)*(2n-2))/((2n+1)(2n-1)) * I_(2n-3) \\ = ((2n)*(2n-2)*...*2)/((2n+1)(2n-1)*...*3) * I_(1) \\\\ I_1=1\\`