Prove that ∫0^2 π cos^2 n θd θ=(1 ·3 ·5 ⋯(2 n-1))/(2 ·4 ·6 ⋯(2 n)) 2 πwhere n=1,2,3, ….

Name: Problem 24, Chapter 5: Schaum's Outlines: Complex Variables
Uploaded: 2020-05-17T20:02:06-08:00
Duration: 6 min 33 s
Channel: Robert Daugherty
Description: Prove that ∫0^2 π cos^2 n θd θ=(1 ·3 ·5 ⋯(2 n-1))/(2 ·4 ·6 ⋯(2 n)) 2 πwhere n=1,2,3, ….

Prove that ∫0^2 π cos^2 n θd θ=(1 ·3 ·5 ⋯(2 n-1))/(2 ·4 ·6...

Key Concepts

Double Factorial Notation

Double factorial notation is used to represent products of integers with a step of two (for example, 1·3·5… for odd integers and 2·4·6… for even integers). This notation emerges naturally when expressing the result of integrals involving even powers of cosine, as it concisely captures the product of successive terms that appear in the simplified integral result.

Trigonometric Integrals

This concept refers to the process of integrating trigonometric functions, particularly powers of cosine, over a complete cycle. Understanding how even powers of cosine behave under integration over an interval such as [0, 2?] involves recognizing symmetry and periodicity properties, which simplify the integration process and often yield expressions that involve products or special functions.

Beta and Gamma Functions

The Beta function is an integral representation that is often used to evaluate integrals of the form ?sin^a(?)cos^b(?)d?. By relating the Beta function to the Gamma function through the identity B(p,q) = ?(p)?(q)/?(p+q), one can transform trigonometric integrals into algebraic expressions involving Gamma functions, which then simplify to yield standard forms of the result.

Transcript

00:01 Section 6 .1, problem 34, they're asking us to show that for even powers of sign, the integral from 0 to power of 2, sine of 2n, is going to be given by 1, 3 ,5, up to 2n minus 1, divided by 246 up to 2n, times pi over 2.

00:20 So in order to do this, let's go back to we were given a formula earlier in this section for the general reduction of the sine function.

00:30 So the sine to the n power of x d x is minus 1 over n, cosine of x, sine to the n minus 1 of x, plus n minus 1 over n, and the integral of the sine of n minus 2 of x d x.

00:53 So let's use that formula to evaluate the value for a sign of to the 2n power.

01:03 So the integral from 0 to pi over 2 of sine to the 2n of x dx is going to be equal to minus 1 over 2n cosine of x, sine to the 2n minus 1 of x.

01:27 This gets evaluated from 0 to pi over 2 plus n minus 1 over n, integral from 0 to pi over 2, the sine to n minus 2 of x d x.

01:46 So what happens when you evaluate the portion of the solution right here? you're going to at minus 1 over 2n.

02:03 And if you look at this, what happens when you evaluate this? when you evaluate at pi over 2, you're going to have a cosine of pi over 2 term.

02:17 That is always going to be 0.

02:19 And when you evaluate this at 0, you're going to have a sign of 0 term, which will always be 0.

02:26 So this term will always account to be 0.

02:29 So it tells me then that the integral from 0 to pi over 2, of sine to the 2n of x d x is equal to and sorry about this this was 2n it was 2n minus 1 over 2n so using i applied that reduction of formula wrong so when sign was equal to 2n here that would be 2n minus 1 over 2n so this is going to be 2n minus 1 over 2 n integral from 0 pi over 2 sign of 2 2n minus 2 of x d x...

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