Prove that, for even powers of sine, ∫0^π/ 2 sin^2 n x d x=(1 ·3 ·5 ·⋯·(2 n-1))/(2 ·4 ·6 ·⋯·2 n)

Prove that, for even powers of sine, ∫0^π/ 2 sin^2 n x d...

Key Concepts

Reduction Formula

This is a method in calculus used to simplify the integration of functions raised to a power by relating an integral of a higher power to that of a lower power. In the context of integrating even powers of sine, a reduction formula systematically reduces the power step-by-step until the integral becomes more manageable or matches a known standard form.

Double Factorial

The double factorial is a notation used to represent the product of numbers in steps of two. It is particularly useful for expressing products of either all odd or all even numbers up to a certain point. In the given identity, the product of odd numbers in the numerator and the product of even numbers in the denominator are both elegantly represented using double factorial notation, providing a compact expression for the integral result.

Beta and Gamma Functions

The Beta and Gamma functions are special functions that generalize the factorial function to non-integer values and are deeply connected to many integrals involving powers of trigonometric functions. When evaluating integrals of the form involving sine raised to an even power, the integral can be expressed in terms of the Beta function, which, in turn, can be rewritten using Gamma functions. This connection provides a powerful method for deriving closed-form solutions for these integrals.

Transcript

00:01 Given any positive integers n, we want to show the integral from 0 to pi over 2, sine 2 to the power n, sine to the 2 to the power n x dx is equal to 1 times 3 times, times 2 minus 1 over 2 times 4 times 2n, times pi over 2.

00:29 Let's begin with n being 1, integral from 0 to pi sine square x dx, can be written as 0 to pi over 2, sine x times d negative cosine x.

00:49 It's d negative cosine x is equal to sine x times dx.

00:55 So use the integration by part, we have negative cosine x sine x, you value pi over 2, 0 minus 0 to pi over 2, negative cosine x d sine x.

01:11 It's easy to say this term is 0 because when x is equal to pi over 2, cosine will be equal to 0, when x is equal to 0, sine will be equal to 0.

01:23 We can really specify, let's define it as i1.

01:26 We have i1, doing some cancellation, will be equal to integral from 0 to pi over 2 cosine square x dx.

01:37 By our previous definition, i1 is equal to 0 to pi over 2 sine square x dx.

01:46 Sum them all together, we have 2i1 is equal to, cosine square plus cosine square is equal to 1.

01:54 So this integral is pi over 2, which is equivalent to say i1 is equal to pi over 2, which just obeys our identity before.

02:10 Then by the induction, we know to prove this identity, we only need to show, so this is to show for any n, we have in plus 1 is equal to 2n plus 1 over 2n plus 1.

02:32 We want to show this identity, and then by the induction, everything can be done.

02:37 By the definition, inplus1 is equal to integral from 0 to pi over 2, x dx...

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