Use the result of Exercise 96 to prove the following...

Use the result of Exercise 96 to prove the following versions of Wallis's Formulas. (a) If $n$ is odd $(n \geq 3)$, then $$ \int_{0}^{\pi / 2} \cos ^{n} x d x=\left(\frac{2}{3}\right)\left(\frac{4}{5}\right)\left(\frac{6}{7}\right) \cdots\left(\frac{n-1}{n}\right) $$ (b) If $n$ is even $(n \geq 2)$, then $$ \int_{0}^{\pi / 2} \cos ^{n} x d x=\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{5}{6}\right) \cdots\left(\frac{n-1}{n}\right)\left(\frac{\pi}{2}\right) $$

Use the result of Exercise 96 to prove the following versions of Wallis's Formulas.
(a) If $n$ is odd $(n \geq 3)$, then
$$
\int_{0}^{\pi / 2} \cos ^{n} x d x=\left(\frac{2}{3}\right)\left(\frac{4}{5}\right)\left(\frac{6}{7}\right) \cdots\left(\frac{n-1}{n}\right)
$$
(b) If $n$ is even $(n \geq 2)$, then
$$
\int_{0}^{\pi / 2} \cos ^{n} x d x=\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{5}{6}\right) \cdots\left(\frac{n-1}{n}\right)\left(\frac{\pi}{2}\right)
$$

Key Concepts

Handling Even and Odd Powers

The integration of even and odd powers of cosine leads to distinct patterns and recursive multipliers. Recognizing and handling these differences appropriately is crucial, as it leads to the formulation of precise product formulas that differ by a multiplicative constant, as seen in the different cases of Wallis’s formulas.

Definite Integrals of Trigonometric Functions

Definite integrals, particularly those involving powers of cosine or sine over specific intervals such as [0, ?/2], are central in establishing relationships like Wallis's formula. These integrals utilize properties of trigonometric functions and symmetry to derive exact product formulas and other important results in calculus.

Reduction Formula

A reduction formula is a recursive relation that expresses an integral involving a function raised to a power in terms of an integral with a lower power. In the context of trigonometric integrals, it simplifies the process of evaluating ?cos^n x dx by reducing it step by step to a base case, ultimately leading to expressions like those in Wallis's formulas.

Wallis's Formula

Wallis's formula is a product representation for ? derived from evaluating integrals of even powers of sine or cosine over the interval from 0 to ?/2. It reveals an interesting connection between infinite products and definite integration and serves as a precursor to more advanced concepts in analysis, such as the gamma function.

Transcript

00:01 In this problem of trigonometric integration, we have to use the reduction formula for the proof of the following version of wally's formula.

00:13 So first is here part a if n is odd, that means n belongs to odd number and also n is greater than 3, then we have to prove that this is then we have to prove integration from 0 to pi divided with 2 of cost to the power n.

00:31 So this is cost to the power n x d x would be equals to 2 divide with 3 multiplied with 4 divide with 5 multiplied with 6 divide with 7 and so on up to n minus 1 divide with n so first we have to the reduction formula so reduction formula say that if we have the integration of say we can say this is an integration of cost to the power n this is cost to the power n, cost to the power nx d x.

01:10 And we have also the integration limit, say, from 0 to pi divide with 2, then this would be equals to cost to the power n minus 1 multiplied with sine x to the power, this is to the whole power.

01:24 And also this has to first divide with n.

01:27 And also we have to put the limit, say 0 to pi divide with 2 and this is plus n minus 1 divide with n an integration from the limit 0 to pi divide 2 and cost to the power n minus 2x d x so this is the required reduction formula now when we proceed this reduction formula so this value would be say after some simplification this value would be like this here put the value say cos, this is here also cos x to the power n minus 1.

02:10 Now putting the value x is equal to pi divide with 2, so this is equals to here 0 and if we put lower limits so this would be again 0, so sine 0 is 0.

02:20 So this term is equal to 0.

02:21 Now we have to integrate only this term because this term would give you 0.

02:27 Now if we integrate this term so here this value would be again n minus 1 divide with n.

02:34 Now we have to integrate this term.

02:36 So similarly as the previous one, so here this value would be cost to the power this time.

02:43 N minus 2 minus 1 so this is n minus 3 so this is n minus 3 and sign x this is also x divided with this is now n minus 2 so n minus 2 and also this value would be integration limit say from 0 to pi divide with 2 plus now here we have n minus 2 so we have to again subtract 1 so n minus 3 this value would be n minus 3 divided with n minus 2 now integration would be from 0 to pi divide with 2 and n minus 2 so this value would be n minus 4 so here n minus 4 x d x and this would be repeated so here n minus 1 minus divide with n and also when we put upper limit say this value would be 0 and if we put lower limit so this value would be 0 that means this term would give you again 0 so integration of this term would be here n minus 1 divide with n and multiplied with n minus 3 divide with n minus 2 and this would be continued up to it comes 1 and also we can reverse the order so if we reverse the order so here we would get say n minus 2 this is n minus 3 divided with n minus 2 so this value would be first this is 2 divided with 3 and this is 1 here and before that 4 divide with 5 and if we reverse the order so we can say integration would be he say integration of cost to the power n x d x from the limit 0 to pi divide with 2 would be equal to this is 2 divide with 3, 4 divide with 5, 6 divide with 7 up to this term, say this is n minus 1 divide with n.

04:46 So we can say hence proved.

04:48 So here this is proved.

04:51 Now in second part, so this is this solution for first part...

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