. (a) Use the reduction formula in Example 6 to show that ∫0^π/ 2 sin^ε x d x=(n-1)/(n) ∫0^π/ 2

. (a) Use the reduction formula in Example 6 to show that...

. (a) Use the reduction formula in Example 6 to show that $$ \int_0^{\pi / 2} \sin ^{\varepsilon} x d x=\frac{n-1}{n} \int_0^{\pi / 2} \sin ^{n-2} x d x $$ where $n \supsetneqq 2$ is an integer. (b) Use part (a) to evaluate $\int_0^{\pi / 2} \sin ^3 x d x$ and $\int_0^{\pi / 2} \sin ^5 x d x$. (c) Use part (a) to show that, for odd powers of sine. $$\int_0^{\pi / 2} \sin ^{2 n+1} x d x=\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot 2 n}{3 \cdot 5 \cdot 7 \cdot \cdots\cdot(2 n+1)}$$

. (a) Use the reduction formula in Example 6 to show that $$ \int_0^{\pi / 2} \sin ^{\varepsilon} x d x=\frac{n-1}{n} \int_0^{\pi / 2} \sin ^{n-2} x d x $$
where $n \supsetneqq 2$ is an integer.
(b) Use part (a) to evaluate $\int_0^{\pi / 2} \sin ^3 x d x$ and $\int_0^{\pi / 2} \sin ^5 x d x$.
(c) Use part (a) to show that, for odd powers of sine. $$\int_0^{\pi / 2} \sin ^{2 n+1} x d x=\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot 2 n}{3 \cdot 5 \cdot 7 \cdot \cdots\cdot(2 n+1)}$$

Key Concepts

Reduction Formula

A reduction formula is a recurrence relation that expresses an integral involving a power of a function in terms of an integral with a lower exponent. In the context of trigonometric integrals, the reduction formula simplifies evaluation by reducing the power of sine or cosine by using integration by parts, ultimately breaking down complicated integrals into simpler, more manageable components.

Recurrence Relations in Integration

Recurrence relations in integration provide a systematic way to relate an integral of a function with a given exponent to an integral with a lower exponent. This approach is useful especially for integrals of the form ? sin^n x dx, where the relation allows one to use previously computed integrals to find the result for higher exponents, significantly simplifying the calculation process.

Definite Integration of Trigonometric Functions

Definite integration of trigonometric functions involves evaluating integrals of sine or cosine raised to certain powers over fixed intervals, often [0, ?/2]. In this setting, the properties of the sine function and its symmetry within the interval can be exploited along with reduction formulas to obtain explicit expressions or closed-form results.

Product Representation for Integrals

For odd powers of sine, the evaluation of the definite integral can be expressed as a finite product of fractions. This product form results from repeatedly applying the reduction formula and represents the integral in terms of a ratio of products of even and odd numbers, akin to factorial expressions. This compact form is particularly useful for understanding the pattern and behavior of the integrals as the power increases.

Transcript

00:02 Hello, welcome to this lesson in the first place we would like to show that the integral from 0 to pi on 2 signs nx dx called to n minus 1 over n integral from 0 to pi on 2 sign n minus 2 d x okay but the whole of the whole of the integral of sine n x d x dx without any restriction without having the bounce is equal to negative one on n cost x sign n on 1 x dx plus n minus 1 on n integral sine n minus 2 x dx okay so here realize that realize that if you're taking the with bounce set that we have this to zero then the first place that we put we put pi on two in it cross pi on two is zero and in a second place where we put zero in it sign zero is equally zero so for whatever values of x that is the pi on two and zero one of these would always result to zero there by making the whole product zero okay so that means that this is ignored outrightly okay then we have only this path that works so we can say that that is n plus one n minus one on n then the integral sign n minus 2 x dx okay so the second part is to use the first part to evaluate the integral 0 to pi on 2, sine 3x d x.

03:12 So this is equal to the end here is 3.

03:20 So we have 3 minus 1 on 3.

03:23 Then integral sign 3 minus 1, which is a 3 minus 2, 3 minus 2, that is 1.

03:40 So we have x d x.

03:44 And this is 2 over 3.

03:51 So that is 2 over 3.

04:02 Negative cost x.

04:13 So you take it from pi on 2 to 0.

04:17 And this becomes 3.

04:21 2 on 3...