Problem

Prove that, for even powers of sine, $$ \int_0^{…

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Problem 49 Hard Difficulty

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

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Video Transcript

Yeah. Hello. Welcome to this lesson in the first place. I would like to show that the integral from zero to pile into science. An ex d ex cult too and minus one of our end integral from zero to pi on to sign and minus two DX. Okay, but the whole of that Okay, The whole of the anti golf sign N x. Yes. Without any restriction without having the bounce is equal to negative one on n course X sign in on one x the x last 10 minus one on and integral sign and minus two x dx. Okay, so he realized that, uh, realize that if you're taking the integral with bounce Yeah, set that we have this 20 then? Yeah, the first place that we put we put down to connect course piling 20 in a second place where we put zero in it. Sign zero is equally zero. So for whatever values of X, that is the pie on two and zero. One of these would always resort to zero, thereby making the whole product zero. Okay, So that means that this is ignored outwardly. Okay, then we have only this path that works. Yeah. Mhm. So we can say that that is n plus one and minus one on end than the integral sign and minus two x dx. Yeah. All right. Okay, So the second part is to use the first party value the integral zero to buy on two Sign three x, the eggs. So this is equal to the end. Here is three. So we have three minus one on three, then integral Sign three minus one, which is the third minus two third, minus two. That is one mhm. So we have X dx, and this is to over three. So that is Yeah. So to over three. Negative, of course. Yeah, X. So you take it from pi on 2 to 0, and this becomes three. 22 on three. They get to Of course, I own two, then minus costs zero because minus minus. Mix up. Last. So that is two on three. Cause pound to zero. Then that is CO zero. Because there is one. Yeah, so last one. And that gives us two on. Sorry. Okay, So let's use the same thing to evaluate. Integral from zero to pi on two. Sign five X, The X. So this would be five minus one on five, then the integral from zero to apply on to sign five minus two. That is three x dx. So that is four on five. Then the whole of the integral from zero to pound to sign three x signed to the past three x. The X has been evaluated US two on three. So this gives us eight on 15. Okay. All right. So, looking at the pattern, looking at the pattern we the CPAP will be next to look at this. Mhm. Okay, so here we have two. And last one or minus 1/2 and last one, then the integral for pie from zero to pi on to sign two n plus one minus two. Mhm. Okay, so this always yelled to n last one, minus 1 to 2 n over two and last one. Yeah. Then it's multiples. Okay, So you always find out that you have the end starting form the end that is greater than or equal to two. Okay, so this test from to us in a case of any calls to three. So this start from two king, then it goes to four. As you saw in ex, uh, And when n was five. And it goes to 61 n is seven. Okay, Was the nature we saw start at three. Then 25 If it goes in, that sequence would have a seven and a nine. So have s three, five, seven and on and on. Okay, Mhm. So this proves what we are looking for. Thanks for your time. This is the end of the lesson, Yeah.

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Problem

Prove that, for even powers of sine, $$ \int_0^{…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kayleah Tsai

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kayleah Tsai

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Kayleah Tsai

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals