7–84. Evaluate the following integrals.
19. ∫ from 0 to ...

Question

7–84. Evaluate the following integrals.

19. ∫ from 0 to π/2 [sin⁷x] dx

Accepted Answer

Welcome back, everyone. In this problem, we want to compute the integral of the sign to the 6th of XDX between the limits of 0 and 1/2 of pi. Now to help us compute this integral, we can use the reduction formula for the integral of sin to the end of XDX. Recall that that formula basically tells us that the integral, OK, are assigned to the end of XDX. is equal to -1 divided by N assigned to the N minus 1 X multiplied by the cosine of X, OK? plus N minus 1 divided by N multiplied by the integral of sin to the N minus 2 of XD X. OK? So in this case, we can use the reduction formula where N equals 6. Now, when N equals 6, OK. Then that means our integral, OK. Between the let me write it properly here. Between the limits of 0 and half of 5 of sin to the 6th of XDX is going to be equal. To -16 multiplied by a sin to the fifth of X multiplied by the cosine of x between the limits of 0 and 1/2 of pi. Plus, N minus 1, so that's 6 minus 15 divided by N6 multiplied by the integral of sin to the 4th of X between the limits of 0 and 1/2 of pi, OK. Now if we think about it here for our first term, it's going to be zero at both limits because the cosine of half of pi is equal to 0. So that means we can focus on the second term. In other words, this is really going to work out to be 5/6 multiplied by the integral of sin to the 4th of XDX between the limits. However, notice that we are in a similar problem that we were before, just like sin to the 6th of XD X. So again, we can recursively use the reduction formula. In other words, applying it when N E equals 4, OK. Then 5/6 of, well, let's not keep our constant here. Let's just focus on the expression. So the integral of sin to the 4th DX between the limits of 0 and 2 of pi by the reduction formula will now be equal to -1/4 of sinq of X plus X, OK, between the limits. Plus 3/4 N minus 1 divided by N multiplied by the integral of sin square xDX between those limits, OK? And now we're in a bit of a better position here because in this case similar to the one before, for the first term, both expressions or sorry, the first term is 0 at both limits rather so that will leave us with 3/4 of sine squared XD X between the limits. However, here we can rewrite the sine squared of x as 1 minus + 2 x divided by 2. So now. If we take that, OK, focusing on again, our expression, let's just leave out a constant for no. If we can try to find the integral between 0 and a half of pi, so between our limits of 1 minus + 2 x divided by 2DX, then we'll be able to find the integral for a Sanskrit XDX and use it to help us recursively compute our integral for assigned to the 6th of X. Now if we think about it here, we're dividing both of these terms by 2. So we can rewrite this as a half of the integral of 1 DX between those limits. OK. Minus a half of the integral. Between the limits of costs to XD X. Now for our first term, OK, that's gonna be equal to 1/2 multiplied by X, OK? With, or sorry, between the limits rather. OK, minus a half of 2 X. Between the limits. I know for our first term. That means we're gonna have a half multiplied by. Half of pie, OK. And for the second term, the term at both limits is going to be 0, OK. So no, or, or rather the term here is yes, it's going to be 0. So in this case, that's gonna be that minus 0, which will be equal to 14th of pi. Now, where do we go from here? Well, remember that this value, 4th of pi is what would be the value that we are going to multiply by a half. Well, no, sorry, we have multiplied it by 2. That's the value that we are now going to multiply by 3/4, OK? So now when we think about it here, this would imply. And keep it in red. This would imply if we substitute back up the chain between the limits are integral for the sign of sign of X to the 4th DX, OK. Is gonna be equal to 3/4 multiplied by the integral of sin squared XDX which we just found to be a 4th of pi. And when we multiply that, we get 3/16 of pi. No, when we take the value for the sine of X to the 4th DX between the limits and plug it into our formula for the sign of sign to the 6th of X, OK. D X. Then no, remember that was 5/6 of the sin of x to the 4th DX, so that's no 5/6 multiplied by 3 pi divided by 16, and this will work out to 5 pi divided by 32. So by using the reduction formula recursively, the correct answer is 5 pi divided by 32. Thanks a lot for watching everyone. I hope this video helped.

7–84. Evaluate the following integrals.
19. ∫ from 0 to π/2 [sin⁷x] dx

Key Concepts

Definite Integral

Trigonometric Functions

Integration Techniques

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