Evaluate the integrals in Exercises 51–56 by making a substitu...

Question

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.

∫ (from 0 to 1/√3) dt / (t² + 1)^(7/2)

Accepted Answer

Welcome back everyone. In this problem, we want to compute the integral by a trigonometric substitution of DT divided by 1 + T2 to the power of 9 halves between the limits of 0 and 1/3 of root 3. Now, let's start by doing our substitution for trigonometric substitution in our integral, OK? So by substitution. Let's set t to be equal to 10 theta. Then this would imply that the derivative of t with respect to theta is going to be equal to sex squared theta. Which means then. OK. Which means that DT. DT is going to be equal to sex square theta d theta. And 1 + t2 will be equal to 1. Plus Time square theater, which we know that identity is sex squared theater. Now we've been able to substitute DT in terms of theta, T in terms of theta, and our expression 1 + T2 in terms of theta. So all we need to do now is to replace our limits. What do we know? Well, not that. OK. Theta is equal to the inverse tangent of T. And based on that, we know then that when T equals 0. Then that means theta will be equal to the inverse tangent of 0, which equals 0. And when T equals. A third of Route 3, OK. Then theta is going to be equal to the inverse tangent of 1/3 of root 3, which equals 1/6 of pi. So these are going to be the replacements for our limits in terms of theta, where since theta equals the inverse tangent oft, we know the principal range of the inverse tangent function is from -12 of pi to 1/2 of pi. So now we have everything we need to substitute it into the integral. So now this means then that the integral, or sorry, or definite integral from the problem statement in terms of T. OK. Is going to be equal to the integral between 0 and 6 of pi. OK, of 6 squad theta, D theta. Divided by 6th grade theater. To the power of 9 halves. And that when we simplify here is going to be equal to. Sex square theta with respect to theta divided by sex theta to the 9th. OK, I know when we simplify that. Then we eventually end up with the integral between 0 and 1/6 of pi. OK, I, I fixed our first integral here. This should have been a 6th of pi, 0 and 6 of pi of the cosine of pi to the 7th with respect to theta. Now where do we go from here? Well, rewriting our integrand to make it a bit simpler to integrate, OK. Then We know that the integral between 0 and 1/6 of pi of the cosine of theta to the 7th with respect to theta can be rewritten as the definite integral of cus theta to the 6th multiplied by cus theta with respect to theta. And that can further be rewritten. As Cos square cus square theta cubed. Multiplied by cost theta with respect to theta. Well, why would we do that? Because we know that costs squared theta is really 1 minus sine square theta by its identity. So this is going to be 1 minus sine squared theta cubed multiplied by cost theta with respect to theta. And now we can further solve this by using some more substitution. So in this case, OK, let's let you, let me put this in black here. Let's, again, let's let you be equal to the sine of theta, OK? Then that means. That the derivative of u with respect to theta will be equal to the cosine of theta, which means then that duu will be equal to cus theta d theta. So now we have expressed uh you in terms of, sorry, sine theta in terms of you, uh DU in terms of the theta, again, like before we need to rewrite the limits. Well, if you equals sine theta, OK, then when you, sorry, when theta equals 0. Then you will be equal to the sin of 0, which is 0. And when theta equals 1/6 of pi. OK, then you will be equal to the sine of 1/6 of pi, which is going to be equal to 1/2. So now that we have this, OK, then we can substitute it into the integral and then expand. So if we take our integral. Which we know it was uh the integral between 0 and 1/6 of pi of 1 minus sin square theta, the cubed multiplied by cost theta with respect to theta, it's going to be equal to the integral between 0 and 5. Let me put this in red here. To show a new substituted integral between 0.5 of 1 minus u 2. Cubed, OK. Multiplied by the u. And now, in this case. We could, if we expand here our cube. Then we could rewrite this. As 1 minus 3U 2 plus 3U 4 minus u 6th with respect to you. I know we can integrate using the power root because if we integrate each term, yeah. Then the integral of 1 with respect to u will be u for -3 u squared, it will be -3 u cubed divided by 3. For 3 u to the 4th, it will be 3 u to the 5th divided by 5. And for u to the 6th, it will be u to the 7th divided by 7th. And all of these are between the bones of 0 and 1/2. If we simplify it even further, it's going to be U minus you, well, we could even resimplify it in this expression here, because 3 cancers, 3, so we're just left with U cubed, OK. And now we know if we substitute 0 into our expression, the entire expression will turn into 0. So we can substitute a half as the upper bone. So this is going to be a 12 minus 12 cubed. Plus 3/5 of 125 minus 1/22. Sorry, minus 12 1/7. Of a half. To the 7th, yeah. I know when we start to simplify here, OK, then we could rewrite this as a 12 minus 1/8, OK, plus 3 divided by 160 minus 1 divided by 896, which combining it into the same denominator would be 2240 minus 560 plus 84 minus 5, all divided by 4480. Which is equal to 1,759 divided by 4480. Therefore, this is the integral of dt divided by 1 + t 2 to the power of 9 halves between the bones of 0 and 1/3 of root 3. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.∫ (from 0 to 1/√3) dt / (t² + 1)^(7/2)

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Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.
∫ (from 0 to 1/√3) dt / (t² + 1)^(7/2)