from the MIT integration bee! 🐝

here we're going to look at an integral from the 2006 mit integration bee we're going to do this via a straightforward substitution but our solution will give hints towards a trick which is a little bit more elegant but it's detailed in math stack exchange post so i'll let you guys look at that if you're interested okay so we've got this integral which i'll call i and it's a definite integral from 0 to pi over 4 of cosine of x plus sine of x over 9 plus 16 sine 2x dx okay so let's maybe get to it i'll copy this thing up here so i've got the integral from 0 to pi over 4 sine x plus cos x over 9 plus 16 sine 2 x dx and like i said we're just going to do straightforward substitutions kind of follow our no's not thinking that we've worked on this problem before and so we don't know what the tricks are so maybe the first thing that we would like to do is make some sort of u substitution for the denominator so that's a common first step so let's go ahead and do that and see where that gets us so let's set u equal to 9 plus 16 times sine 2x now let's calculate the derivative because we'll need a du component if u is the denominator so let's see d u is going to be well the derivative of 9 is 0 a 2 is going to come out here and we'll have 32 times cosine because the derivative of sine is cosine and then 2x dx okay great but notice this is not at all what looks like we have in the denominator so this is not super helpful at the moment but perhaps we can use some trig identities to help us we've got two trig identities down here those are the double angle identities for sine and cosine we've got cosine here so let's maybe go ahead and use that we've got cosine of 2x equals cos squared of x minus sine squared of x so that means we can rewrite d u as 32 times cosine squared of x minus sine squared of x dx but notice we've got a difference of squares now so we might as well factor that difference of squares that's going to give us 32 and then we'll have cosine of x plus sine of x times cosine of x minus sine of x dx but that's actually pretty helpful because notice we've got a sine x plus cosine x up here that can be gobbled up into this guy right here okay so let's see where we can go from there well maybe we could also write something in terms of u using this thing right here um [Music] so maybe we'll like square this and take the square root and i want to notice that doing that will not really mess anything up because we're on the interval 0 to pi over 4 and since we're on the interval 0 to pi over 4 the cosine is always bigger than the sine so we don't need to get any absolute values involved so let's see we've got this is equal to 32 and then we have cos x plus sine x and then we'll have the square root of well this cosine squared well cosine minus sine squared but multiplying that out will give us cosine squared plus sine squared minus 2 cosine times sine so let's maybe write that down cosine squared plus sine squared that's 1 minus 2 cos x sine x like that and then we'll have dx okay good but let's see we can go a little bit further because we know that this 2 cosine x sine x is actually sine 2x so let's maybe go ahead and write that we have this is equal to 32 cos x plus sine x and now we're going to have the square root of 1 minus 2 sorry 1 minus sine 2x good but now let's look at our substitution that we have originally so our substitution that we have originally has sine 2x in terms of u essentially so notice maybe we could plug that in to our substitution way down here by solving for sine of 2x so if we solve this for sine of 2x let's see what we're going to get we're going to get 1 over 16 so that's what we get after dividing by 16 which will kind of be the second step and then we'll have times u minus 9. great so now let's see how this works so by our substitution this entire denominator is going to become u but then our sine x plus cosine x dx which i'll underline over here in blue can be replaced with this du divided by all of these things that i have over here in purple okay so let's see what that gives us so this is going to be a bit of a mess for a second so i've got an integral well i'll talk about the bounds in just a second and then i'll have 1 over u so that become that comes from the u that's in the denominator and then i'll also have a 1 over 32 so that comes from dividing this thing by 32 and then i'll have this 1 over the square root of well let's see what we have here we've got 1 minus this 1 over 16 u minus 9. so let's see if we can make that look a little bit better that's going to end up being 25 over 16. minus u over 16 and all of that is under the radical so let's talk our way through that we've got here one which is 16 over 16. and then we're subtracting a negative 9 over 16. so that gives us this 25 over 16. okay so now it looks like we're in good shape we can just put a d u here and now we need to talk about the bounds of integration so let's see when x is equal to zero well u is going to be equal to nine good and now when x is equal to pi over four u is going to be equal to well sine of pi over 4 times 2 is sine of pi over 2 sine of pi over 2 is 1 so we've got 9 plus 16. we've got 25 up here so that's the kind of thing that we have going on at the moment okay great so now let's maybe simplify this as much as we can so how much can we simplify well first let's maybe go ahead and take a 16 out of this denominator so that's going to give us the integral from 9 to 25 we can take a 16 out of that denominator it becomes a 4 but then that 4 in the denominator will cancel the 32 and give us an 8. i can bring that out of the integral so i've got 1 over 8 and then i have 1 over u times the square root of 25 minus u d u so now we'd really like the more complicated thing to be outside of the radical instead of inside of the radical so notice we've got u outside of the radical but 25 minus u inside of the radical that's a little bit more difficult than if we had the opposite so let's make another substitution let's say we'll let t equal 25 minus u that means d u is equal to minus dt like that and then that also means that u is equal to 25 minus t so when we do our change here let's see what we get we'll get 1 over 8 and then we'll have the integral of so it's going to be minus dt from the minus sign that we picked up from d u and then next we'll have 25 minus t times the square root of t now let's see what we have for our bounds of integration so our lower bound of integration sorry our upper bound of integration will be zero because if we plug in u equals 25 we get 0 and our lower bound of integration will be 16 25 minus 9 is 16. okay good now i'm going to go ahead and take this minus sign and let it flip the bounds of the integration so we'll have 1 over 8 and then the integral from 0 to 16 of 1 over 25 minus t times the square root of t dt good now let's maybe bring this to here and we'll continue on so we've worked ourselves down to the following integral so i which was our original goal is now 1 8 the integral from 0 to 16 of 1 over 25 minus t times the square root of t d t now we're going to do one last substitution before we kind of start finishing this thing off and the substitution will be for this square root of t the motivation here is that the derivative of the square root of t puts a square root of t into the denominator then we can express this t as the square root of t squared so we'll be good to go there so let's say maybe y is equal to the square root of t notice that's equal t to the half that's going to make d y equal to one half t to the minus half but notice that's going to be equal to 1 over 2 times the square root of t like that and then i should put a dt here okay so let's see we can take this 8 and rewrite it as 4 times 2 and then this 2 along with this root t dt those are all gobbled up my by d y and then this t right here is going to be y squared okay so let's see what effect that has we're going to have one quarter so the quarter comes from canceling out the two it gotten eaten up by the d y and then next we're going to have the integral we'll do the bounds of integration in just a second then we'll have 1 over 25 minus y squared d y like that now let's see when t is 0 y is 0 when t is 16 y is 4. so this is the integral from 0 to 4 like that okay but now it looks like we could maybe do a partial fraction decomposition so let's maybe outline the partial fraction decomposition over here so we would like to take 1 over 25 minus y squared and probably rewrite it as a over y minus 5 plus b over y plus 5. so standard partial fraction decomposition for something like this so let's multiply both sides by 25 minus y squared so we can cancel both sides of the equation or maybe by 25 or y squared minus 25 that'll leave us with a minus 1 on the left-hand side of the equation it'll leave us with an a times y plus 5 on the right and a b times y minus 5 on the right as well so let's see now we'll have constant term equation and an equation from our linear term in other words our y term so now the constant terms on the right hand side of the equation are 5 a minus 5 b and so that's going to be equal to negative 1 because that's what we've got on the left hand side of the equation then the coefficients of y on the right are a plus b and on the left is zero okay cool so that's going to tell us immediately that b is equal to negative a but now we can plug b equals negative a back up here and let's say that means we're going to have 10 a equals negative 1 which means a equals negative 1 over 10 which means b equals 1 over 10 positive because they're of opposite sign so that means we have a way to decompose this integral using this partial fraction decomposition so let's see what that gives us we're going to have a quarter like that then we'll have the integral from 0 to 4 and then i've just rewritten this 1 over 25 minus y squared using these parts right here i can go ahead and factor a 1 over 10 out of the whole thing that'll give us a 1 over 40 in front of the whole thing 4 times 10 and then let's see that's going to leave us with a negative 1 over this y minus 5 and a positive 1 over y plus 5. then we'll have d y here now the anti-derivative of that is pretty straightforward it involves natural logarithms so let's see we'll have one over 40 and then we'll have the ln of the absolute value of y plus 5 minus the natural log of the absolute value of y minus 5. we need to go ahead and evaluate that from 0 to 4. so let's see what we get when we evaluate that thing at 0 we'll have the natural log of 5 minus the natural log of 5 that's going to cancel off when we evaluate that at 4 we're going to have the natural log of 9 from this term minus the natural log of 1 but the natural log of 1 is 0. so that ends us up with 1 over 40 natural log of 9. but now 9 is 3 squared so we can use some logarithm rules to rewrite this as 1 over 20 natural log of three and that's a good place to stop

Channel: Michael Penn

Views: 28,387

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Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: 0_Hun_qiwsQ

Length: 14min 48sec (888 seconds)

Published: Fri Nov 20 2020