A great integral calculus review in one problem!!

and we're going to do a classic integral video this one's going to be pretty tricky but only use elementary results that you might find in a calculus 2 class or an integral calculus class which i think makes this problem a great review for a course like that so what we want to do is find the antiderivative of 1 over the cube root of tangent x dx and we're going to start with like a standard u substitution so we want to look at this and see well what kind of u-substitution should we make well we'd probably like the stuff inside here to be something like a perfect cube so maybe we we would want u cubed to be equal to something in terms of tangent and obviously the first thing you would try is just tangent so u cubed equals tangent that's going to simplify everything nicely but i'll let you guys play with it and what you will see is that that turns this thing into a pretty gnarly integral it doesn't really help very much so what will make a nice simplification is if we let u cubed equal tangent squared of x so that'll have everything cancel off very nicely okay so let's see if u cubed equals tangent squared of x what does that tell us about tangent of x well it's not too hard to see that tangent of x is going to be u to the three halves so we're going to keep that around because we'll need it eventually okay so next what we want to do is maybe calculate the d u component in order to figure out what to do with this dx term so let's see taking the differential of both sides of this substitution equation will give us 3 u squared d u equals we need to take the derivative of this with respect to x we need to use the chain rule for that so that'll be 2 times tangent of x times secant squared of x again the tangent because of the power rule and then secant squared is the derivative of tangent and then we have dx but what we need is to know what dx is in terms of all of our u stuff so let's maybe make a simple solution for that by solving this equation and then we'll start to resubstitute so initially we'll have dx equals three halves then we'll have a big fraction with u squared in the numerator we have tan x in the denominator and secant squared of x also in the denominator and then a d u but now we'll need to substitute the tangent term and the secant squared term back into u's so let's see what we get for that so we'll have three halves u squared and like we said tangent of x is u to the three halves so that's that's good to know and then secant squared well what's secant squared well i want to recall that secant squared is the same thing as 1 plus tangent squared but since it's 1 plus tangent squared that's going to be 1 plus u cubed so i'll write this as 1 plus u cubed d u so now we can see that we'll substitute all of this in everywhere we see a dx up there so that'll turn this integral which has only x's into it into an integral that has only u's i guess we need one more thing here but that's not too hard to see and that is that the cube root of tangent here so let's write this the cube root of tangent of x is going to be equal to the square root of u by this equation right here or by that equation right there just as well okay great so now let's go ahead and make our substitution so i'll take this three halves out front and then we'll have the antiderivative of well in the numerator we've got a u squared and then in the denominator we'll have a square root of u i'll write that as u to the half we've got a u to the three halves and then we have finally a one plus u cubed d u so that's where we get all of our parts from our substitution next we can see that this u to the half and u to the three halves multiply together to get u squared so these two will cancel with this thing leaving us a one in the numerator further if you want to recall the sum of cubes formula this 1 plus u cubed can be factored like 1 plus u times 1 minus u plus u squared giving us some motivation for a partial fraction decomposition so this thing will be able to be decomposed via partial fractions as follows so we'll have the antiderivative of a over 1 plus u d u plus the antiderivative of b u plus c over 1 minus u plus u squared d u so that means we need to figure out how to take the partial fraction decomposition of 1 over 1 plus u cubed in terms of those two over there so let's maybe get rid of this calculation involving the substitution and we'll make this partial fraction computation so as a reminder i've left our original substitution on the board so we've got u cubed equals tangent squared of x now we want to do the calculation for our partial fraction decomposition so in other words we want to find a b and c that makes the following equation true so we have 1 plus u cubed is a over 1 plus u plus b u plus c over 1 minus u plus u squared we're going to do this calculation the standard way so i'll take this and multiply by something that will cancel out all of the denominators that'll be 1 plus u cubed then that gives us a polynomial equation then essentially we're actually using a fact from linear algebra about what is a basis for the space of polynomials so but you don't really need to know all those details so let's see if we multiply that 1 plus u cubed through we'll have a 1 on the left hand side then on the right hand side we'll have a times 1 minus u plus u squared because 1 plus u cubed factors like that so obviously the 1 plus u term will cancel and then we'll have b u plus c times 1 plus u that's what's left over for that second term so we've got a quadratic polynomial on the right hand side and well trivially a quadratic polynomial which has zero is the quadratic term and the linear term on the left hand side now what we'll do is extract coefficients from every power of u on both sides of the equation so i'll do that like this i'll say 1 for extracting the constant term on both sides of the equation u for extracting the u term and then finally u squared for extracting the u squared term so extracting the one term on the right hand side so let's see if we can eyeball that we'll have a from a times one and then we'll have also c from c times one so i'll have a plus c equals one so that gives us our first equation and then the one comes from the one on the left hand side so let's do that for u so let's see from this object we'll have minus a then here we'll have plus b and then plus c so bu times 1 and then c times u so here that's going to give us 0 because there's no u on the left hand side now we need to do the same thing for the u squared term so here we'll have a like that and then the u squared term will be b so a plus b equals zero so we've got a system of three equations and three unknowns so there's about a million ways to solve that let's just hack it together so this equation tells us that b equals negative a okay so that tells us that this equation is going to become let's see minus a plus minus a so that's going to be minus 2a then let's see if we can maybe use this as well so this tells us that c equals 1 minus a so we've got minus 2a plus 1 minus a equals zero like that okay so that's good but notice that gives us the equation three a equals one in other words a equals one third okay then notice that means that b equals negative one third from this equation right here oh and then finally we see that c is one minus a third so c equals two thirds so those are our three values of a b and c that make this partial fraction decomposition possible so that means over here we can go and replace this a with one third this b with negative one third and then this c with two thirds okay great but now we can do some simplification on that so first off notice that this 3 in the numerator can be cancelled down to a 1 by canceling this to a 1 this to a negative 1 and then this to a 2. so we've got something a little bit nicer okay well let's maybe clean that up and we'll have a fresh problem to work on okay so we're in a pretty good position we've got our goal integral is equal to one-half and then the difference of these two integrals that are in the u domain keeping in mind that this was our original substitution so now this one has a fairly simple anti-derivative so we just need to worry about this guy over here so let's see if we can do that now i'm going to split this up into two integrals that we can calculate in different ways and i'm going to do this like in kind of a trick but you'll see where it comes from so this is going to be minus one-half and then we have the anti-derivative of two u minus one over u squared minus u plus one so i flip that around so here notice that we've got minus half times two u well that's the same thing as minus u but now we need to worry about this part right here so at the moment i have a plus 2 overall in the numerator but i've only accounted for a plus one half so that means in order to have the rest of it i need a plus three halves so that means left over i have plus three halves and then the antiderivative of one over u squared minus u plus 1 d u and now i want to notice that if we carefully look at this this guy right here will break up into these two parts so the coefficient of u in the numerator is minus one and then the constant term in the numerator well that's going to be half plus three halves but that's that'll be two so that's good now next we can do another substitution on this guy right here keeping in mind that the derivative of the denominator is exactly the numerator so let's maybe make that substitution over here we'll let t equal u squared minus u plus 1 and then we'll see that dt equals 2u minus 1 d u so check it out this whole thing and the denominator will be t and then here we have our dt earmuffs in the numerator like that okay so let's maybe bring those first two integrals down this will be one half antiderivative of one over one plus udu minus half anti-derivative of dt over t okay those are fairly straightforward now let's see what we can do over here we have one over an irreducible quadratic which means we'll probably like to complete the square on that denominator in order to write it as an inverse tangent so that's the standard trick whenever you have an irreducible quadratic in the denominator so i'm going to take this u squared minus u and then split this one up into plus one quarter plus three quarters so you might think well how do i know the plus a quarter well to complete the square i take half the coefficient of this that'll be minus half i square it and i get plus a quarter so that means i need to split a quarter off of that one now obviously there's a bunch of ways of doing that but that's the way we'll explain it okay but now if we group these two we can see that we have this nice simplification u minus half squared plus three quarters now let's bring that down using this uh rewriting about our denominator so we have three halves anti-derivative of one over u minus half squared like that and then plus i'm going to write this three quarters in a fancy way i'm gonna write that as the square root of three over 2 quantity squared and then finally i have a d u out here now this might be like a standard anti-derivative but let's maybe do all of the steps just for completeness so i'm going to factor a square root of 3 over 2 squared out of the denominator and let's see what that gives us so just bringing the rest of these down i have one-half anti-derivative of 1 over one plus u d u minus one half antiderivative of t dt over t and then plus three quarters and then times one over well the this squared which will be 3 over 4 so we have that that's what we took out of the denominator so that means we've got that thing in the denominator and then the antiderivative of 1 over so this is going to be u minus half over the square root of 3 over 2 all of this squared plus 1 d u like that okay so finally i can take this guy right here and rewrite it a little bit so that'll give us what 2 u minus 1 over the square root of 3. so that's what we have inside of that square okay and then next we can simplify what we have here keeping in mind that 1 over 3 over 4 is the same thing as 4 over 3. so this 3 and this 3 will cancel and then this 2 and this 4 will cancel down to a 2 in the numerator then next i'll bring this line to the top and we're ready to finish it off so i brought this final integral we had at the bottom of the board to the top now we can finish this last one off with one last substitution i think we've suitably got the first two in terms that they are just like elementary anti-derivatives so let's see what we can do for this last bit so let's maybe say y will be equal to two u minus one over the square root of three well that is going to make d y equal to 2 over root 3 d u but that means d u is equal to root 3 over 2 d y so these are the two substitutions that will simplify this last integral so let's see i'll go ahead and bring this down we have a half square root of one over one plus u d u minus half antiderivative of dt over t plus 2 and then we can bring this square root of 3 over 2 out so this is the square root of 3 over 2. again that's from our d u component and then we have the anti-derivative of 1 over y squared plus one d y and now we're good to go this two cancels this two and we're left with one half natural log of one plus u minus half natural log of t but let's go ahead and substitute back in for t this so we've got u squared minus u plus one and then finally plus root 3 and this is going to be the arc tan of y but now we'll use this value for y that substitution so we've got 2 u minus 1 over root 3. okay great and then maybe plus a constant okay so finally we have to substitute back in for our original variable which is given up here so we have u cubed equals tangent squared of x so what does that tell us u is so u is going to be equal to tangent to the two thirds of x so that's kind of a bummer but that that's just the situation here right so let's see what we've got we've got that this is going to be equal to one half natural log of tangent to the two thirds of x plus one that's this first term minus half then we have the natural log of so that's going to be tangent to the four thirds of x minus tan two-thirds x plus one plus root three and then finally the inverse tangent so arctan of well so we're gonna have twice tangent to the two thirds x minus one over the square root of three and then here we've got plus a constant so that would be the final answer and that's a good place to stop

Channel: Michael Penn

Views: 41,100

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

Id: w27ZetqoRmg

Length: 19min 25sec (1165 seconds)

Published: Thu Nov 26 2020