Euler Substitutions of the Third Kind – Summary
Krystian Karczyński
Founder and General Manager of eTrapez.
Graduate of Mathematics at Poznan University of Technology. Mathematics tutor with many years of experience. Creator of the first eTrapez Courses, which have gained immense popularity among students nationwide.
He lives in Szczecin, Poland. He enjoys walks in the woods, beaches and kayaking.
Euler Substitutions of the First, Second, and Third Kind – More Than Enough
In previous posts, I showed how to use Euler substitutions in integrals of the type:
- Euler Substitutions of the First Kind (when a>0)
- Euler Substitutions of the Second Kind (when c>0)
In this post, we’ll deal with the third and final type of Euler substitution, which we can use when in the integral:
the quadratic polynomial
But before we get to the practical stuff, let’s notice that these three cases:
- First kind, when a>0
- Second kind, when c>0
- Third kind when there are two distinct roots
will allow us to solve any integral of the type:
In fact, even just the first and third kinds are enough.
Why?
The case when
But what about the case when a<0 (doesn’t fit the first kind) and the quadratic polynomial has one or no roots at all (doesn’t fit the third kind)?
Then its graph would look like this (remember from middle school – arms down):
or, if it had no roots at all, like this:
What’s the takeaway? That in both cases the quadratic polynomial would take negative values (except for at most one point), and I remind you, we’re calculating the integral:
Meaning, in the integrand, the quadratic polynomial is under the square root, and you can’t take the square root of negative values (we’re playing with real numbers, of course). So the domain of such a function would be at most one point, which is pointless, and we certainly won’t get such an example. Unless the professor is really sleep-deprived when making the test.
So, the case when a<0 and the quadratic polynomial
So, let’s get to the third kind of Euler substitution.
Euler Substitutions of the Third Kind
We have an integral:
where
where
The substitution we use here is:
We square both sides of this substitution, write the quadratic on the left in factored form (we know we can), divide both sides by
Finally, we substitute everything into the original integral and end up with a usually tedious rational integral.
Let’s get started.
Example
Our
We first calculate
We use the third kind of Euler substitution:
We square both sides:
We write the quadratic on the left in factored form (remember
We divide both sides by
We determine
We have
Returning to our first substitution, we have:
We substitute the determined
We have nicely determined
So we have determined:
, everything using variable
We simplify:
As expected, we get a really complex rational integral, which I won’t calculate.
Finally, it’s worth noting that…
Note on Euler Substitutions
We have an integral:
where:
- First kind, when a>0
- Second kind, when c>0
- Third kind when there are two distinct roots
It’s obvious that it can often be solved using one of the two Euler substitutions, or even any of them (when a>0, c>0 and at the same time
No problem, although for ease of calculation I would recommend using the first kind first, if that doesn’t work, then the second, and if that doesn’t work, then finally the third.
That’s all on using Euler substitutions, I hope it helps you in your studies, and as always, feel free to comment below the post.
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