Jim Coroneos’ 100 Integrals (021-040)

Both the numerator and denominator of this function strongly suggest that the way ahead is to make a trigonometric substitution. More than that, both would become considerably simpler if x = cosu ... the numerator, because that is the inverse of cosˉ¹x, and the denominator, because we would then be utilising the trigonometric identity sin²u + cos²u = 1.

Before making this substitution, however, I decided to use this video to discuss the importance of always noting the domain of a function (and the limits of a definite integral). I had thought to leave this as an exercise for you to do yourself but, upon reflection, this particular integral provided too good an opportunity to be missed! So, if you wish to bypass all my discussion concerning domains, you might jump to about 6:10 in the video (which is where I actually begin to evaluate this integral).

You will readily see that the original function only makes sense if x lies between -1 and 1 (believe it, or not, YouTube does not allow "less than" symbols in the description!). When we substitute x = cosu , or u = cosˉ¹x, we note that u must lie between 0 and π. This becomes very relevant when, in simplifying the integral after substitution, √sin²u = |sinu|. Normally, in dealing with an indefinite integral, we would need to take account of the domain of u that would cause sinu to be positive, and that which would cause it to be negative ... and present the two separate cases. In this integral, however, because we now know that u must lie between 0 and π, we observe that sinu must therefore always be positive! This means that we do not have to use the absolute value sign (nor take account of any negative value for sinu).

The rest of the integration proceeds very smoothly indeed. This is quite a lovely little integral, and I hope that my discussion of domains has inspired you to be very careful in noting them in your work. Sadly, in order to keep my videos shorter, I have opted NOT to discuss domains for most of the integrals in this series. I say this is sad because, if you are to integrate properly/thoroughly, then you should ALWAYS be watching out for this detail! This is certainly true at university level. In school courses, however, we are often more concerned that you simply learn the mechanics of HOW to integrate without also demanding that you check the CONDITIONS under which your integral is valid!

You should realise that some of our key standard integrals have the sum of or difference between squares in the denominator, (and some of them are inside a radical as well). By multiplying the numerator and denominator by √(x+1) we achieve something approaching a standard integral form ... because we will have √(x²-1) as our denominator. After separating the two parts of the numerator, the rest of the integration proceeds swiftly.

At this stage, let me recommend that you don't simply rely on learning or keeping a list of standard integrals (as useful as that may be). You will learn some very good skills if you set study time aside to derive each of the standard integral forms. They are often not trivial! That is why we memorise the result, and use it! However, you will learn some good mathematics and develop enhanced integration skills if you take the trouble to do this.

At first glance, the prospects of integrating the cube of a logarithm IN THE DENOMINATOR would not excite us, but we then notice the 1/x term. This, of course, is the derivative of lnx. A clue! We pounce on it ... and the integral reveals its secrets almost without effort!

Normally, we would evaluate an integral such as this using 'substitution.' In this video, however, I take the opportunity to share with you a different (but related) way to show your working. I hope you like it. Please let me know if it is new to you.

There is a variety of ways to evaluate integrals of powers of trigonometric functions. This is one of them.

We remember that the derivative of tanx is sec²x. We also remember that tanx and secx are linked by a trigonometric identity ... sec²x = 1 + tan²x. Separating the fourth power into two squares allows us to use both of these tools, and the rest of the integration proceeds quite smoothly.

We have now reached the 24th integral in Jim Coroneos' list of 100 integrals! One more, and we will have evaluated one quarter of them. I hope, by now, you can see why the list has been such a challenge for better students. It is a significant test of stamina, skill and speed/finesse/insight!

With this video, we are now just 25% of the way through Jim Coroneos' list of 100 integrals. This time, we are challenged to evaluate ∫1/[x²(1-x)].dx.

I hope you noticed (almost immediately) that we have, here, a ratio/fraction composed of two polynomials. I.e. there are no trigonometric or exponential or logarithmic or other 'strange' functions. I hope you also quickly notice that the polynomial in the numerator (1) is of lower degree than that in the denominator. This indicates that this expression can be simplified using 'partial fractions' IF the polynomial can be factorised (we are thinking real numbers here).

In this case, the task is made easier for us, because the polynomial is already factorised! Once that step has been completed, the rest of the integration proceeds without incident.

By now, you can see what a challenge it is for young mathematicians at school to evaluate all 100 integrals. It is not a trivial task!

The twenty-sixth problem is to evaluate ∫1/[x²(1+x²)].dx

We immediately observe that we have a fraction made up of two polynomials and that the degree of the denominator (4) is considerably greater than the degree of the numerator (0). More than this, the denominator has already been factorised for us! This strongly suggests that the 'way ahead' is to use partial fractions ... at least, I hope you automatically thought "partial fractions!"

As I demonstrate, by separating this expression into two, simpler, fractions we are able to reduce the expression to much smaller, and more manageable, expressions. Once that step has been completed, the rest of the integration proceeds without incident.

We do rely on a "standard integral" but I decline to evaluate it "step-by-step" here in order to save time. I will devote a later series of videos to deriving many of those standard integrals. If you have special requests, please let me know.

Here is the solution to Jim Coroneos' 27th integral, ∫1/(1+x²)².dx.

The solution is a little involved but, for experienced mathematicians, relatively straight-forward. I will explain why.

Since we have a denominator that contains the sum of squares (1+x²), our first step in evaluating this integral will be to use the trigonometric substitution, x = tanθ. This is because we will be using the trigonometric identity 1 + tan²θ = sec²θ to compress the (1+x²) into a single square (sec²θ). This substitution has the added benefit that the derivative of tan²θ is also sec²θ. In the video, I take time to explain what to look for to decide whether a trigonometric substitution is likely to be of use.

Jim's 28th Integral (∫tan³x.dx) is a delightful one.

It looks deceptively difficult, but it can be evaluated in just a few lines of work when we realise that tanx has a remarkable property! Not only is d/dx(tan x) = sec²x, but we also have an identity that links tanx and sec²x ... namely, 1 + tan²x = sec²x.

Watch the video and see how this property of tan x allows us to 'unlock' this integral very rapidly.

Here we encounter, for the first time, an integral that we cannot readily evaluate unless we use "t-formulae" or "half-angle formulae."

As a general "rule of thumb," we should use t-formulae when we integrate a function that is in the form of a fraction containing a simple trigonometric ratio (and sometimes more than one). That is precisely what we have here in Jim's 29th integral ... a fraction containing a cosine ratio ... ∫1/(5 + 3cosx).dx

In a later video series I will be deriving and explaining how to use t-formulae in considerably more detail. In this video I simply use them to evaluate the integral. I hope you find it interesting.

Jim Coroneos' 30th integral looks similar to his previous integral. The only difference is that the 3 and the 5 have changed positions ... i.e. we now have ∫1/(3 + 5cosx).dx

Because this is also in the form of a fraction containing a simple trigonometric ratio, we first simplify its structure using "t formulae" or "half angle formulae."

Interestingly, this time we get a denominator which is a DIFFERENCE between squares. This means that, unlike the previous integral (which had a sum of squares that gave rise to an inverse tangent function), we need to create two terms in the integral using partial fractions. These produce two logarithmic functions that can then be combined into one rather interesting solution.

Jim Coroneos had a mischievous sense of humour!

This 31st integral provides us with an example of this. In the midst of his set of quite challenging integrals, he has inserted one that looks just as difficult as the others, but which can actually be evaluated in just two simple steps! This "keeps us on our toes" and reminds us that very tiny changes to functions/integrals can greatly alter their outcomes and can also change their difficulty level substantially.

I hope you enjoy evaluating his thirty-first problem ... ∫sinx/(5 + 3cosx).dx

You could be excused for thinking that Jim Coroneos' 32nd integral [∫1/(1 + cos²x).dx] would produce an inverse tangent function, given the sum of squares in the denominator.

You would, in fact, be correct! But, what havoc is that curious cos²x term going to wreak?

It turns out that the solution is quite elegant. We remember that 1/cos²x = sec²x and that, dividing numerator and denominator by cos²x will produce a sec²x term in both numerator and denominator. Because tanx has a delightful property, that it is associated with sec²x via its derivative and via an identity, we can use both these facts to simplify and evaluate the integral very rapidly.