In the next two videos I will present some methods
for integrating products of trigonometric functions.
Most calculus textbooks contain a section
that explains how to integrate functions of this type:
a product of a power of sine times a power of cosine,
or a power of secant times a power of tangent,
and sometimes even others.
I want to try to explain where the different methods and tricks come from,
so you don't have to memorize any of them,
and so that every time you're faced with integrating
a function that looks like this, or even of all the types,
you can figure out on the spot how to do it without having to resort to memory.
And I'm going to begin with one example.
I want to compute the antiderivatives of this function: sin⁵x cos²x dx.
I'm going to try to do this with substitution.
If I don't think much about it, there are at least two substitutions I could try.
Since the function involves sines and cosines,
perhaps I can try u = sinx, or perhaps I can try u = cosx.
In this video I'm going to attempt both and see which one is better.
Now, if I use sinx, of course I also need to transform the differential.
du will be cosx dx.
And luckily I do have a cosine, I have two.
So one of them will be used for the differential,
and then I have a second cosine left,
and I need to transform it back into sines to be able to complete the substitution.
I can use the main trig identity: sin²x + cos²x is always 1.
So I will use one cosine for the differential,
and with this identity I convert the other cosine into sine
so I can complete the substitution.
In my second attempt,
I'm going to start using cosine as my new variable for the substitution,
and then I need to change the differential. du will be -sinx dx.
And again,
I will need one sine for the differential.
We have five, so I will be left with four.
The other sines will be transformed into cosines using this identity.
Which one will work best? Well let's find out.
At this moment, I invite you to pause the video
and try to compute this integral using both substitutions separately.
If you do that, you will very quickly find out
that one of them works much better than the other.
See which one it is and then continue watching.
First, I will attempt the substitution u = sinx.
I'm going to need a cosine for the differential, and I have two.
So I'm going to separate the one for the differential
and leave the rest in the integrand.
This is the cosine I need for the differential.
Now everything else in the integrand I have to write it in terms of sine,
because that's my substitution.
The only thing I can think of here is let's use the main trig identity
to convert this cosine into a sine, and that will look like this:
cos²x is 1 - sin²x, so when I have a single cosine,
this will be +/- √(1 - sin²x),
and I still have this extra term here.
Okay, there, now I can complete the substitution.
This is +/-∫u⁵√(1 - u²)du.
And I'll pause here for a moment. Let's forget the plus/minus for now.
We would have to figure out what to do with it, but let's ignore it for now.
I have introduced a square root, and the purpose of integration methods is
always to transform the original integral into one that is easier to compute.
And in this case, I've done the opposite.
I don't think that ever introducing a square root is a good idea.
So I'm going to declare this not helpful.
Perhaps there is a way to complete the problem this way,
but it doesn't look promising,
because I ended up with something worse than what I started with.
Since I have a second substitution I can try, I'm going to do that first
and hope that the second substitution doesn't require me
to include a new square root.
I will now attempt the substitution u = cosx on the same function.
I need to separate a sine for the differential, so let's do that.
And to complete the substitution,
I now need to make all the remaining sines to become cosines.
Well, in this case it's easier than before.
I now have sin⁴x, and sin⁴x can be written as (sin²x)².
And now using the main trig identity, I can rewrite sin²x as 1-cos²x.
And there.
Here I can substitute now,
because everything except for what I need for the differential
is in terms of cosine.
So after substituting, I get -∫(1-u²)²·u²·(du).
And this was much better than in the previous example,
because in order to substitute I didn't need to include any square roots.
This new function is just a polynomial. Now I'm going to be able to finish it.
I can take this polynomial and expand it, multiply it out,
and then integrate it term by term.
And after integrating term by term, I will be able to undo the substitution.
So I won't finish it, but it should be clear
that this is going to work better than my first attempt.
We can summarize what we have learned, and we can get a little rule that tells us
when each one of the two substitutions is useful.
Let's say I want to compute an arbitrary integral of the form
a certain power of sine times a certain power of cosine.
When will the substitution sine and when will the substitution cosine help us?
Well, if I try u = sinx,
I'm going to need to get a cosine for the differential.
So I separate one cosine out of here,
and the remaining cosines I have to transform into sines
using the main trig identity.
For that to be easy, I want to have an even number of cosines, a multiple of 2.
So after separating one cosine, I want an even number of cosines left.
In other words, this is a good idea whenever m is odd.
Similarly, if I want to try the substitution u = cosx,
I'm going to need to separate one sine for the differential.
And the remaining sines have to be converted back to cosines
with this identity.
For that to be simple,
I want that after I separate one sine I have an even number of sines left.
In other words, I need n to be odd.
To finish this video, two comments.
First, what happens if we want to perform an integral like this
and both m and n are even?
Well in that case, none of these substitutions is very helpful;
none of them is a good idea.
And we need to try something else.
In the next video I will suggest a trick that can help in that situation.
And second, this is not a formula to be memorized.
If you understand the process I went through,
every time you're faced with computing an integral like this,
you can quickly do this in your head
and determine which one of the substitutions, if any,
is going to be helpful.
And if you can do that,
then now you can compute many more integrals then just these ones
because there are other families of integrals
for which the same ideas are going to apply.
For example,
let's say that we want to compute these three integrals.
They're all functions of the form a power of secant times a power of tangent.
Secant and tangent are related the same way that sine and cosine are
because their derivatives can be written in terms of each other.
So following the same ideas in this video, to compute these three integrals
there are two substitutions I may want to try.
Perhaps I want to use a substitution u = secx,
or perhaps I want to use the substitution u = tanx.
The derivative of secant is secx·tanx, and the derivative of tangent is sec²x.
So when will the first substitution help,
and when will the second substitution help?
Like for sine and cosine, there's a trig identity here I need to use as well,
sec²x = 1 + tan²x,
that allows me to convert secants into tangents and vice versa.
So can you use the same methods of this video
and figure out which one of these two substitutions, if any,
will help with each one of those three integrals?
That is your challenge.
In the next video, I will continue with integrals of this form,
and explain what to do when both n and m are even.