0:09

In our lesson on integration by substitution, the question remains,

which substitution should I make?

Well, that's not always an easy question to answer.

It's a bit of an art form to know exactly what to substitute.

In this lesson, we'll focus on a class of integrals that are amenable

to a trigonometric substitution.

0:34

Recall from a previous lesson that when we apply the substitution formula

to compute an integral, we are typically going from the more complicated

looking expression to the simpler looking expression.

However, any time you have an equality in mathematics,

it's a road that runs both ways, and

we can try exploiting it in the opposite direction.

That is what we're going to do in this lesson.

This type of substitution is especially useful with trigonometric functions.

We'll look at a trivial example,

consider the integral of dx over 1 + x squared.

You may know this one already.

Whether you do or whether you don't, consider the following substitution.

We're going to let x be equal to tangent of theta.

Notice we're not defining theta in terms of x, we're defining x in terms of theta.

Then, dx would be secant squared theta d theta.

Unlike what we do when we perform use substitution,

here we directly substitute these in, and we get

the integral of secant squared theta d theta over 1 + tangent squared theta.

2:01

Now of course, recalling that tangent squared + 1 is secant squared,

we obtain a secant squared in both the numerator and the denominator.

These cancel, revealing the integral of d theta, which is simply theta + C.

Now, we have to substitute back in for x,

which involves inverting that substitution,

we obtain simply arctan of x.

2:33

Now the various trigonometric identities that you may have encountered

in the past are what dictate these trigonometric substitutions.

Some of these you're going to want to have memorized.

Cosine squared + sine squared = 1, and the version with tangent and secant.

Not all trigonometric identities need to be memorized, but the more you have

in your head, the easier it will be to see which substitutions to use.

Hyperbolic trigs are also useful, and you should at least have seen

some of the basic simplification formulae for hyperbolic trigonometric functions.

3:20

Let's compute some examples.

Consider the integral of the square root of 1- x squared dx.

Let's substitute for x the function sine theta.

So that dx is cosine theta d theta.

This yields the integral, the square root of 1- sine squared theta,

times cosine theta d theta.

And now we see why this is the right choice for x.

Because using the fact that sine squared + cosine squared = 1,

we can replace 1- sine squared with cosine squared.

Hence, eliminating the square root to yield the integral of cosine squared.

This too, after a trigonometric identity,

gives the integral of 1/2 + 1/2 cosine 2 theta.

This is something we can integrate.

1/2 integrates to 1/2 theta,

1/2 cosine 2 theta integrates to 1/4 sine 2 theta.

4:33

Now, to substitute back in and obtain a function of x, what do we do?

1/2 theta becomes 1/2 arcsine of x.

What about that second term?

Well, we have a 1/4 times sine 2 theta.

The double angle formula for sine allows us to replace this

with 2 times 1/4, is 1/2, times sine theta,

which is x, times cosine theta.

What is cosine theta in terms of x?

Well, if we set up a right triangle with an angle theta whose sine

is equal to x, then we can read off the cosine

from the Pythagorean theorem as square root of 1- x squared.

This yields our answer, 1/2 arcsine of x + (1/2)x times square root 1-x squared.

5:51

I think that there was something involving tangents and

secants that might allow us to get rid of that square root.

So, let's try x = tangent of theta.

Substituting in that, and secant squared theta d theta for

dx yields something that looks a bit complicated.

6:15

Now I hoped to use the fact that tangent squared

+ 1 = secant squared to eliminate that square root.

However, I don't have tangent squared + 1, I have tangent squared + 4.

And I can't eliminate that square root.

6:35

What I should have done was chosen my coefficients more wisely.

If instead of letting x be tangent theta, if we let x be 2 tangent theta,

and replacing dx with 2 secant squared theta, that

puts a 4 in front of the tangent squared terms.

And now, I can factor that 4 out from under the square root,

obtaining a 2, cancelling, simplifying a little bit.

And now I can replace tangent squared + 1,

with secant squared, and eliminate that square root.

When I do so, I get 1/4 times the integral of secant squared d theta,

over tangent squared times secant.

I can eliminate the secant in the numerator and the denominator, and

simplify what remains, rewriting in terms of sines and cosines.

And we obtain 1/4 the integral of cosine over sine squared.

Now this integral is not gonna be so bad, we've done something just like it before.

We get after substituting u = sine theta,

1/4 the integral of du over u squared.

This is -1/(4u) + C.

Substituting back in for theta,

we get -1/4 cosecant theta + C.

Oh wait, we're still not done.

We need our final answer in terms of x.

So again, let's set up a right triangle,

something for which the tangent of theta is x/2.

Then, we can compute the cosecant of theta.

As square root of x squared + 4 over x,

putting the minus sign in front and adding a constant gives us our answer.

8:47

It is sometimes the case that a bit of algebraic simplification can

reveal the appropriate substitution to be used.

Consider the integral of dx over the square root of 3 + 2x- x squared.

This seems completely opaque.

However, if we use the method called completing the square,

something wonderful happens.

We're going to rewrite the term under the square root,

-x squared + 2x + 3, as -(x squared- 2x) + 3.

Then we're going to turn the x squared- 2x into a perfect square,

rewriting that as x squared- 2x + 1.

Of course, we're not allowed to stick in that + 1 without

compensating in what is left over and keeping track of that minus sign.

We get the equivalent expression 4- (x- 1) squared.

Now, rewriting our integral in this form gives us something

a bit easier to work with.

The first thing we're going to do is replace that x- 1 with a u.

10:15

du is dx, and so we get simply the integral of du

over square root of 4- u squared.

And now, the appropriate trigonometric substitution reveals itself.

If we let u be 2 sine theta,

then substituting in 2 cosine theta and d theta for du,

and performing a little bit of simplification yields a nice integral.

We can use the fact that cosine squared + sine squared = 1 and

factor out that conveniently located 4 under

the square root to obtain the integral of 2

cosine theta d theta over 2 cosine theta.

Those, of course, cancel and yield a trivial integral.

Our answer is theta + C.

What is that in terms of u?

That's just arcsine of u.

What is that in terms of x?

That is arcsine of (x- 1).

That is a very simple integral

that doesn't look anything at all like that integrand.

11:40

Hyperbolic trigonometric functions are also very helpful.

Consider an integral of the form dx over square root of 1 + x squared.

Now we could have tried using tangent and secant method.

If x is tangent of theta and dx is secant squared of theta d theta,

then using the fact that 1 + tangent squared = secant squared,

and eliminating the square root gives the integral of secant theta d theta.

12:18

And here, I'm stuck, I don't remember how to integrate secant.

It always seemed to me that it was a tricky sort of integral.

Let's try a different approach.

Consider x as the hyperbolic sine of u,

so that dx is the hyperbolic cosine.

Then this integral is equivalent to the integral of cosh

u over square root of 1 + sinh squared u.

12:51

Now, keeping in mind that sinh squared + 1 is cosh squared, this hyperbolic identity,

we can eliminate the square root entirely, and

we get cosh u over cosh u, which simplifies to the integral of du.

And that is, u + C, or substituting back in for

x, we get the arc hyperbolic sine of x.

Wasn't that a simple integral to do?

13:27

Now, if you pay attention, you see something a bit remarkable here.

Namely, that the integral of secant,

something that seems kind of difficult to derive, is,

in fact, the arc hyperbolic sine of the tangent substituting in tan theta for x.

That's a non-trivial result.

Now of course,

that's not the integral that you usually see in the back of the book.

You usually see a log of secant + tangent.

Both of these are correct and indeed equivalent.

14:08

As a general bit of advice,

when you see certain forms having to do mostly with square roots,

then taking advantage of the appropriate trigonometric or

hyperbolic trigonometric identity prompts a certain substitution.

And these will often simplify your integral to something that is very doable.

You don't have to necessarily memorize this table, but

it's good to have done enough homework problems so

that you get an intuition for which substitution is appropriate.

14:46

Trigonometric and hyperbolic trigonometric functions

can be wonderfully useful in substitutions for solving integrals.

However, some complexities can arise as you'll see in the bonus material for

this lecture.

In our next lecture, we'll consider some methods of integration that

are based more on algebraic manipulations.