0:09

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

Â which substitution should I make?

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

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

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

Â to a trigonometric substitution.

Â 0:34

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

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

Â looking expression to the simpler looking expression.

Â However, any time you have an equality in mathematics,

Â it's a road that runs both ways, and

Â we can try exploiting it in the opposite direction.

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

Â This type of substitution is especially useful with trigonometric functions.

Â We'll look at a trivial example,

Â consider the integral of dx over 1 + x squared.

Â You may know this one already.

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

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

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

Â Then, dx would be secant squared theta d theta.

Â Unlike what we do when we perform use substitution,

Â here we directly substitute these in, and we get

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

Â 2:01

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

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

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

Â Now, we have to substitute back in for x,

Â which involves inverting that substitution,

Â we obtain simply arctan of x.

Â 2:33

Now the various trigonometric identities that you may have encountered

Â in the past are what dictate these trigonometric substitutions.

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

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

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

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

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

Â some of the basic simplification formulae for hyperbolic trigonometric functions.

Â 3:20

Let's compute some examples.

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

Â Let's substitute for x the function sine theta.

Â So that dx is cosine theta d theta.

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

Â times cosine theta d theta.

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

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

Â we can replace 1- sine squared with cosine squared.

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

Â This too, after a trigonometric identity,

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

Â This is something we can integrate.

Â 1/2 integrates to 1/2 theta,

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

Â 4:33

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

Â 1/2 theta becomes 1/2 arcsine of x.

Â What about that second term?

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

Â The double angle formula for sine allows us to replace this

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

Â which is x, times cosine theta.

Â What is cosine theta in terms of x?

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

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

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

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

Â 5:51

I think that there was something involving tangents and

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

Â So, let's try x = tangent of theta.

Â Substituting in that, and secant squared theta d theta for

Â dx yields something that looks a bit complicated.

Â 6:15

Now I hoped to use the fact that tangent squared

Â + 1 = secant squared to eliminate that square root.

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

Â And I can't eliminate that square root.

Â 6:35

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

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

Â and replacing dx with 2 secant squared theta, that

Â puts a 4 in front of the tangent squared terms.

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

Â obtaining a 2, cancelling, simplifying a little bit.

Â And now I can replace tangent squared + 1,

Â with secant squared, and eliminate that square root.

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

Â over tangent squared times secant.

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

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

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

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

Â We get after substituting u = sine theta,

Â 1/4 the integral of du over u squared.

Â This is -1/(4u) + C.

Â Substituting back in for theta,

Â we get -1/4 cosecant theta + C.

Â Oh wait, we're still not done.

Â We need our final answer in terms of x.

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

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

Â Then, we can compute the cosecant of theta.

Â As square root of x squared + 4 over x,

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

Â 8:47

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

Â reveal the appropriate substitution to be used.

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

Â This seems completely opaque.

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

Â something wonderful happens.

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

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

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

Â rewriting that as x squared- 2x + 1.

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

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

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

Â Now, rewriting our integral in this form gives us something

Â a bit easier to work with.

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

Â 10:15

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

Â over square root of 4- u squared.

Â And now, the appropriate trigonometric substitution reveals itself.

Â If we let u be 2 sine theta,

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

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

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

Â factor out that conveniently located 4 under

Â the square root to obtain the integral of 2

Â cosine theta d theta over 2 cosine theta.

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

Â Our answer is theta + C.

Â What is that in terms of u?

Â That's just arcsine of u.

Â What is that in terms of x?

Â That is arcsine of (x- 1).

Â That is a very simple integral

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

Â 11:40

Hyperbolic trigonometric functions are also very helpful.

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

Â Now we could have tried using tangent and secant method.

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

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

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

Â 12:18

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

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

Â Let's try a different approach.

Â Consider x as the hyperbolic sine of u,

Â so that dx is the hyperbolic cosine.

Â Then this integral is equivalent to the integral of cosh

Â u over square root of 1 + sinh squared u.

Â 12:51

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

Â we can eliminate the square root entirely, and

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

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

Â x, we get the arc hyperbolic sine of x.

Â Wasn't that a simple integral to do?

Â 13:27

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

Â Namely, that the integral of secant,

Â something that seems kind of difficult to derive, is,

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

Â That's a non-trivial result.

Â Now of course,

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

Â You usually see a log of secant + tangent.

Â Both of these are correct and indeed equivalent.

Â 14:08

As a general bit of advice,

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

Â then taking advantage of the appropriate trigonometric or

Â hyperbolic trigonometric identity prompts a certain substitution.

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

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

Â it's good to have done enough homework problems so

Â that you get an intuition for which substitution is appropriate.

Â 14:46

Trigonometric and hyperbolic trigonometric functions

Â can be wonderfully useful in substitutions for solving integrals.

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

Â this lecture.

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

Â are based more on algebraic manipulations.

Â