Welcome to Calculus. I'm Professor Ghrist. And we're about to begin Lecture
23 on trigonometric substitutions. 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. 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. 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. 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. 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. 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. For another example, consider dx
over x squared times the square root of x squared + 4, which trig
identity is gonna be helpful here? 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. 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. 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. 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. 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. 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. 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. 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? 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. 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. 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.