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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Zach Pino*

Sometimes a simple substitution can make life a lot easier. Imagine how nice it would be if you could replace your federal tax form with a 'Hello, my name is...' name badge! In this lesson, we review how you can use trigonometry to make substitutions to simplify integrals.

When I was learning how to drive, I always asked my mom if I could drive to the mall. She said I was not allowed to drive or get my driver's license until I learned how to drive a stick shift. Now I didn't think this was particularly fair, because my goal was simply to get to the mall and I wanted to do this by driving a car. I knew that a stick shift was the hard way to go; instead, I wanted to take my stick shift car and turn it into an automatic because I knew that driving an automatic would be a lot easier.

This is a lot like a **trig substitution** in math. The goal with trig substitution is to use substitution based on trig identities. We're going to use substitution based on right triangles to make integration easier. So here, your goal might be to evaluate an integral, but you want to do that by finding an anti-derivative. Before you use the right substitution, you might have a complicated mess on your hands, but after using trig substitution, life might be a little simpler.

So let's take an example. Let's say you have the integral of (1/ the square root of (1 - *x*^2))*dx*. I can map this to a right triangle, that is, one where I've got *a*, *b* and *c*, where *c* is the hypotenuse. By the Pythagorean theorem, I know that *a*^2 + *b*^2 = *c*^2. One right triangle that looks a little bit like this equation is if I have 1 on the hypotenuse, *x* on one of the sides and then by the Pythagorean theorem, I know that the other side is the square root of (1 - *x*^2).

So why do I map it like this? Well, I want to keep this triangle as simple as possible but still make it look like my integral. So here I've got a 1 (that's for this value here), and I've got an *x* for this value here and on this third side, as a result of the other two, I have the square root of (1 - *x*^2). I draw out my triangle like this so that this complicated-looking term is just one side of my triangle.

I'll be honest. In general, trig substitutions are very difficult. It's hard to see them, and usually when you see a trig substitution, you might want to look at doing your problem differently. But sometimes you can't avoid them. So what are some rules that might help us find substitutions that make a little bit of sense? Well, if you have a function that depends on some constant squared, *C*^2 + *x*^2, then you should consider using the substitution *x* equals *C* times the tangent of *theta*, which written in symbol form is *x*=*C* * tan(*theta*). Here, you're going to replace *x* in your integrand with a new variable, *theta*. When you use this particular substitution, keep in mind that 1 plus the tangent squared of *theta* is equal to the secant squared of theta (1 + tan^2(*theta*) = sec^2(*theta*)). This is a trig identity.

If on the other hand, you have a function that depends on some constant squared minus *x*^2, you might want to consider the substitution *x* equals *C* times the sine of *theta* (*x*=*C* * sin(*theta*)). When you use this substitution, you might want to remember the trig identity 1 minus the sine squared of *theta* equals the cosine squared of *theta* (1 - sin^2(*theta*) = cos^2(*theta*)).

Finally, if you have *x*^2 - *C*^2 - so some constant squared - you might want to consider using the substitution *x* equals *C* times the secant of *theta* (*x*=*C* * sec(*theta*)). If you use this trig substitution, keep in mind the trig identity secant squared of *theta* minus 1 equals the tangent squared of *theta* (sec^2(*theta*) - 1 = tan^2(*theta*)).

Let's go back to our example of (1 / the square root of (1 - *x*^2))*dx*. Because we have 1 - *x*^2, that's like 1^2 - *x*^2. So according to our little cheat sheet of substitutions, we might want to consider using the substitution *x*=*C* * sin(*theta*), where *C* is just 1.

So let's do a substitution, *x*=sin(*theta*), and if we take the derivative of that, we get *dx* = cos(*theta*) * *d* * *theta*. If I plug those into my integral, my integral becomes 1 / the square root of (1 - sin^2(*theta*)) * (cos(*theta*) * *d* * *theta*), where this is from *dx* and this is from *x*. Now initially you might think, 'Well, that's not really helping me.' I mean, fine, we went from *x*^2, this square root of (1 - *x*^2) to the square root of (1 - sin^2), but we also added this cosine up on the top here. So what did this do? Here we use our knowledge of trigonometry. Remember that 1 = sin^2(*theta*) + cos^2(*theta*). If I subtract sin^2(*theta*) from both sides, I get 1 - sin^2(*theta*) = cos^2(*theta*). So this bottom here becomes the square root of cos^2(*theta*); that's just cos(*theta*). And so I end up with my integrand: cos(*theta*)/(cos(*theta*) * *d* * *theta*), and that just cancels out so that my entire integral reduces to *d* * *theta*.

I know how to integrate *d* * *theta*. That's like integrating 1, and the integral of 1 is just, in this case, *theta* plus some constant of integration. Following our rules of substitution, we need to put *x* back into this equation by putting *theta* back in. So what is *theta*? Well *theta* we defined as being *x*=sin(*theta*). If I want to solve for *theta*, I take the inverse sin(*x*) = the inverse sine of sin(*theta*), which means that the inverse sin(*x*) equals *theta* (sin^-1(*x*)=*theta*). I can plug that in, and I find that my integral of (1 / the square root of (1 - *x*^2))*dx* is equal to sin^-1(*x*) plus some constant of integration *C*. We can also call this the arcsine of *x* + *C*.

Let's do a slightly different example. Let's say we want to find the integral of (1/(4 + *x*^2))*dx*. Well, now I'm looking at a triangle with sides *x* and 2 and a hypotenuse that is the square root of 4 (that's 2^2) + *x*^2. So I see this 4 + *x*^2 and I'm thinking I might want to use some trig substitution here. So what substitution should I think about?

Well, I've got this *C*^2 + *x*^2, where my *C* is 2 because I've got 4 + *x*^2, which is like saying I've got 2^2 + *x*^2. So I'm going to use the substitution *x*= 2 * tan(*theta*). If *x*= 2 * tan(*theta*), then *dx* = 2sec^2(theta) * *d* * *theta*, because the derivative of tangent is the secant squared. If I plug those in, I get 1/4 + 4tan^2(*theta*) - because that's *x*^2 - times 2sec^2(*theta*) * *d* * *theta*, where this is my *dx* from before. Well 4 + 4tan^2(*theta*) is like 4 times 1 + tan^2(*theta*), and I can use a trig identity where 1 + tan^2(*theta*) = sec^2(*theta*), and I can rewrite this term as 4sec^2(*theta*).

So now I have 2sec^2(*theta*) / (4sec^2(*theta*) * *d* * *theta*). All of these sec^2(*theta*)s cancel out, and I can factor 2 out of the top and bottom, so I end up with the integral of 1/2(*d* * *theta*). Well once again, that's a pretty easy integral. I know that that equals 1/2(*theta*)+ a constant of integration. Again, what is *theta*? Let's get rid of that and put *x* back in. Well if *x* is 2tan(*theta*), then I can solve this for *theta* by dividing both sides by 2 and taking the inverse tangent of both sides. So *theta* is tan^-1(*x*/2). My integral then is 1/2(tan^-1(*x*/2))+ a constant of integration *C*.

So the goal with **trig substitutions** is to use substitutions based on trig identities, and you do this to make your integration easier. This is just another way to suggest what kind of *u* substitution will simplify an integral. So the suggestions are, if you have some number *C*^2 + *x*^2, use *x* = *C* * tan(*theta*), where theta is now the variable that you're integrating over. If you have *C*^2 - *x*^2, use *x*=*C* * sin(*theta*). Lastly, if you have *x*^2 - *C*^2, use *x*=*C* * sec(*theta*).

If you use these substitutions and you remember a few trig identities, trig substitutions can simplify your integrals and make them as easy as driving an automatic car to the mall.

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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

- Go to Continuity

- Go to Series

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