Trig Substitution... How? (NancyPi)

Hi, guys. I'm Nancy, and I'm going to show you how to use trig substitution to integrate. Listen guys, trig sub is kind of intense. It's one of those ones where it's probably better to just jump in and figure it out. But don't worry, I'll be right by your side. So let's do it. All right, so say you're integrating something that looks like this with a radical in it, and you've tried other things, and it's not working. That is a dead giveaway to try something called trigonometric substitution, or trig sub, as the cool kids say. Basically, it let's you get rid of the root, so eliminate the radical so that you can integrate. And what you're really going to want to know is how do you know which kind of trig sub to use, which one to choose. So, in general, if you have a number - an x^2 term in your radical, it's going to be a sine substitution. But if you have an x^2 term - the number, so the reverse, in your radical, this is a radical, just in a disguised form, it's going to be a secant substitution. And finally, if you have a number + your x^2 term, it will be a tangent substitution. So let me show you the first kinds. All right, sine substitution. I know I said if you have a number - an x^2 term in your radical, it's going to be a sine substitution. But wait. Why can't we use u-substitution, or something like an inverse trig integration rule? Very smart! You should check to see if you can. Like if we had instead just x on top, we could have done a u-substitution and power rule, but we don't. We have this x^2. If we had just a 1 on top, it would have fit perfectly the inverse sine integration rule. That would have be nice, but we don't have that. Or like this, 1 over 36 - x^2, that is something called the inverse hyperbolic integration rule, so that would have been an inverse hyperbolic cotangent or tangent. Yeah, those exist. We don't have that. We can't use those. You should check first. If you can't use those, and you can't use something from the table of integrals, or integration by parts, or partial fractions, or any other technique, what can you do? We can do a sine substitution. So if you take what's under the radical and rewrite it as a number^2 - x^2, or here, a number^2 - x^2 under the roots, and compare it to the formula for sine substitutions, you'll see that a is 6 for us. And, by the way, if you ever saw a 1 - x^2, 1 is 1^2, so a would be 1 in that case. But our a is 6, so our substitution is going to be x = 6 sine theta. That's our substitution. Not that you would know that right off the bat. I mean, I know this is coming out of nowhere right now, but you'll see how it helps in a second. The next step is to take the radical out on its own, and plug in this for x. And then see how it simplifies, because it's going to break down the radical and simplify the roots through the wonder of Pythagoren identity. So let's try that. So we take the radical and plug in our substitution, so plug in 6 sine theta for x, and simplify. So the 6 gets squared, we have a 36 from both terms we can factor out. This can be factored out front as a 6. And the 1 - sine^2 theta is equal to cosine^2 theta. That's a Pythagoren relationship, Pythagoren identity. So use that. That's very important in all of these. That's the key, the real link that makes this all work. Any time you have the square root of something squared, really, the answer is the absolute value of that something. So, really, the right answer here is 6 absolute value of cosine theta. But the idea is to take this and put it back in our integral, and we're not going to be able to integrate it if we have an absolute value like that. So what we do for our trig sub

is if we have an indefinite integral with no limits, we just assume that cosine theta will be positive, and we drop the absolute value bars. So let's do that. So we got rid of the root. With our substitution, radical. It's a clever trick, this trig sub, the way it uses Pythagoren identities and gets rid of the radical. And I'm sure that if you had a ton of free time and the desire, you could come up with it on your own. I know I enjoy driving new integration techniques in my free time. So we have something we can replace the radical with. We want to replace the whole integral so that it's in terms with theta, not x. So we want to replace this x^2. We can do that using what we know x is, and just square that. What else do we need to do? We need to replace the dx. So let's get dx. So this is our dx we got from the x expression. I know this looks similar to the radical expression we got, that's just a coincidence. So we have our dx, we have everything we need to rewrite this so it's all in terms of theta, it's just the theta versions of everything. So we have replaced the dx, radical, we have replaced the x^2 factor. Whatever leftover x factors you have, replace them, so the x^2... even if the x^2 were down here. Whatever you have, replace those Xs and clean it up. So when we clean it up, we get 36 times the integral of sine^2 theta d theta. And, listen, guys, at this point, when you have trig substituted, and clean it up as much as you can, you need to pull out your best integration skills that you've been buffing up, hopefully. And use them to attack the integral with everything you know about integrating trig functions. Because this is no longer unfamiliar territory, trig sub territory. This is general integration territory. So use everything you know about integrals, trig identities, algebra tricks to integrate it. But because I care, and I feel your pain, really quickly, let me help you out a bit. All right, these are all really common outcomes of trig sub. So I know this is a bit of a cluster-fuck. This is more than you would ever ask for, but these are all really common ways it could turn out when you do trig sub, where you will need an integral rule from the table of integrals to integrate. It's just good to see them so that you're familiar with what you could do. It all depends on what you were given. The specifics of your original integral, instead of getting the integral sine^2, you could just as easily have got one of these, and you will run into all of these if you do enough trig sub problems. But save it instead of the integral of sine^2 theta, after you substitute it and clean it up. Instead of that, if you got the integral of secant, for instance, you can use the integral rule from the table of integrals, the secant integral rule. Or cosecant has one, an integral, secant^2 has one, cosecant^2 has one. Integral tangent^2, there's no integral rule for that, but you can replace 10^2 with secant^2 - 1, the Pythagoren identity, and then use the integral rule for secant^2. This is a common one you might see. Here's the cotangent^2 version, there's not a rule straight away, but you can replace it with the Pythagoren identity and then use the integral rule for cosecant^2. This is ours, this is what we're going to need. Integral of sine^2, is there a rule on the table for that? Maybe, if you're lucky. It depends on your table. You might be able to use this straight away. More likely, you have to replace sine^2 with a half-angle identity. It's a trig identity, also known as a power-reducing identity. And once you've done that, you can integrate that. This is the cosine^2 version of that with a half-angle identity for it. And if you get one that's different trig functions multiplied and divided, but with powers, there's a good chance you might need something

called trig integral rules. More on that later, if you don't already know what I mean by that. But let's use this one for ours, the integral of sine^2 theta. Okay, so we use the half-angle identity, and then we integrate it. We have an answer for the integral, which is great. But it's in terms of theta, it's not in terms of Xs. We have to get back to all Xs for the answer. So how do we convert it back to all Xs? We can use our substitution, x = 6 sine theta. Right now, this is sine of 2 theta. If it were sine of theta, we could use this. Sine of theta would be x over 6, then we could substitute that. We have sine of 2 theta, so what do we do with that? When you see that, use the double-angle identity. Sine of 2 theta is, 2 sine theta cosine theta. So rewrite it with that to simplify. So now let's rewrite this in terms of x. Substitute in x expressions where we can. For sine theta, we do know what that is in terms of x because of our substitution. Sine theta would be x over 6. So we can use that for sine theta to replace that in terms of x. But how do we replace cosine theta? We don't have that already. How do we get it? We have to do a little bit of right triangle trig. So with the right triangle, we can get any trig function we need. So because we needed sine theta, is x over 6, and sine theta is opposite of our hypotenuse. If we draw a little right triangle, and sine of theta is opposite over hypotenuse, x over 6, we can label those sides. And once we've labeled the sides using sine what we know for that, we can find the third side with the Pythagoren theorem, your friend the Pythagoren theorem. And it ends up being the square root of 36 - x^2, which should look familiar. Now that we have all the sides, you can get any trig function that you want, like cosine, it's adjacent over our hypotenuse. So it will be this root over 6. So that's our cosine that we can plug in. It's an x expression for cosine, we can plug in for that. Seems like we should be done, but there's this theta left here by itself, and we can't just leave it like that. So how do we do it? Well, if you see one like this, you can take any trig function you have, like sine theta is x over 6, and just use the inverse trig version of it to get theta alone. Okay, so if you plug all those in, plug in for cosine theta, sine theta, and simplify, you'll get this. It's all in terms of x. That's the answer for the integral. So that was sine substitution. It's a lot of work, huh? work to explain everything. It'll be faster next time. Keep in mind, if you had one that was actually x^2 - 36 under the radical, like the reverse of this, that form would be secant substitution, a different type of trig sub. That's coming up. Also, if you see one that's like this, has a radical, looks like this, but you know you can do u-substitution, or an inverse trig integration rule, even then, you can do trig sub if you want to. If you have to, or if you just want want to be different, go rogue, do your own thing, and not be part of the lockstep masses, you can if you want. It may not be the easiest way to do it, but it will work out. Sometimes you have a choice that way. And trig substitution is just a technique to get the integral into a form that you can integrate, it's not the law. What would you do if you had one with a radical up top in the numerator, though? Or if you had a coefficient on x^2, like 4 x^2, 9 x^2, 25 x^2, how would you do that? Or, when you have both. If you have a root up top in the numerator, and some coefficient on the x^2 term, and, if you have addition under the root. So if you have a plus here under the radical, what do you do then?

I know this seems a little disastrous but don't worry, I will show you how to do it. Just first, though, you should technically check to see if you can do something like u-substitution, or use an inverse trig rule to integrate. You can't. It won't work out here. So let's see if we can get rid of this root using trig substitution, and then see if we can integrate. So I know I said earlier that when you have an x^2 term + a constant^2 under your root, that version, that it will be the tangent type of trig substitution, and it will be. But what do we do with this coefficient here, the 25? So when you have a coefficient on x^2 in the radicand, instead of thinking of it as x^2 + a^2, you can think of it as u^2 + a^2. So you might see a u, some people use u. There are a few different ways of doing this kind, but no matter how you do it, the idea is to rewrite this as two things squared and added together. So when we rewrite this as one thing squared plus another thing squared, we can see that, definitely, a is 2, and u will be 5 x, because 5 x all squared was 25 x^2. So u is 5 x. For this kind, when you have a coefficient on x^2, basically what you do is instead of using x = a 10 theta for your substitution, you use 5 x, the u, the thing getting all squared, = a 10 thetafor your substitution. So, for us, it's 5 x = 2 tangent theta. That's our substitution. That's great. We have that. And now, just like before, take just the radical out, the radical, and plug in your substitution and see how it simplifies. So now just use algebra and plug in, use the substitution, plug in for 5 x, simplify, use a Pythagoren identity to connect this to secant squared. The Pythagoren identity magically breaks down the roots and leaves you with just 2 times the absolute value of secant theta. And just like before, we can drop the absolute value bars, because we're doing an indefinite integral, and we'll just assume everything is positive. So, really, this radical is just = 2 secant theta. So let's use that to replace the radical up here, and replace everything else in the integral, including the dx. Don't forget the dx. Okay, we plugged everything in. Plug in for the radical, plug in for x to the 4th, use what we know x is, x would have to be x 2/5 10 theta, so that gets all raised to the 4th power. This is dx. And then clean up the debris of that trig substitution, and you get this, which you now have to integrate. So how do you do that? We can try u-sub, we can try a table rule. They won't work. So what do you do? Any time you have secants and tangents with powers, divided or multiplied, or, sines and cosines it's a really good idea to try something called trig integral rules. Okay, let me help you out. All right, so if you're doing trig substitution and you wind up with an integral that looks like one of these, it is a really good idea to check trig integral rules, especially like these here. And what those rules will tell you is, generally, to save one trig function, and then convert the rest of it using a Pythagoren identity. So for instance, for this one, it will tell you to keep one sine function, and then take the sine^2 that's left, and convert that using Pythagoren identity, replace that with 1 - cosine^2, expand that, and integrate, and it will all work out. That's the idea for that type. So a lot of these fit that. I'm showing this to you because there's a really good chance you'll run into one of these, especially if you've just learned about trig integral rules. And sometimes, there are odd ones like, you end up having to use integration by parts

for this one. If specific trig integral rules don't work, you could end up just needing to plain replace everything with sines and cosines and see what happens. Like here, in general, if you have an odd secant power, and an even tangent power, like here or here, replace everything with sines and cosines, here, replace with cosines. And then simplify it, and hopefully you'll be able to use power rule, or e-sub, or a table integral rule and see if it works out. This is the one that we need, here, that will help us. So let's replace it all with sines and cosines that it's equal to, and simplify it, and see how it turns out. All right, so replace everything with sines and cosines, clean it up, use the power rule to integrate, and we get this answer. Which, looks legit and all, but it has thetas in it. And we need everything to be back in terms of Xs, to convert back to Xs, so we have to do the side trig work. Don't be scared of the trig. side trig nonsense that we have to do at the end of these problems. I'm sure you've done much harder things at this point, so you've got this. Don't worry. And we're almost done. Okay, so the substitution that we have is the tangent one, and we know that tangent is 5 x over 2. 2 here. And since tangent is the opposite side over adjacent side, we can label the right triangle with the opposite value over the adjacent value 5 x and 2. And on the right triangle, since we have two sides, then can easily get the third side with the Pythagoren theorem. This should look very familiar to you. So that was the tangent information that we used, but what we really need now in the end is cosecant information. And since cosecant is hypotenuse over opposite, that will be equal to the hypotenuse value over the opposite value, so it's this root over 5 x, just kind of a mess, but that is what we will now put in place of cosecant theta so that it's all in terms of x. Clean it up, and that's the answer for integral. So that is tangent substitution. All right, don't be thrown off by weirder, different-looking forms. Like if you have a rational power, a fraction power, like three halves. But first of all, notice in this one, this x^2 - a number, x^2 - 16. This might look similar to the very first one we did, but that one was the sine substitution, and that was for when we have the form of a number - x^2. This is the opposite, the reverse. When we have x^2 - a number, it's going to be a secant substitution, the last kind of trig substitution. And we will have a = 4, our substitution will be x = 4 secant theta. But what about this three halves fractional power? Can you even do trig substitution when you have something like that? Well, you can probably tell already that the answer is yes. I've already written out the work for it. But the trick to see it is to rewrite it as a root raised to a power. In the power, the bottom two means a second root, square root, and the top three means raise to the third power. So write it like that so that you can really see what it is. It will be easier to do the trig substitution. I took the whole denominator and substituted into that. If you plug it in, simplify, use a Pythagoren identity, that secant^2 - 1 is tangent^2, you will get this. We can drop the absolute value bars, plug everything in, simplify, ends up being one where you replace with cosines and sines, then you can do straight up power rule, and you get this answer. After you back-substitute in to get all in terms of Xs with your right triangle trigonometry. That's the answer. So this is how you do secant substitution, and also how you handle a rational power like that. Let's fiddle with the forms a little bit more. I'm going to show you some slightly even weirder kinds where you can still use trig substitution.

Finally, here are some kinds that might not look like you can use trig sub, but you can. Like if there's no radical in the integral at all, turns out you can use trig sub if it still looks like one of these forms. Even though it might not be the fastest way, like this one, it would be a lot faster to use the arctan integral rule, or the inverse tangent integral rule. Here it would be a lot faster to use u-substitution, but you can use trig sub if you want to, or if you have to. Here, if you have just a radical in the integral, you can do trig substitution. If you're really lucky, you might get special integration formulas that tell you right away what these specific kinds equal. If you're not so lucky, you won't get those. But either way, you can do trig sub. And then, finally, something like this does not, right away, look like a trig sub form that we saw. But you can use a little algebra, completing the square to make it look like that, and do some magic, we get it into this form of a number - an x^2 term, and then it does not turn into forms. These are just some oddball kinds that you might run into. That's trig substitution. Basically, it helps you get rid of the radicals so that you can integrate, hopefully. And I know that there are so many parts to these problems, that finishing one of them may seem like nothing short of a miracle, but if it doesn't work out right away, don't worry just yet. Try checking your work, because there are so many places where some little detail could be off, that threw you off, but overall, your method might be right. And I know that there is a lot that can be covered with trig sub, but hopefully this helps you know where to start. It's kind of a mess, I know. You don't have to like it, but you can like my video. So if you did, please click Like or Subscribe.