AAHL 5.16.1 Integration by substitution (HL only)
Learn AAHL 5.16.1 Integration by substitution (HL only) in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 5.16—Integration by substitution, parts and repeated parts.
Learn AAHL 5.16.1 Integration by substitution (HL only) in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 5.16—Integration by substitution, parts and repeated parts.
Learn AAHL 5.16.1 Integration by substitution (HL only) in this free IB Mathematics Analysis and Approaches (AA) video lesson for AHL 5.16—Integration by substitution, parts and repeated parts.
The video focuses on integration by substitution, specifically targeting both standard and higher-level students. It begins with a straightforward example, integrating the function by letting , leading to the substitution .
Key points include:
The lesson emphasizes the necessity of practice with various types of substitution problems to build proficiency and confidence in tackling challenging integrals.
00:00Hi guys. So this is
00:02the second lesson I'm doing
00:03on integration by substitution. The
00:05first lesson that I did
00:06was, well, it is what
00:08standard level students are expected
00:10to know and obviously higher
00:11level students as well. But
00:13for those substitutions, it's always
00:17in a particular form where
00:19it's a product of a
00:23function and the derivative of
00:25the inner function kind of
00:26thing, which I'm going to
00:28been example to explain what
00:30I mean by that. So
00:31say you have x times
00:34sine of x squared. You
00:38can integrate that by using
00:43integration by substitution where you
00:45let u equals u is
00:47equal to x squared, then
00:48du dx equals to x.
00:51And these x's cancel and
00:53it always works out quite
00:54nicely. It's not
00:56easy, certainly for standard level.
01:00But there's a process to
01:02follow and it's always kind
01:03of works out pretty much
01:04the same integration by substitution
01:06in higher level, which is
01:07basically this lesson here, is
01:10not quite as simple. However,
01:14if it's not of that
01:16form, they will give you
01:17the substitution. So that's at
01:18least one. That's a big
01:20plus because you don't have
01:21to figure out what substitution
01:22to use. They will give
01:24it
01:24you. In this case, they're
01:25going to say, say, let
01:26u equals x plus two.
01:27And in this case, they
01:28say that x equals a
01:29sec, theta. So the first
01:31one is a fairly straightforward
01:32example. The second one is
01:35pretty much the hardest question
01:36I could find. I tried
01:37to find a really, really
01:38hard question. These integration with
01:40substitution questions can be really,
01:42really difficult. However, I actually
01:45find them really, really fun
01:46on a lot of my
01:47students to do that they're
01:50maybe not the types of
01:51questions you want to do.
01:52under kind of high stakes
01:55exam pressure, but they're certainly
01:57fun to do when you're
02:00practicing at home. And obviously
02:01by doing that kind of
02:04practice and by enjoying doing
02:05them and practicing lots of
02:06different types of them, you
02:07get better and the exam
02:09becomes easier, etc. Okay, let's
02:13let's go straight into part
02:15A. So it says that
02:16you equal x plus two
02:17find the integral of x
02:20the integral of x times
02:22the square root of x
02:24plus 2 dx. So we
02:27say let over the side
02:29u is equal to x
02:31plus 2, that's very kind
02:32of them. And it's always
02:35the same, I differentiate this,
02:36so the same d u
02:37dx is equal to, well,
02:391, so that's nice, and
02:41I can just go straight
02:41into du equals dx. So
02:46this becomes
02:48the integral of, now remember
02:50I'm changing the x's into
02:52u. So x, well, x
02:55plus 2 is u. Now
02:59I don't have the situation
03:00where this x is gonna
03:02cancel nicely as I said
03:03before, but I can still
03:05write x in terms of
03:07u. Look, if I'm gonna
03:10draw a line here, if
03:12u is equal to x
03:13plus 2, then x is
03:15equal to u
03:16minus two. So I can
03:19now write this as u
03:21minus two, that's my x
03:23times the square root of
03:27u dx. Now I should
03:31maybe have said at the
03:31start, the reason why, so
03:35the reason why I even
03:36need to use integration for
03:39substitution is because what could
03:40I do with this? I
03:41mean I can't, like you
03:42can't obviously can't multiply the
03:44x
03:44into the square root of
03:45the root of the root
03:45of the root of the
03:46root of the root of
03:47the root of the root
03:49of the root of the
03:50root of the root of
03:50the root of the root
03:50of the root of the
03:51root of the root of
03:52the root of the root
03:52of the root of the
03:53root of the root of
03:56the root of the root
03:57of the root of the
03:57root of the root of
03:57root of the root of
03:57the root of the root
03:57of the root of the
03:58root of the root of
03:58the root of the root
03:58the root of the root
03:59of the root of the
04:01root of the root of
04:02the root of the root
04:02of the root of the
04:03root of the root of
04:04the root of the root
04:05of the root of the
04:08root of the root of
04:09the root of the root
04:09of the root of the
04:09root of the root of
04:12the root of the root
04:12of the root of the
04:12of the root of the
04:12root of the root of
04:12the root of the root
04:12of the root of the
04:12root of the root of
04:12the root of the root
04:12of
04:12changed, I can change the
04:14dx into du because they're
04:15just equal to the same
04:16thing. But now, yeah, this
04:20one, I can multiply them
04:21out because it's just a
04:24u on its own to
04:25the bar of a half.
04:26So u times u to
04:27the bar of a half,
04:28that's 1 plus a half,
04:29which is u to the
04:30bar of 3 over 2,
04:32minus 2 times u to
04:34the half du. Now I'm
04:37ready to integrate. So look,
04:38I've had to write 4
04:40integrals before I'm actually going
04:42to integrate. So this is
04:43going to become u to
04:45the power of, so I'm
04:46going to raise the power
04:47by a 1, so I'm
04:49going to 5 over 2,
04:50and I divide by the
04:51new power, divide by 5
04:52over 2, minus 2 u
04:56to the power of 3
04:56over 2, divide by the
04:59new power, 3 over 2,
05:01and then, let's keep going
05:04down, I then say this
05:07is equal to 2,
05:08two times u to the
05:11five over two divided by
05:13five. So because I'm dividing
05:14by the fraction, dividing by
05:16over two, I flip it
05:17and multiply so it becomes
05:19two over five. And then
05:21minus similar thing here, I'm
05:23gonna flip this, which will
05:24make two times two, which
05:26is four u to the
05:28three over two, all over
05:32three. And look, notice I
05:34forgot my plus c, that's
05:35twice a dollar, plus
05:36C. And then finally he
05:40doesn't want it in terms
05:42of u, he wants in
05:43terms of x, he has
05:44to be defined in terms
05:44of this, there's no u
05:46in that, so I need
05:46in terms of x. But
05:48u is x plus 2,
05:49so finally I can write
05:512 times x plus 2
05:54to the power of 5
05:55over 2, all over 5,
05:59minus 4 times x plus
06:022 to the power of
06:043 over 2,
06:04All over 3 plus c
06:10done. Now you may say,
06:12well hang on Mr. finished,
06:13that's the easy question. What's
06:15the hard question going to
06:16be like, well, it's hard.
06:19Trust me. But as I
06:20said, it's fun. And this
06:21is it. So, okay. Use
06:28a substitution x equals this
06:30to show that this equals
06:31this fine. So,
06:33I'm gonna say, like this
06:35is my answer, so I'm
06:36not gonna use this, this
06:37is what I'm hoping to
06:38get at the end. I'm
06:39gonna say the integral, let's
06:42write this out, the integral
06:43from A root two to
06:46two A of dx all
06:51over x cubed square root
06:58of x squared minus A
07:01squared.
07:01So this equals, so what
07:04I'm going to do is,
07:06well, he tells me what
07:06to do, he says let
07:08x equal a sec theta.
07:12Fine. So x is equal
07:14to a sec theta. Now
07:16we always, always differentiate the
07:19substitution. So I'm going to
07:20say dx d theta dx
07:22d theta equals this, the
07:24derivative of sec theta is
07:26in the formula booklet. It's
07:28sec theta
07:29Pan theta, so that's a
07:30sec theta tan theta. Straight
07:36from the formula, we'll put
07:37it. Fine. And I can
07:41rearrange this and say dx
07:42is equal to a sec
07:46theta tan theta d theta.
07:52Fine. Now, this, I'm going
07:56to write
07:57down here is equal to
07:58the integral of. So instead
08:02of dx, I'm going to
08:03write, this is my numerator
08:04A, sec theta tan theta
08:12d theta. The denominator is
08:18going to be x cubed.
08:21Now, a x cubed is
08:22this cubed, which is a
08:24cubed.
08:25Theta cubed, theta. Fine, square
08:32root. X squared, this is
08:36going to be a squared,
08:38sec squared. Theta minus a
08:45squared. Okay, nice, so everything
08:49has been turned into theta,
08:52except now this root
08:53really really careful. These limits,
08:57let me move these over
08:58here a little bit. These
09:00limits here to a, and
09:04to a, and a root
09:07two, these are x limits.
09:09So these are, these are
09:11x values if you like.
09:13I want to find the
09:16equivalent value for theta, not
09:17the equivalent value, but I
09:18want to find what is
09:19theta equal when x equals
09:21equals 2a or when x
09:22equals root a root 2.
09:25Now, I have what x
09:27is in terms of theta,
09:28but let's find what theta
09:30is in terms of x.
09:31So, x equals a sec
09:35theta, which means sec theta
09:39is equal to sec theta
09:41is equal to x over
09:43a. Now, sec theta, remember,
09:46is one over cos theta.
09:49that's what sec theta is.
09:51So I can actually just
09:52rearrange this and say cos
09:54theta. So cos theta equals
09:59a over x. So when
10:08x equals to a, I'm
10:11going to get theta of
10:11x is 2a and when
10:12x is a root 2.
10:14So when x is 2a,
10:17cos
10:17theta equals a over 2a
10:22which is equal to a
10:24half and when cos theta
10:30is a half theta is
10:33equal to pi over, let
10:36me get this straight, I'm
10:38visualizing my triangles, it's equal
10:40to pi over 3 and
10:44when
10:45X equals what was the
10:48other limit, A root two.
10:50Next it was A root
10:51two. Cos theta is equal
10:55to A over A root
10:58two, which is actually equal
11:00to one over root two.
11:04And again, visualizing my triangles,
11:08that means theta is pi,
11:11theta is pi over phi.
11:13So now I have my
11:15limits, it goes from pi
11:16over 4 to pi over
11:183. So look, for A
11:21root 2 to 2A, pi
11:23over 4 to pi over
11:243. So look, every part
11:26I'm not going to pretend
11:27this is easy. Every part
11:29of this, every little step
11:31I do is hard or
11:33tricky, but as I say,
11:34it's fun. Okay, so now
11:37I have my integral. Nice.
11:41This equals, okay, I want
11:43to simplify this. Now, how
11:45can I simplify this? Well,
11:48see this thing here? This
11:51is equal to, well, it's
11:54integral of pi over four,
11:56pi over four, to pi
11:58over three, and I'm gonna
11:58write this out again, a
12:00sec theta tan theta d
12:04theta, all over a cubed,
12:09sec cubed theta and then
12:16the square root now the
12:18square root of, I'm going
12:26to take out the a
12:26squared, take out the a
12:28squared and I'm going to
12:29say sec squared theta minus
12:33one. Now you might say
12:36why have you done that?
12:38And how did I even
12:39know to do that? Well,
12:40look, again, that's what makes
12:43this tricky. There is an
12:48identity. So I'm trying to
12:50simplify all this. Now this
12:50looks horrible, but there is
12:52an identity that, well, if
12:54you remember your, your, um,
12:58let me just write it
12:59here, cos squared theta plus
13:03sine squared theta equals one.
13:061 plus tan squared theta
13:10equals, and let me move
13:13that over here. Sec squared
13:20theta. So this is in
13:22the formula booklet. I like
13:23to just kind of derive
13:24these from there. I remember
13:26this one and figure these
13:27ones out for that. But
13:28look, if you have your
13:29formula booklet, you can go
13:30straight to your formula booklet.
13:31So I can actually say
13:33that
13:3410 squared theta equals 6
13:37squared theta minus 1. Now,
13:41yeah, again, you can say,
13:42how did I know to
13:43do that? Well, look, I
13:46saw, I kind of saw
13:47there's a, I don't know,
13:49I saw there's a 6
13:50squared theta minus 1 hidden
13:51in there. I knew that
13:53if I could change that
13:53to 10 squared theta, and
13:55then get the square root
13:56of 10 squared, that would
13:57give me 10, and my
13:58tens will cancel, and I'll
13:59be left with this. So
14:01I
14:02I could just kind of
14:03see it was going to
14:03work out quite nicely, but
14:05don't get me wrong. Not
14:06many people would have gotten
14:08this right in the exam.
14:10So it goes pi over
14:12three. pi over four. Same
14:15thing. A sec theta tan
14:18theta d theta. This now
14:21becomes a. This is a
14:27cubed sec cubed theta.
14:30into now, watch this. This
14:34a squared, this is, it's
14:38like a squared tan squared
14:39theta. But the square root
14:41of a squared tan squared
14:43theta is just a tan
14:47theta. Okay, hopefully that step
14:51makes sense because now what
14:52I'm gonna do is that
14:53cancels with that and that
14:55cancels with that and I'm
14:57now left with
14:58I am left with the
15:01integral from pi over 4
15:04to pi over 3. A
15:09cubed is at the bottom,
15:10so I can actually take
15:12out that A cubed. So
15:13it's 1 over A cubed
15:14at the bottom. And then
15:17it's 1 over sec cubed
15:22theta. Now sorry, sec squared
15:24theta. So it's 1 over
15:26sec
15:26x squared theta, which is
15:28actually equal to, let me
15:31write this step one over
15:33sec squared theta d theta.
15:37That equals one over a
15:39cubed times the integral from
15:41pi over four to pi
15:43over three. One over sec
15:46squared is actually cos squared
15:49theta because that's slipped by
15:53definition. That's what sec is
15:54The second is one over
15:55cos, so one over the
15:56second is cos, one over
15:57the second squared is cos
15:59squared. And just when you
16:01think you're nearly done, you
16:03have to integrate cos squared
16:04theta, which actually isn't that
16:06straightforward, because we can't just
16:08integrate that straight away. We
16:09have to use the double
16:11angle identity. Now straight from
16:14the formula booklet we have
16:16cos two theta, this is
16:18our double angle identity, cos
16:19two theta, is equal to
16:21two
16:22Cos squared theta minus one.
16:24This is how we integrate
16:25cos squared theta and how
16:26we integrate sine squared theta
16:27as well. Actually, then I
16:29need to say cos 2
16:31theta plus one equals 2
16:33cos squared theta. And then
16:36cos squared theta is equal
16:39to this divided by two,
16:41which is a half, cos
16:432 theta, I'm dividing it
16:44by two, cos 2 theta
16:47plus a half. So,
16:50This now becomes 1 over
16:56a cubed times the integral
16:58from pi over 4 to
17:01pi over 3 of a
17:05half cos 2 theta plus
17:10a half and all of
17:12this theta. This equals 1
17:16over a cubed now and
17:18only
17:18Now can I actually integrate
17:20it? So I'm going to
17:21integrate it. This becomes a
17:23half cost two theta. So
17:25this would be sign. This
17:28would be sign two theta,
17:29sign two theta. But I
17:31have to divide by two
17:32because of this two and
17:33there's a divided by two
17:34here as well. So I'm
17:35going to divide by four
17:36plus a half theta just
17:39because that's a constant. I
17:40have theta closed my brackets
17:42and I'm going from pi
17:43over three to pi over
17:45four. Now
17:46I can just sub those
17:48in. So I've won over
17:50a cubed into, that's a
17:53big bracket and I will
17:54do two smaller brackets. So
17:57I'll sub in pi over
17:58three. So that's sine of,
18:02this is going to be
18:04sine of two pi sine
18:08of two pi over three
18:10all over four plus a
18:13half
18:14times pi over three. Close
18:19this bracket and I'm going
18:20to subtract. I'm going to
18:22now I'm going to sub
18:22in pi over four. So
18:24this is sine brackets sine.
18:30Two times pi over four
18:31is actually just going to
18:32be pi over two and
18:34that's all over four plus
18:36a half times pi over
18:41four. Close this
18:42this and close this big
18:45bracket. Right at this point
18:48I should probably look at
18:48what they want me to
18:51find but let's just let's
18:53keep going because this is
18:55fun. One over a cubed.
18:58Now what is sine of
19:032 pi over 3 so
19:07every bit of this question
19:08is hard. So 2 sine
19:09of 2 pi over 3
19:10sine
19:11I like to use my
19:13cast diagram, C -A -S
19:16-T, so it's gonna be
19:17up here. So it's gonna
19:18be the same as the
19:19sine of pi over three,
19:21and the sine of pi
19:22over three is root three
19:24over two. So this is
19:25root three over two, and
19:27it's divided by four, so
19:28it's actually root three over
19:29eight, plus pi over six.
19:35And then I'm going to
19:36minus, so sine of pi
19:38over two is one,
19:39minus a quarter and then
19:41it's minus pi over 8.
19:46Okay, at this point I
19:48want to see what do
19:49they actually want. So they
19:51want, okay, I'm actually going
19:53to copy and paste this.
19:57Copy, because I want to
20:00see what I'm trying to
20:01get. So I want to
20:03turn this into this
20:07that's kind of my answers.
20:09So here's my, okay, I'm
20:11getting there. Here's my a
20:14cubed find. So this is
20:16one over a cubed. And
20:20now I'm going to change
20:20that to normal brackets. So
20:22I need a common denominator
20:24here and that's 24 nice
20:26because that's a 24 here.
20:27So this is going to
20:29be, well, let's put the
20:3124. Let's put it here
20:34for now 24.
20:35and then this is 3
20:39root 3. This is 4
20:45pi, because this times 4,
20:47this times 4 obviously. This
20:49is 6 and this is
20:54minus 3 pi. Okay, great,
21:00because I think this is
21:01now good. So this becomes
21:03over 24 a cubed open
21:08brackets I have 3 over
21:113 4 pi minus 3
21:14pi is pi so it's
21:16plus plus pi and my
21:18minus 6 and there we
21:20go I'm gonna write Q
21:22E D to show I
21:25just to say that I
21:26have shown that okay fine
21:28look no denying that's one
21:30of the most difficult
21:31questions I've seen on an
21:33exam paper in any topic.
21:35But genuinely, as I said
21:37before, I actually quite enjoyed
21:39doing that and I hope
21:40you did too. Definitely, definitely,
21:46definitely practice lots and lots
21:48of those types of questions.
21:51That's one example. And this
21:53is another example, but there's
21:55many different types of examples
21:57they can ask you obviously,
21:58but with the
21:59thing to note is they
22:01always give you the substitution
22:02unless it is in that
22:04form that I showed you
22:06where you see the derivative
22:08of the inner function and
22:10it's a product of that.
22:13Okay, that's it. Hope you
22:14enjoyed that very fun lesson.
22:18I will see you in
22:19the next one.