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
00:28give
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.