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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 5

Lesson 6: Trigonometric substitution- Introduction to trigonometric substitution
- Substitution with x=sin(theta)
- More trig sub practice
- Trig and u substitution together (part 1)
- Trig and u substitution together (part 2)
- Trig substitution with tangent
- More trig substitution with tangent
- Long trig sub problem
- Trigonometric substitution

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# Long trig sub problem

More practice with a hairy trig sub problem. Created by Sal Khan.

## Want to join the conversation?

- correct me if im wrong but didnt he get the problem wrong because he stopped carrying over the +C?(15 votes)
- No-It's kind of understood to be there (during the problem), but your final answer HAS to have the constant of integration, or C.(25 votes)

- So, Sal shows sample problems for the forms (a^2 - x^2) as well as (a^2 +x^2); however, I did not see any problems with the form (x^2 - a^2). One of the practice problems I was given took this form, and I thought that hyperbolic trig substitution would be appropriate since we can use the identity cosh^2(theta) - 1 = sinh^2(theta). I arrived at a reduced answer to the problem in terms of inverse hyperbolic trig functions; however all of the multiple choice answers were in terms circular trig functions. Is there an easy substitution I'm missing?(6 votes)
- Here is another solution, but with an example (it might even be a question from Khan).

This is my very first LaTeX document - so, 'scuse any weirdness.

http://bajasound.com/khan/khan1.pdf(3 votes)

- Is integration just a process of trial and error and remembering types or are there any things to look out for and do to bring a expression into a expression that can be easily integrated ?

As there are no formulas like quotient rule or chain rule for it as in differentiation if a complicated expression to integrate is given how should it be approached ?(7 votes)- There are tips and tricks for integration. For example, the power rule is (I think) the simplest integration rule. It is really the reverse of the power rule for derivatives: d/dx (x^n) = nx^(n-1)

The power rule for integrals says: ∫ x^n dx = ( x^(n+1) ) / (n+1)

There are also methods of integration like trig sub, u sub, integration by parts, partial fraction decomp...

Knowing what methods to use when just requires a lot of practice. You can probably find practice problems if you search google (Khan Academy does not have integration practice modules).(5 votes)

- I got the same solution as the video from doing it manually, but how come wolfram's integration calculator got (1/2)(x-3)(sqrt(-x^2+6x-5)-2arcsin((3-x)/2) instead?(3 votes)
- There are different ways of solving these problems. You can get very different-looking, but mathematically equivalent answers depending on how you solved the problem. This is especially the case when you have trigonometric functions involved.(5 votes)

- Some practice exercises will be added for u-substituiton, trig substituiton and other integrals? Thanks.(3 votes)
- There is always more videos than exercises, because they are produced much faster (as Sal said in one of his videos). But I am sure your patience will pay off, its just a matter of time before we get exercises of all kinds :D(4 votes)

- Can you just solve this and these types of problems with normal substitution? You could just set 6x - x^2 - 5 = u and go on from there. Trig substitution seems unnecessary and long. Or am I wrong and you can't use regular substitution?(3 votes)
- Try out the substitution you've suggested to see if it works. :)

The challenge you'll run in to is with du and dx.(4 votes)

- Sir, you have replaced (x-3)/2 with sin theta. Does that not alter the values x can have? Replacing it with sin function makes x lie between 1 and 5 only.. Is that acceptable??(1 vote)
- Look at the original problem.
`∫√(6x-x²-5)dx`

Any value inside the square root that's less than 0 gives you a problem already. So`6x-x²-5 ≥ 0`

already meaning`1 ≤ x ≤ 5`

. So replacing it with sin doesn't actually change anything.(7 votes)

- sir I am student of XI standard from Calcutta,India...and I have been watching your videos since class 8....

I am really a great fan of yours......

Actually I have a question that has been disturbing for a couple of weeks and i really can't solve it by own...

Is it possible to integrate this function: x^x (with respect to dx)..?

Actually I have a question(3 votes)- I entered this into wolframalpha.com and confirmed that the integral cannot be written with standard mathematical functions. The integral can be written only in terms of an infinite series, and the series is quite complicated.

Have a blessed, wonderful day!(2 votes)

- I kind of thought you could integrate something like ∫(cosx)^2 dx by parts, as ∫(cosx)(cosx)dx. But when I do that, I come up with a tautological statement. Am I doing something wrong, or can it just not be solved that way?(1 vote)
- I do that too sometimes when integrating by parts! The problem is that if you change which one is f'(x) and which one is g(x) (or u dv whatever) then you will get a tautology, which is TRUE, but not useful at all. If you see you get a tautology after integrating by parts twice, go back to the second IbP and switch the terms, make the other function f' etc.(4 votes)

- Why doesn't he use u substitution in this problem? I find that it is easier than to do that (x-3/2)^2 business (sorry if that expression is incorrect).(3 votes)
- Some problems can be solved using both u substitution and trig substitution, but in my experience u substitutions are often easily identifiable, you can usually see a function and its derivative in the problem or at least manipulate it to get the function and its derivative. (In most cases)

So even though you may find a u substitution easier, in this case that might not be true or even possible as far as I can see. So look at the problem and think what you can possibly do to change it up a bit, for a second order polynomial such as this you can complete the square, and often completing the square leads to a constant and a (function)^2. Once you have that you can apply a trig substitution because that allready looks like the standard forms a^2-x^2 in the above example.

Doing enough examples of these kinds of problem is the only way to get to know what they want you to do, In 90% of the cases the way the question is asked, hints at how you should answer it.(1 vote)

## Video transcript

Let's say we have the
indefinite integral of the square root of 6x minus
x squared minus 5. And obviously this is not
some simple integral. I don't have just, you know,
this expression and its derivative lying around, so
u-substitution won't work. And so you can guess from just
the title of this video that we're going to have to
do something fancier. And we'll probably have to do
some type of trig substitution. But this immediately doesn't
look kind of amenable to trig substitution. I like to do trig substitution
when I see kind of a 1 minus x squared under a radical sign,
or maybe an x squared minus 1 under a radical sign, or
maybe a x squared plus 1. These are the type of things
that get my brain thinking in terms of trig substitution. but that doesn't quite
look like that just yet. I have a radical sign. I have some x squared, but it
doesn't look like this form. So let's if we can get
it to be in this form. Let me delete these guys
right there real fast. So let's see if we can maybe
complete the square down here. So let's see. If this is equal to,
let me rewrite this. And if this completing the
square doesn't look familiar to you, I have a whole
bunch of videos on that. Let me rewrite this as equal to
minus 5 minus-- I need more space up here-- minus
5 minus x squared. Now there's a plus 6x, but
I have a minus out here. So minus 6 x, right? A minus and a minus
will becomes plus 6x. And then, I want to make
this into a perfect square. So what number when I add it
to itself will be minus 6? Well, it's minus 3
and minus 3 squared. So you take half of this
number, you get minus 3, and you square it. Then you put a 9 there. Now, I can't just
arbitrarily add nines. Or actually, I didn't
add a 9 here. What did I do? I subtracted a 9. Because I threw a 9 there, but
it's really a minus 9, because of this minus sign out there. So in order to make this
neutral to my 9 that I just threw in there,
this is a minus 9. I have to add a 9. So let me add a 9. So plus 9, right there. If this doesn't make complete
sense, what I just did, and obviously you have the dx right
there, multiply this out. You get minus x squared plus
6x, which are these two terms right there, minus 9, and then
you'll have this plus 9, and these two will cancel out, and
you'll just get exactly back to what we had before. Because I want you to realize,
I didn't change the equation. This is a minus 9
because of this. So I added a 9, so I
really added 0 to it. But what this does, it gets
it into a form that I like. Obviously this, right here,
just becomes a 4, and then this term right here becomes, what? That is x minus 3 squared. x minus 3 squared. So my indefinite integral now
becomes the integral, I'm just doing a little bit of algebra,
the integral of the square root of 4 minus x minus
3 squared dx. Now this is starting to look
like a form that I like, but I like to have a 1 here. So let's factor a 4 out. So this is equal to, I'll
switch colors, that's equal to the integral of the radical,
and we'll have the 4, times 1 minus x minus 3 squared over 4. I just took a 4 out of
both of these terms. If I multiply this out, I'll
just get back to that, right there, dx. And now this is starting to
look like a form that I like. Let me simplify it even more. So this is equal to the
integral, if I take the 4 out, it becomes 2 times the square
root of 1 minus, and I can rewrite this as x minus 3,
let me write this way. 1 minus x minus 3
over 2 squared dx. And where did I
get that 2 from? Well, if I just square both
of these, I get x minus 3 squared over 2 squared. Which is x minus 3 over 4. So I have done no
calculus so far. I just algebraically rewrote
this indefinite integral as this indefinite integral. They are equivalent. But this, all of a sudden,
looks like a form that I recognize. I showed you in the last video
that cosine squared of theta is just equal to 1 minus
sine squared of theta. You could actually do
it the other way. You could do sine squared
is equal to 1 minus cosine squared. No difference. But they both will work out. But this looks an
awful lot like this. In fact, it would look exactly
like this if I say that that is equal to sine squared of theta. So let me make that
substitution. Let me write that x minus
3 over 2 squared is equal to sine squared of theta. Now, if we take the square root
of both sides of that equation, I get x minus 3 over 2 is
equal to sine of theta. Now we're eventually, you know
where this is going to go. We're eventually going to have
to substitute back for theta. So let's solve for
theta in terms of x. So theta in terms of x, we
could just say, just take the arc sine of both sides of this. You get theta is equal to,
right, the arc sine of the sine is just theta. Theta is equal to the arc
sine of x minus 3 over 2. Fair enough. Now, to actually do the
substitution, though, we're going to have figure out what
dx is, we're going to have to solve for x in terms of theta. So let me do that. So we get, if we multiply both
sides of the equation by 2, we get x minus 3 is equal to 2
times the sine of theta, or that x is equal to 2
sine of theta plus 3. Now, if we take the derivative
of both sides with respect to theta, we get dx d theta, is
equal to 2 cosine of theta, derivative of this is just 0. Or we can multiply both sides
by d theta, and we get dx is equal to 2 cosine
of theta d theta. And we're ready to substitute
back into our original indefinite integral. So this thing will now be
rewritten as the integral of 2 times the radical, if I can get
some space, of 1 minus, I'm replacing this with
sine squared of theta. And all that times dx. Well, I just said that dx is
equal to this, right here. So dx is equal to 2
cosine of theta d theta. What does this simplify to? This action right here, this is
just cosine squared of theta. And we're going to take
the square root of cosine squared of theta. So this, the square root of
cosine squared of data, this whole term right here, right? That becomes the square root of
cosine squared of theta, which is just the same thing
as cosine of theta. So our integral becomes, so our
integral is equal to, 2 times the square root of cosine
squared of theta, so that's just 2 times cosine of theta,
times 2 times cosine of theta. That's that one, right there. This is this, and all of
this radical sine, that's this, right there. 1 minus sine squared was cosine
squared, take the radical, you get cosine squared. And then everything
times d theta. Now this is obviously equal
to 4 times cosine squared of theta d theta. Which, by itself, is still not
an easy integral to solve. I don't know, you know, I
can't do U-substitution or anything like that, there. So what do we do? Well, we resort to our
good old trig identities. Now, I don't know if you have
this one memorized, although it tends to be in the inside cover
of most calculus books, or inside cover of
most trig books. But cosine squared of theta
can be rewritten as 1/2 times 1 plus cosine of 2 theta. And I've proven this
in multiple videos. So let's just make
this substitution. Let me just replace this
thing with that thing. So this integral becomes,
it equals, 4 times cosine squared of theta, but cosine
squared of theta is this. 4 times 1/2 times 1 plus
cosine of 2 theta d theta. Now, this looks
easier to deal with. So what is it? 4 times 1/2, that's 2. So my integral becomes the
integral of 2 times 1, so it's 2, plus 2 times 2 cosine of 2
theta, all of that d theta. Now, this antiderivative is
pretty straightforward. What is this, right here? This is the derivative
with respect of theta of sine of 2 theta, right? This whole thing. What's the derivative
sine of 2 theta? Take the derivative of the
inside, that's 2, times the derivative of the outside,
cosine of 2 theta. And this, of course, is the
derivative of 2 theta. So this is equal to, the
antiderivative of 2 with respect to theta, is just 2
theta, plus the antiderivative of this, which is just sine of
2 theta, and then we have a plus c. And of course, we can't forget
that we defined theta, our original antiderivative
was in terms of x. So we can't just leave
it in terms of theta. We're going to have to
do a back substitution. So let's just remember,
theta was equal to arc sine of x minus 3 over 2. Let me write that
on the side here. Now, if I immediately
substitute this theta straight into this, I'm going to get a
sine of 2 times arc sine of x minus 3 over 2, which
would be correct. And I would have a 2 times arc
sine of x minus 3 over 2. That would all be fine,
and we would be done. But that's not satisfying. It's not a nice clean answer. So let's see if we can simplify
this, so it's only in terms of sine of theta. So when you take the sine of
the arc sine, then it just simplifies to x minus 3 over 2. Let me make that clear. So if I can write all of this
in terms of sines of theta, because the sine of theta is
equal to the sine of the arc sine of x minus 3 over 2. Which is just equal
to x minus 3 over 2. So if I can write this in terms
of sines of theta, then I can just make this substitution. Sine of theta equals
that, and everything simplifies a good bit. So let's see if we can do that. Well, you may or may not know
the other identity, and I've proven this as well, that sine
of 2 theta, that's the same thing as sine of theta plus
theta, which is equal to sine of theta cosine of theta plus,
the thetas get swapped around, sine of theta plus cosine of
theta, which is equal to, this is just the same thing written
twice, 2 sine of theta cosine of theta. Some people have this memorized
ahead of time, and if you have to take an exam on trig
substitution, it doesn't hurt to have this memorized
ahead of time. But let's rewrite
this like this. So our indefinite integral
in terms of theta, or our antiderivative, became 2 theta,
plus, instead of sine of 2 theta, we could write, 2 sine
of theta cosine of theta, and of course we have a plus c. Now, I want to write everything
in terms of sines of theta, but I have a cosine if theta there. So what can we do? Well, we know we know that
cosine squared of theta is equal to 1 minus sine squared
of theta, or that cosine of theta is equal to the square
root of 1 minus sine squared of theta. Which seems like we're adding
complexity to it, but the neat thing is, it's in terms
of sine of theta. So let's do that. Let's make the substitution. So our antiderivative, this is
the same thing as 2 data plus 2 sine of theta times cosine of
theta, which is equal to this, times the square root of 1
minus sine squared of theta, and all of that plus c. Now we're in the home stretch. This problem was probably
harder than you thought it was going to be. We know that sine theta is
equal to x minus 3 over 2. So let's make that
reverse substitution. So we have 2 times theta. This first term is just 2
times theta, right there. So that's 2 times, we can't
escape the arc sine. If we just have a theta, we
have to say that theta is equal to arc sine of
x minus 3 over 2. And then we have plus,
let me switch colors, plus 2 sine of theta. Well, that's plus 2 times sine
of theta is x minus 3 over 2. So 2 times x minus 3 over 2,
and then all of that times the square root of 1 minus
sine of theta squared. What's sine of theta? It's x minus 3 over 2 squared. And of course we have a plus c. Let's see if we can simplify
this even more, now that we're at the home stretch. So this is equal to 2 arc sine
of x minus 3 over 2, plus these two terms, this 2 and this 2
cancel out, plus x minus 3, times the square root of, what
happens if we multiply everything here by, let's see,
if we take a, so this is equal to 1 minus x minus 3 over 4. x
minus 3 squared over 4. This simplification, I should
write that in quotes, is taking me longer than I thought. But let's see if we can
simplify this even more. If we multiply, let me just
focus on this term right here, if we multiply the outside,
or say, let's multiply and divide this by 2. So I'll write this as--
so let's just multiply this times 2 over 2. You might say, Sal, why
are you doing that? Because I can rewrite this, let
me write my whole thing here. So I have 2 arc sine of x
minus 3 over 2, and then I have, I could take this
denominator 2 right here. So I say, plus x
minus 3 over 2. That 2 is that 2 right there. And then I could write
this 2 right here as a square root of 4. Times the square root
of 4, times the square root of all of this. 1 minus x minus 3
squared over 4. I think you see
where I'm going. I'm kind of reversing
everything that I did at the beginning of this problem. And maybe I'm getting a little
fixated on making this as simple as possible, but I'm
so close, so let me finish. So I get 2 times the arc sine
of x minus 3 over 2, which I'm tired of writing, plus x minus
3 over 2, and if we bring this 4 in, right, the square root of
4 times the square root of that is equal to the square root
of 4 times these things. So it's 4 minus x minus
3 squared, all of that that plus c. And we're at the home stretch. This is equal to 2 times the
arc sine of x minus 3 over 2 plus x minus 3 over 2 times the
radical of 4 minus, let's expand this, x squared
minus 6x plus 9. And then this expression right
here simplifies to minus minus, it's 6x minus x squared, and
then you have a 4 minus 9 minus 5. Which is our original
antiderivative. So finally, we're
at the very end. We get the antiderivative is 2
arc sine of x minus 3 over 2 plus x minus 3 over 2 times,
time the radical of 6x minus x squared minus 5. That right there is the
antiderivative of what this thing that we had at the very
top of my little chalkboard, which is right there. So that is equal to the
antiderivative of the square root of 6x minus
x squared minus 5 dx. And I can imagine that you're
probably as tired as I am. My hand actually hurts. But hopefully you find that
to be vaguely satisfying. Sometimes I get complaints
that I only do easy problems. Well, this was quite a hairy
and not-so-easy problem.