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### Course: Integral Calculus (2017 edition) > Unit 6

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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# Trig and u substitution together (part 2)

More of all of the substitution! Created by Sal Khan.

## Want to join the conversation?

- Is there anything wrong with just plugging in cosθ for all the u's and then plugging in arcsin(x/3) for all the θ's?(56 votes)
- I did it with the arcsin and i think its correct. Either way its not going to be a pretty answer and i try to avoid having to do all the trig and geometry stuff, especially if its on a test and im rushed for time.(24 votes)

- so, in general, when given an integral where 2 functions are multiplied or divided one should:

1) Check to see if Trig substitution is applicable, if not

2) Check to see if U Substitution is applicable, if not

3) Integrate by parts

would this be the optimal algorithm for calculating integrals of this format or is there a better way?(16 votes)- Plan A should be look for obvious u sub. Next, try integration by parts. Then see if it can be turned into u sub by trig identities, then try trig substitution, finally try partial fraction decomposition.(26 votes)

- Where did the -243 go? You copied it as 243 in the following line.(4 votes)
- He distributed the -1 inside the parentheses.

-243(a - b) = 243(b - a)(16 votes)

- How did the 1/3 become 1/9 when it was put in the radical?(4 votes)
- It became that because we're inserting that value into a square root. If you undo the 1/9 from the square root you should be having the original value 1/3. Similar to have an original value of 4, now when you put that in a square root it should be 16 simply because when you take the square root of that you should be able to get your original value 4. I hope that helps someway for people with similar question.(1 vote)

- We can use the Pythagorean identity, right? Here's what I've done:

cos θ = u, sin θ = x/3

cos^2 (θ) + sin^2 (θ) = 1

u^2 + (x^2)/9 = 1

u^2 = 1 – (x^2)/9

u = sqrt (1 – (x^2)/9)

which is the same as sir Sal's result.(5 votes) - you can just to u-sub and avoid trig sub. change x^3 to x and x^2 where x^2 = 9-u...and du/dx = -2x. It saved my sanity.(3 votes)
- At2:21, Sal says that u=cos(theta)=sqrt(1-sin^2(theta)). I thought that sqrt(1-sin^2(theta)) = cos^2(theta)(2 votes)
- You're forgetting to take the square root. As you know, sin²Θ + cos²Θ = 1. Then cos²Θ = 1 - sin²Θ. Now take the square root of both sides of the equation to get Sal's result.(3 votes)

- Couldn't you just solve for θ by taking the inverse sine of both sides of (x/3)=sin(θ) to get θ=arcsin(x/3) and then u=cos(arcsin(x/3))?(1 vote)
- That creates problems because the domain of arcsin(x/3) is just [-3, 3]. We want this to hold in general, not just in that interval.(2 votes)

- At1:42, would it be okay to say that cos θ = sin 90-θ? So if x/3 = sin θ, then u = 90 - x/3, right?(2 votes)
- Where did the part "cos θ = 90 - sin θ" come from? I think it is incorrect, actually I'm pretty sure, because the range of cos and sin is the set [-1, 1], and your equation implies that either sin or cos have values way beyond that interval.(0 votes)

- could you use:

sin(x) = cos(x - pi/2)(2 votes)- Someone might find this useful many years later, but it seems like you cannot. If u = cos (theta) does that mean that u = cos (theta - pi/2) is also true? Obviously not since the input for cosine changed.(1 vote)

## Video transcript

In the last video,
in order to evaluate this indefinite integral, we
first made the substitution that x is equal to 3 sine theta. And then this got us to
an integral of this form. Then we were able to break
up these sines and cosines and use a little bit
of our trig identities. To get it into the form where
we could do u substitution, we did another
substitution where we said that u is equal
to cosine of theta. And then finally, we
were able to get it into a form using that
second round of substitution. And this time, it
was u substitution. We were able to
get it into a form that we could actually
take the antiderivative. And we got the final
answer here in terms of u. But now we've got to
go and undo everything. We have to undo
the substitutions. So the last substitution
we had done-- we're now going to
go in reverse order-- was that u was equal
to cosine theta. So you might just want to
substitute u with cosine theta here. But then we're going
to have everything in terms of cosine theta, which
still doesn't get us to x. So the ideal is if I can
somehow express u in terms of x. So let's think how we can do it. We know that u is
equal to cosine theta. We know the relationship between
x and theta is right over here. x is equal to 3 sine theta. So let's write that over here. So we know that x-- let
me write it over here-- we know that x is
equal to 3 sine theta. So if we could
somehow write cosine-- let me rewrite this a
little bit differently. Or we could also say that x
over 3 is equal to sine theta. I just divided both sides by 3. So if we could
somehow re-express this in terms of
sine theta, then we can replace all the sine
thetas with x over 3's, and we are done. So how can we do that? And I'll actually show you
two techniques for doing it. So the first one is to
make the realization, OK, u is equal to cosine of theta. If I want to write this
in terms of sine of theta, I can just say that this
is equal to-- straight up, this is the most fundamental
trigonometric identity. Cosine theta is the square root
of 1 minus sine squared theta. And we see sine of theta
is equal to x over 3. So this is the square root
of 1 minus x over 3 squared. So this is u in terms of x. So everywhere we
see a u up here we can replace it with
this expression. And we are essentially done. We would have written
this in terms of x. Now, there's another
technique you might sometimes see in a calculus class
where someone says, OK, we know that u is equal
to cosine theta. We know this relationship. How can we express
u in terms of x? And we'll say, let's
draw a right triangle. They'll draw a right
triangle like this. They'll draw a right triangle,
and they'll say, OK, look, sine of theta is x over 3. So if we say that this is theta
right over here, sine of theta is the same thing as
opposite over hypotenuse. Opposite over hypotenuse
is equal to x over 3. So let's say that this is x and
then this right over here is 3. Then the sine of theta
will be x over 3. So we look at that first
substitution right over here. But in order to figure out
what u is in terms of x, we need to figure out
what cosine of theta is. Well, cosine is adjacent
over hypotenuse. So we have to figure out
what this adjacent side is. Well, we can just use the
Pythagorean theorem for that. Pythagorean theorem
would tell us that this is going
to be the square root of the hypotenuse
squared, which is 9, minus the other side
squared, minus x squared. So from this, we fully
solved the right triangle in terms of x. We can realize that
cosine of theta is going to be equal to the
adjacent side, square root of 9 minus x squared,
over the hypotenuse, over 3, which is the same thing
as 1/3 times the square root of 9 minus x squared,
which is the same thing if we square 1/3 and
put it into the radical. So we're essentially going
to take the square-- 1/3 is the same thing as
the square root of 1/9. So can rewrite this as the
square root of 1/9 times 9 minus x squared. Essentially, we just brought
the 1/3 third into the radical. Now it's 1/9. And so now this is going
to be the same thing as the square root of 1
minus x squared over 9, which is exactly this
thing right over here. x squared over 9 is the same
thing as x over 3 squared. So either way, you
get the same result. I find using the trig
identity right over here to express cosine of theta
in terms of sine theta and then just do
the substitution to be a little bit
more straightforward. But now we can just substitute
into the original thing. So either of these-- I can
write it as either way-- this thing right over here,
this is the same thing as 1 minus x squared over
9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just
substitute it with this thing. So our final answer
in terms of x is going to be equal to 243
times u to the fifth, this to the fifth power over 5. This to the fifth power is
1 minus x squared over 9. It was to the 1/2, but if we
raise that to the fifth power, it's now going to
be to the 5/2 power over 5 minus this to the
third power, 1 minus x squared over 9 to the 3/2 raising this
to the third power-- that's this right over here-- over 3,
and then all of that plus c. And we're done. It's messy, but using first
trig substitution then u substitution, or trig
substitution then rearranging using a couple of our
techniques for manipulating these powers of trig functions,
we're able to get into a form where we could use
u substitution, and then we were able to
unwind all the substitutions and actually evaluate
the indefinite integral.