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## Calculus, all content (2017 edition)

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

Lesson 2: u-substitution- 𝘶-substitution intro
- 𝘶-substitution: rational function
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: logarithmic function
- 𝘶-substitution: challenging application
- 𝘶-substitution: special application
- 𝘶-substitution: indefinite integrals
- 𝘶-substitution warmup
- 𝘶-substitution: definite integral of exponential function
- 𝘶-substitution: double substitution
- 𝘶-substitution: definite integrals
- u-substitution challenge

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# 𝘶-substitution: double substitution

Finding the indefinite integral of cos(5x)/e^[sin(5x)]. To do that, we need to perform 𝘶-substitution twice. Created by Sal Khan.

## Want to join the conversation?

- Okay, so I tried solving the problem my own way and the answer I got puzzles me.

I did the same as Sal up until2:15where I just decided to take the indefinite integral of 1/e^u, and this gave me (1/5)*(ln|e^u| + C). By the properties of the natural logarithm, wouldn't this simplify to (1/5)*(u + C)? When substituting back the expression instead of u, I don't get the same result as Sal did. I assume the absolute value messes up my reasoning somehow, and my question is: Why? I can't think of any possible situation where |e^u| would not be equal to (e^u).(14 votes)- The indefinite integral of 1/e^u isn't ln |e^u| + C, it's -1/e^u + C.

Anything of the form f´(x)/f(x) integrates to ln |f(x)| + C, 1/e^u clearly isn't of that form.(32 votes)

- Can't we just make a general property that integration of e^-x is -e^x?(0 votes)
- The integral of e^(-x) is DEFINITELY not -e^x. The integral of e^(-x) is -e^(-x). The x in the exponent is negative as well.

As for making "general properties," in these integral things, usually they only make general properties out of the basic things you need. Since we already know that the integral of e^x is e^x, we can figure out the integral of e^(-x) using skills we already know.

In short, it's not made into a general property since you already are able to solve it using other general properties. That being said, if you want to memorize it to be able to use it more quickly, feel free to do so.(36 votes)

- Why substitute again what's wrong with e^-u ??(10 votes)
- You could integrate e^-u in your head, but it may not be clear to some people what the antiderivative of e^-u is. Therefore, we must use u-substitution to change the integral into one we can solve, namely integral of e^u, which is very simple.(1 vote)

- How do you know when to do u substitution twice?(4 votes)
- I think it's just pretty much that if you have a problem that's seems to be workable with u-sub, but you don't quite "get there" on the first go. Then you do it again. :)

My guess is that most of the time you will only notice that you need to do it double towards the end of the problem.

Hope that helps!(7 votes)

- Why e^-u at2:33.I dont get it!(4 votes)
- Basic property of powers in a fraction: those on the bottom are negatives of those on the top. 1/x = x^(-1) Therefore, 1/e^u = e^(-u) to get rid of the fraction.(8 votes)

- I don't quite get why Sal had to do the u-sub twice. 1/5 int e^-u du would be simply -1/5 e^-u + c?(4 votes)
- Yes, you are correct. You don't have to do that second substitution, but you can do it. When you learn how to integrate by the reverse chain rule (if you haven't already), you can do this problem without any substitution.(4 votes)

- Why d(sin5x)/dx = 5cos5x ? Please explain.(1 vote)
- You have to use the chain rule here

d/dx (sin x) = cos x so if there is a function inside instead of just an x you do

d/dx (sin (f(x))) = cos (f(x))*d/dx(f(x))

in other words find the derivative of sin then multiply by the derivative of the inside function so...

d/dx (sin (5x)) = cos(5x)*d/dx(5x)= cos(5x)*5 = 5cos(5x)(8 votes)

- Instead of multiplying the whole integral by 5/5, could you not just make the substitution 1/5 du = cos(5x) dx ?(3 votes)
- You could make the substitution that way as well. It is merely a stylistic preference.(4 votes)

- Why did he have to do a double substitution? Can't I just stop at 1/5 int( du/e^u ) and say that this is 1/5 ln(e^u) + C ?(4 votes)
- Have you figured out why? I did the same thing as you, but I don't understand why it's not right.(1 vote)

- Why did Sal use a double substitution? He ultimately got the same result, that the antiderivative of e^-u=e^-u, much like the antiderivative of e^x = e^x.(1 vote)
- Not quite. The antiderivative of e^-u is -e^-u (plus c).(4 votes)

## Video transcript

Let's see if we can take the
integral of cosine of 5x over e to the sine of 5x dx. And there's a crow squawking
outside of my window, so I'll try to stay focused. So let's think about
whether u-substitution might be appropriate. Your first temptation
might have said, hey, maybe we let u equal sine of 5x. And if u is equal
to sine of 5x, we have something that's
pretty close to du up here. Let's verify that. So du could be equal to--
so du dx, derivative of u with respect to x. Well, we just use
the chain rule. Derivative of 5x is 5. Times the derivative of sine
of 5x with respect to 5x, that's just going
to be cosine of 5x. If we want to write this
into differential form, which is useful when we do
our u-substitution, we could say that du is
equal to 5 cosine 5x. And we look over here, we
don't have quite du there. We have just cosine of 5x dx. Cosine of 5x dx just like that. So when you look over here,
you have a cosine of 5x dx, but we don't have a
5 cosine of 5x dx. But we know how to solve that. We can multiply by
5 and divide by 5. 1/5 times 5 is
just going to be 1, so we haven't changed the
value of the expression. When we do it this way, we see
pretty clearly we have our u and we have our du. Our du is 5. Let me circle that. Let me do that in that blue
color-- is 5 cosine of 5x dx. So we can rewrite
this entire expression as-- do that 1/5 in purple. This is going to
be equal to 1/5-- I hope you don't hear
that crow outside. He's getting quite obnoxious. 1/5 times the integral of all
the stuff in blue is my du, and then that is
over e to the u. So how do we take the
antiderivative of this? Well, you might be tempted to
well, what would you do here? Well, we're still not
quite ready to simply take the antiderivative here. If I were to rewrite
this, I could rewrite this as this is equal to 1/5
times the integral of e to the negative u du. And so what might
jump out at you is maybe we do
another substitution. We've already used the letter
u, so maybe we'll use w. We'll do some w-substitution. And you might be able
to do this in your head, but we'll do
w-substitution just to make it a little bit clearer. This would have been really
useful if this was just e to the u because we know the
antiderivative of e to the u is just e to the u. So let's try to get it
in terms of the form of e to the something and not e
to the negative something. So let's set. And I'm running
out of colors here. Let's set w as
equal to negative u. In that case, then
dw, derivative of w with respect
to u, is negative 1. Or if we were to write that
statement in differential form, dw is equal to du times
negative 1 is negative du. So this right over
here would be our w. And do we have a dw here? Well, we have just a du. We don't have a
negative du there. But we can create a negative
du by multiplying this inside by a negative 1, but then also
by multiplying the outside by a negative 1. Negative 1 times
negative 1 is positive 1. We haven't changed the value. We have to do both of these
in order for it to make sense. Or I could do it like this. So negative 1 over here and a
negative 1 right over there. And if we do it in that form,
then this negative 1 times du, that's the same
thing as negative du. This is this right over here. And so we can
rewrite our integral. It's going to be equal to now
it's going to be negative 1/5. Trying to use the
colors as best as I can. Negative 1/5 times the
indefinite integral of e to the-- well,
instead of negative u, we can write w-- e to the w. And instead of du times
negative 1, or negative du, we can write dw. Now this simplifies
things a good bit. We know what the antiderivative
of this is in terms of w. This is going to be equal
to negative 1/5 e to the w. And then we might have
some constant there, so I'll just do a plus c. And now we just have to do
all of our unsubstituting. So we know that w is
equal to negative u. So we could write that. So this is equal
to negative 1/5. I want to stay
true to my colors. Negative 1/5 e to the negative
u, that's what w is equal to, plus c. But we're still not
done unsubstituting. We know that u is
equal to sine of 5x. u is equal to sine of
5x, so we can write this as being equal to
negative 1/5 times e to the negative u, which is
negative u is sine of 5x. And then finally,
we have our plus c. Now, there was a simpler way
that we could have done this by just doing one substitution. But then you would have had
to look ahead a little bit and realize that it was not
trivial to take your-- not too bad to take your antiderivative
of e to the negative u. The insight that you might have
had-- although you shouldn't really hold yourself,
feel too bad if you didn't see that insight-- is
that we could have rewritten that original integral--
and let me rewrite it. It's cosine of 5x over
e to the sine of 5x dx. We could have written this
entire original integral as being equal to
cosine of 5x times e to the negative sine of 5x dx. And in this situation,
we could have set u to be equal to negative
5x and say, well, if u is equal to
negative sine of 5x. If u is equal to
negative sine of 5x, then du is going to be equal
to negative 5 cosine of 5x. And we don't have a
negative-- oh, dx. We don't have a negative 5
here, but we could construct one by putting a negative 5
there and then multiplying by negative 1/5. And then that would have
immediately simplified this integral right over here
to be equal to negative 1/5 times the integral of
well, we have our du. Let me do this in
a different color. We have our du. That's the negative 5. Let me do it this way. Negative 5 cosine of 5x dx. So that is our du. I'm just changing the
order of multiplication. Times e to the u. This whole thing now is u
this second time around. So if we did it this way,
with just one substitution, we could have immediately gotten
to the result that we wanted. You take the
antiderivative of this. I'll do it in one color
now just because I think you get the idea. This is equal to negative
1/5 e to the u plus c. u is equal to
negative sine of 5x. So this is equal
to negative 1/5 e to the negative sine of
5x plus c And you're done. So this one is faster. It's simpler. And over time, you
might even start being able to do
this in your head. This top one, you
still didn't mess up by just setting u
equal to sine of 5x, we just have to do an
extra substitution in order to work it through all the way. And I was able to do this
video despite the crowing crow outside, or squawking crow.