# Indefinite Integral without Trig Substitution: integrate (dx/(sqrt(-4x - x^2)))

#### debased

##### New member
So I'm trying to understand how to integrate: (dx/(sqrt(-4x - x^2))) without using trig substitution as I'm in a Calc 1 class and we're not supposed to use anything other than u-substitution. I get stuck after completing the square:

indefinite integral of (dx/(sqrt(-4x - x^2)))
indefinite integral of (dx/(sqrt(-(x + 2)^2 + 4)))

Let u = x + 2

indefinite integral of (dx/(sqrt(4 - u^2)))

Perhaps this is an algebra question, but I'm not sure how you get that 4 to be a 1 as it's inside a radical and part of a sum. What do I do from here?

#### Jomo

##### Elite Member
So I'm trying to understand how to integrate: (dx/(sqrt(-4x - x^2))) without using trig substitution as I'm in a Calc 1 class and we're not supposed to use anything other than u-substitution. I get stuck after completing the square:

indefinite integral of (dx/(sqrt(-4x - x^2)))
indefinite integral of (dx/(sqrt(-(x + 2)^2 + 4)))

Let u = x + 2

indefinite integral of (dx/(sqrt(4 - u^2)))

Perhaps this is an algebra question, but I'm not sure how you get that 4 to be a 1 as it's inside a radical and part of a sum. What do I do from here?
Until you integrate or differentiate any questions you have will be either algebra or arithmetic.

Note that 4 =2^2.
4-u^2 = 4(1 - u^2/4) = 4(1-(u/2)^2). So sqrt(4-u^2) = sqrt(4(1-(u/2)^2)) =sqrt(4)*sqrt((1-(u/2)^2) = 2*sqrt((1-(u/2)^2).

Now what???

#### Dr.Peterson

##### Elite Member
So I'm trying to understand how to integrate: (dx/(sqrt(-4x - x^2))) without using trig substitution as I'm in a Calc 1 class and we're not supposed to use anything other than u-substitution. I get stuck after completing the square:

indefinite integral of (dx/(sqrt(-4x - x^2)))
indefinite integral of (dx/(sqrt(-(x + 2)^2 + 4)))

Let u = x + 2

indefinite integral of (dx/(sqrt(4 - u^2)))

Perhaps this is an algebra question, but I'm not sure how you get that 4 to be a 1 as it's inside a radical and part of a sum. What do I do from here?
You say you can only use u-substitution. I would call a trig substitution a kind of u-substitution, but I won't quibble over that.

The question is, what basic forms have you learned, that you can handle after you substitute? Are you implying that you have been given a formula for the integral of, say, du/sqrt(1 - u^2)?

#### debased

##### New member
Until you integrate or differentiate any questions you have will be either algebra or arithmetic.

Note that 4 =2^2.
4-u^2 = 4(1 - u^2/4) = 4(1-(u/2)^2). So sqrt(4-u^2) = sqrt(4(1-(u/2)^2)) =sqrt(4)*sqrt((1-(u/2)^2) = 2*sqrt((1-(u/2)^2).

Now what???
Then I get to 1/2 * arcsin(1/2x + 1) + C which doesn't work.

#### Dr.Peterson

##### Elite Member
Then I get to 1/2 * arcsin(1/2x + 1) + C which doesn't work.
I presume you mean arcsin((1/2)x + 1).

So you started with INT du/[2 sqrt(1-(u/2)^2)], and ended up with (1/2) arcsin((1/2)x + 1) + C. That's what I got, though I wrote it differently.

What do you mean by saying it doesn't work? Do you mean you took the derivative and didn't end up with the integrand?

#### debased

##### New member
I presume you mean arcsin((1/2)x + 1).

So you started with INT du/[2 sqrt(1-(u/2)^2)], and ended up with (1/2) arcsin((1/2)x + 1) + C. That's what I got, though I wrote it differently.

What do you mean by saying it doesn't work? Do you mean you took the derivative and didn't end up with the integrand?
Yeah if I differentiate that I get 1/(2 sqrt(-x (4 + x)))

#### Dr.Peterson

##### Elite Member
Yeah if I differentiate that I get 1/(2 sqrt(-x (4 + x)))
I don't.

#### debased

##### New member
I don't.

d/dx [ (1/2) * arcsin((1/2)x + 1) ]

(1/2) * d/dx [ arcsin((1/2)x + 1) ]

(1/2) * ( 1/(sqrt(1 - ((x+2)/2)^2)) * (1/2) )

(1/4) * (1/(sqrt(1 - ((x+2)/2)^2)))

1/(4 * sqrt(1 - ((x+2)/2)^2))

1/(4 * sqrt((-x^2 - 4x)/4))

1/(2 * sqrt(-4x - x^2))

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#### Dr.Peterson

##### Elite Member
d/dx [ (1/2) * arcsin((1/2)x + 1) ]

(1/2) * d/dx [ arcsin((1/2)x + 1) ]

(1/2) * ( 1/(sqrt(1 - ((x+2)/2)^2)) * (1/2) )

(1/4) * (1/(sqrt(1 - ((x+2)/2)^2)))

1/(4 * sqrt(1 - ((x+2)/2)^2))

1/(4 * sqrt((-x^2 - 4x)/4))

1/(2 * sqrt(-4x - x^2))
Sorry about that -- there are too many 1/2's floating around. The result I got for the integral didn't have the 1/2 on the outside; it was just arcsin((1/2)x + 1).

So we'll have to look at the last steps of your integration, rather than your check.

#### debased

##### New member
Sorry about that -- there are too many 1/2's floating around. The result I got for the integral didn't have the 1/2 on the outside; it was just arcsin((1/2)x + 1).

So we'll have to look at the last steps of your integration, rather than your check.
No worries, I've been staring at this for hours and I can't figure out where I went wrong. Can you show me the steps you did to integrate?

#### Dr.Peterson

##### Elite Member
No worries, I've been staring at this for hours and I can't figure out where I went wrong. Can you show me the steps you did to integrate?
It would be more instructive for you to write out your work (which might lead to you seeing the problem yourself); but I understand the frustration.

We substitute z = u/2, so that u = 2z and du = 2 dz. This gives INT 2 dz/[2 sqrt(1-z^2)].

The 2's cancel, so this is INT dz/sqrt(1-z^2) = arcsin(z) = arcsin(u/2) = arcsin((x+2)/2) = arcsin((1/2)x + 1).

So, what did you do differently?

#### debased

##### New member
It would be more instructive for you to write out your work (which might lead to you seeing the problem yourself); but I understand the frustration.

We substitute z = u/2, so that u = 2z and du = 2 dz. This gives INT 2 dz/[2 sqrt(1-z^2)].

The 2's cancel, so this is INT dz/sqrt(1-z^2) = arcsin(z) = arcsin(u/2) = arcsin((x+2)/2) = arcsin((1/2)x + 1).

So, what did you do differently?
I didn't do the second substitution with z. So you have to get it in the exact form of the integral i.e. 1/(sqrt(1 - x^2))? Because I did the first substitution and reached

(1/2) * integral of du/(sqrt(1 - (u/2)^2)) and saw the (u/2)^2 and thought that is good to go to fit as the arcsin. But I guess like the chain rule for differentiation you have to keep substituting until you get it in that exact form?

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#### Dr.Peterson

##### Elite Member
I didn't do the second substitution with z. So you have to get it in the exact form of the integral i.e. 1/(sqrt(1 - x^2))? Because I did the first substitution and reached

(1/2) * integral of du/(sqrt(1 - (u/2)^2)) and saw the (u/2)^2 and thought that is good to go to fit as the arcsin. But I guess like the chain rule for differentiation you have to keep substituting until you get it in that exact form?
Exactly. In fact, substitution is the chain rule in reverse. The reason we have to do substitution is to make the integral you are working on exactly match a known integral, not just look generally like it.

Incidentally, many tables of integrals give a more general form than the one you presumably know. For example, see #16 and #17 here. What they have done here is to "pre-digest" the substitution for you in #17. (Both can be done by trig substitution, which is no different from substitution in general except that you are replacing a variable with an expression rather than an expression with a variable, and you have to learn to recognize which one will work.)

#### debased

##### New member
Exactly. In fact, substitution is the chain rule in reverse. The reason we have to do substitution is to make the integral you are working on exactly match a known integral, not just look generally like it.

Incidentally, many tables of integrals give a more general form than the one you presumably know. For example, see #16 and #17 here. What they have done here is to "pre-digest" the substitution for you in #17. (Both can be done by trig substitution, which is no different from substitution in general except that you are replacing a variable with an expression rather than an expression with a variable, and you have to learn to recognize which one will work.)
Wow that seems rather important and my teacher didn't cover that. In fact, knowing that it's the chain rule in reverse makes this a lot easier to look at for me. Thanks for your help!