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

1. Originally Posted by debased
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?

2. Originally Posted by Dr.Peterson
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?

3. Originally Posted by debased
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.)

4. Originally Posted by Dr.Peterson
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!

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