Question

Integration question: does this require the use of arctan's derivative? (Equation posted below)

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{4-x} }dx\]

Answer 2

After slogging through a good deal of integration-by-substitution problems, I can't get my head around this one...

Answer 3

No... we don't need trigo sub. here :|

What would you get when you differentiate \(-\sqrt{4-x}\)?

Answer 4

Presumably we'd end up with

Answer 5

I apologize for the drawing and the messiness, but I'm running low on brain power right now and I can't deal with the Equation tool.

Answer 6

Check the sign again?!

Answer 7

Isn't it the same as deriving -((4-x)^(1/2)?
I see now that I forgot to use the chain rule and that there's an extra - there, making it positive. But you'll have to forgive me if how the problem needs to be solved isn't quite clicking yet.

Answer 8

Do you need to use two substitutions?

Answer 9

\[\frac{d}{dx} (-\sqrt{4-x})= -\frac{-1}{2\sqrt{4-x}}= \frac{1}{2\sqrt{4-x}}\]Does that make sense to you?

Answer 10

Yes, it does.

Answer 11

So then would \[2\int\limits_{}^{}\frac{ 1 }{ 2\sqrt{4-x} }dx = \int\limits_{}^{}\frac{ 1 }{ \sqrt{4-x} }dx\]
If so, would it make sense to specify our du term as being multiplied by 2, in order to end up with the above result?

Answer 12

First question : Yes
Second question: Sub u = \(-2 \sqrt{4-x}\)

Answer 13

I understand why that works and how to get there and I will put that down... and look at it again tomorrow when maybe I can wrap my brain around it fully enough to be able to apply that workaround automatically next time. Thank you very much for being so patient.

Answer 14

You're welcome :)