When to use what the u and du substitutes

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Homework Help Overview

The discussion revolves around the integral ∫2/(√(1-9x^2))*dx, focusing on the appropriate choices for u and du in the context of u-substitution. The subject area pertains to calculus, specifically integration techniques.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different choices for u and du, with one suggesting u = 3x and discussing the derivative. There is also a mention of recognizing a shortcut to the answer, prompting questions about the nature of that shortcut.

Discussion Status

The discussion has progressed with some participants offering guidance on potential substitutions and derivatives. There is an acknowledgment of the need to show work, and one participant expresses satisfaction after receiving help with identifying u and du.

Contextual Notes

Participants are navigating the constraints of needing to demonstrate their work while grappling with the complexities of u-substitution in integration. There is an emphasis on understanding the process rather than simply arriving at the answer.

neshepard
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Homework Statement



∫2/(√(1-9x^2))*dx

Homework Equations





The Attempt at a Solution



I just want to know what to use for the u and du substitutes. I know the final and I can see a short cut to the answer, but I will need to show my work and trying the find the correct u and du is proving hard. The correct answer is 2/3sin^-1(3x)+C
 
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You said you saw a shortcut to the answer. What shortcut is that? There isn't much to this u-substitution, so I'm curious if you might just have the answer and not quite recognize it.
 


It's not a short cut in the sense of ∫sinx=-cosx, just from working the problem, I see that in the equation I take the numerator,2, and pull it out, plus I know that 1/(1-x^2) = sin^-1(x). So I can take the √9 and get 3, so the answer is 2/3sin^-1(3x)+C. But like I said, I need to show the work, and when I take (1-9x^2) for u I have no other x to account for in du. So what would my u and du be?
 


I don't think you fully understand what you're doing to get from the original question to the solution.

I'll take a shot at what I think u and du are, but can you please finish the question showing your work from start to finish for us? I want to make sure you fully understand what you're doing.

Why not try,

[tex]u = 3x \ldots \frac{du}{dx} = 3[/tex]

?
 


neshepard said:
It's not a short cut in the sense of ∫sinx=-cosx, just from working the problem, I see that in the equation I take the numerator,2, and pull it out, plus I know that 1/(1-x^2) = sin^-1(x). So I can take the √9 and get 3, so the answer is 2/3sin^-1(3x)+C. But like I said, I need to show the work, and when I take (1-9x^2) for u I have no other x to account for in du. So what would my u and du be?

How about sin(u)^2=9*x^2? You have to figure out the du.
 


That did it. Thanks

∫2/(√(1-9x^2))*dx

u^2=9x^2 u=3x & du=3*dx 1/3du*dx

2*1/3∫1/√(1-u^2)*du

2/3sin^-1(u)+C

2/3sin^-1(3x)+C

and I had another one that I could see the answer, but couldn't find u and du till this help and can solve on paper.
 

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