When to use what the u and du substitutes

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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,

u = 3x \ldots \frac{du}{dx} = 3

?
 


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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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