- #1

- 147

- 0

C is a constant

any ideas as to how I shoud start to integrate this one?

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- Thread starter LocationX
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- #1

- 147

- 0

C is a constant

any ideas as to how I shoud start to integrate this one?

- #2

LeonhardEuler

Gold Member

- 859

- 1

[tex]\int x\sqrt{C^2-x^2}dx[/tex]

Then just make the substitution [itex]u=x^2[/itex].

- #3

- 147

- 0

so then how would I integrate

[tex]\int (C^2-u)^{0.5}[/tex]

[tex]\int (C^2-u)^{0.5}[/tex]

- #4

- 50

- 0

[tex]du = -2x dx[/tex]

[tex]\frac{du}{-2x} = dx [/tex]

then you just use your normal integration rules.

- #5

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,025

- 134

Just a minor correction.LeonhardEuler said:

[tex]\int x\sqrt{C^2-x^2}dx[/tex]

Then just make the substitution [itex]u=x^2[/itex].

We should have:

[tex]\int\sqrt{C^{2}x^{2}-x^{4}}dx=\int|x|\sqrt{C^{2}-x^{2}}dx[/tex]

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