- #1

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## Homework Statement

[tex]\int xarcsin2xdx[/tex]

**2. The attempt at a solution**

Can someone explain to me what is happening at step 2? I understand how the integration by parts was done, but where does the (1/8) or (2x) come from?

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- Thread starter wvcaudill2
- Start date

- #1

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[tex]\int xarcsin2xdx[/tex]

Can someone explain to me what is happening at step 2? I understand how the integration by parts was done, but where does the (1/8) or (2x) come from?

- #2

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[tex]\frac{x^2}{\sqrt{1-4x^2}}=\frac{8x^2}{8\sqrt{1-4x^2}}[/tex]

and then they rewrote this slightly.

- #3

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[tex]\frac{x^2}{\sqrt{1-4x^2}}=\frac{8x^2}{8\sqrt{1-4x^2}}[/tex]

and then they rewrote this slightly.

I still dont see how this gets (1/8), and even then, what is the purpose of doing this, and where does the (2x) come from?

- #4

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And the numerator is rewritten as [itex]8x^2=2(2x)^2[/itex].

The purpose of doing that is to do a substitution t=2x.

- #5

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And the numerator is rewritten as [itex]8x^2=2(2x)^2[/itex].

The purpose of doing that is to do a substitution t=2x.

Oh, ok. Thanks again!

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