Integration by parts & inv. trig fxn

  • Thread starter wvcaudill2
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  • #1
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Homework Statement


[tex]\int xarcsin2xdx[/tex]

2. The attempt at a solution
Image1.jpg


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?
 

Answers and Replies

  • #2
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They multiplied the numerator and denominator by 8. Thus

[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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They multiplied the numerator and denominator by 8. Thus

[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
22,129
3,297
Yoi get the 1/8 by bringing the denominator out of the integral.
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
54
0
Yoi get the 1/8 by bringing the denominator out of the integral.
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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