Integration: completing the square and inverse trig functions

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  • #1
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Homework Statement



Find [tex]\int[/tex](x+2)dx/sqrt(3+2x-x2)

Homework Equations



[tex]\int[/tex]du/sqrt(a2 - u2) = sin-1u/a


The Attempt at a Solution



I began by completing the square:
3+2x-x2 = 4 -(x2-2x+1)

So, 4-(x-1)2 = a2-u2 and a=2 and u=(x-1)
Further, since x=(u+1), dx=du and (x+2)=(u+3)

Substituting, I got:

[tex]\int[/tex](x+2)dx/sqrt(3+2x-x2) = [tex]\int[/tex](u+3)du/sqrt(a2-u2)

= [tex]\int[/tex]udu/sqrt(a2-u2) + 3[tex]\int[/tex]du/sqrt(a2-u2)

This is where I get confused. I know that 3[tex]\int[/tex]du/sqrt(a2-u2) = 3sin-1u/a

But, according to my book, [tex]\int[/tex]udu/sqrt(a2-u2) = -sqrt(a2-u2) and I don't understand why.

Can someone offer an explanation? Thanks.
 

Answers and Replies

  • #2
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This is where I get confused. I know that 3[tex]\int[/tex]du/sqrt(a2-u2) = 3sin-1u/a

But, according to my book, [tex]\int[/tex]udu/sqrt(a2-u2) = -sqrt(a2-u2) and I don't understand why.

Can someone offer an explanation? Thanks.


using substitution

cos [tex]\theta[/tex] = [tex]\frac{u}{a}[/tex]
 
  • #3
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using substitution

cos [tex]\theta[/tex] = [tex]\frac{u}{a}[/tex]

Are you saying that I should substitute u=a sin theta to obtain:

[tex]\int[/tex]udu/sqrt(a2-u2) = [tex]\int[/tex](a sin theta)(a cos theta)(d theta)/(a cos theta) = a cos theta

So, u = a cos theta and cos theta = u/a, but u/a doesn't equal -sqrt(a2-u2)

What am I missing?
 
  • #4
1,719
151

This is where I get confused. I know that 3[tex]\int[/tex]du/sqrt(a2-u2) = 3sin-1u/a

But, according to my book, [tex]\int[/tex]udu/sqrt(a2-u2) = -sqrt(a2-u2) and I don't understand why.

Can someone offer an explanation? Thanks.


see the difference :
3[tex]\int[/tex]du/sqrt(a2-u2) = 3sin-1u/a

and

[tex]\int[/tex]u du/sqrt(a2-u2) = -sqrt(a2-u2)

to obtain the second one, use : v = a2-u2
 
  • #5
Mentallic
Homework Helper
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Just take the derivative of both answers and you'll see why.
 
  • #6
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I see it now. Thanks for your help.

[tex]\int[/tex](a2-u2)-1/2udu

Substitute v=a2-u2, dv=-2udu, udu=-dv/2

= -1/2[tex]\int[/tex]v-1/2dv = -v-1/2 = -sqrt(a2-u2)
 

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