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Integration With Completing the Square

  • Thread starter Bashyboy
  • Start date
  • #1
1,421
5
The problem is: [itex]\int\frac{1}{\sqrt{1-4x-x^2}}dx[/itex]

I took the expression under the radical and I completed the square, yielding: [itex]\int\frac{1}{\sqrt{5-(x+2)^2}}dx[/itex]

Then I figured that I could apply the arcsin formula, where [itex]a^2=5[/itex] and[itex]u^2=(x+2)^2[/itex]

But by solving for "a" and "u," I would be left with two roots, wouldn't I?
 

Answers and Replies

  • #2
6,054
390
What exactly would you solve for a and u?
 
  • #3
123
0
Forget about the "formula". If you're unsure, do some integration by substitution and try to get to something familiar.

[tex]\int \frac{1}{\sqrt{5-(x+2)^2}}dx=\int \frac{1}{\sqrt{5}}\frac{1}{\sqrt{1-\frac{(x+2)^2}{5}}}dx=\frac{1}{\sqrt{5}} \int \frac{1}{\sqrt{1-\left(\frac{x+2}{\sqrt{5}}\right)^2}}dx[/tex]
What should you do next? The integrand is essentially [itex]\frac{1}{\sqrt{1-x^2}}[/itex], you just need the right substitution.
 
  • #4
1,421
5
Thank you, Christoff, that was rather clever.
 

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