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Trig Integration By Substitution

  • Thread starter Rachael95
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  • #1
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Mod note: Moved from technical math section
∫(2x+6)/sqrt(5-4x-x^2)

I have 2/3(ln|tan(theta)+sec(theta)|-3|cos(theta)|) where x=sin^-1((x+2)/3)
 
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  • #2
SteamKing
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∫(2x+6)/sqrt(5-4x^2-x^2)
Are there supposed to be two x^2 expressions under the SQRT?
 
  • #3
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Apologies no it should be 5-4x-x^2
 
  • #4
HallsofIvy
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Complete the square in the square root: [tex]5- 4x- x^2= 5- (x^2+ 4x+ 4- 4)= 5- (x+ 2)^2+ 4= 9- (x+ 2)^2.

Now make the substitution u= x+ 2, du= dx, x= u- 2 so 2x+ 6= 2u+ 2. The integral becomes [tex]\int \frac{2u+ 2}{\sqrt{9- u^2} du[/tex]

Now let [itex]u= 3 sin(\theta)[/itex].
 
  • #5
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Homework-type problems should be posted in the Homework & Coursework section. I have moved this thread.
 

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