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Integration by Trig Substitution

  • Thread starter MathHawk
  • Start date
  • #1
9
0
I've attempted solutions at this in two different manners and found myself stuck both ways. I'll show you the way that seems to make progress. The other way involves not factoring the junk under the radical in the denominator.

Homework Statement


[tex]\int[/tex][tex]^{1}_{0}[/tex] [tex]\frac{3x^2 -1}{\sqrt{x-x^3}}[/tex] dx




Homework Equations



For case [tex]\sqrt{a^2-x^2}[/tex], use substitution a*sin[tex]\theta[/tex].



The Attempt at a Solution



[tex]\int[/tex][tex]^{1}_{0}[/tex] [tex]\frac{3x^2 -1}{\sqrt{x-x^3}}[/tex] dx

= [tex]\int[/tex][tex]^{1}_{0}[/tex] [tex]\frac{3x^2 -1}{\sqrt{1-x^2}\sqrt{x}}[/tex] dx

Substitute: x = sin[tex]\theta[/tex], dx = cos[tex]\theta[/tex] d[tex]\theta[/tex], bounds of 0 - 1 remain the same as sin 0 = 0 and sin 1 = 1.

[tex]\int[/tex][tex]^{1}_{0}[/tex] [tex]\frac{3sin^2\theta -1}{\sqrt{cos^2\theta}\sqrt{sin\theta}}[/tex] cos[tex]\theta[/tex] d[tex]\theta[/tex]


= [tex]\int[/tex][tex]^{1}_{0}[/tex] [tex]\frac{3sin^2\theta -1}{cos\theta\sqrt{sin\theta}}[/tex] d[tex]\theta[/tex]



At this point I no longer know where to go. Factoring the numerator appeared to bear no fruit, no u substitutions make any sense to me at the moment.
 

Answers and Replies

  • #2
33,646
5,313
Trig substitution is the wrong way to go. A simple substitution will work here: u = x - x3. Note that the numerator is very close to du.
 
  • #3
9
0
Oh my gosh, thank you. Made that way too hard on myself ;). I even thought about substituting using the 1-sin^2x inside the radical (instead of changing it to cos^2x) and it never occured to me to look at the original question. Your help is very appreciated.
 
  • #4
9
0
This changes my bounds to 0 and 0. Can I stop here and claim that the answer is 0?
 
  • #5
rock.freak667
Homework Helper
6,230
31
This changes my bounds to 0 and 0. Can I stop here and claim that the answer is 0?
Yes you can

EDIT: Actually, on second thought for

[tex]\frac{3x^2-1}{\sqrt{x-x^3}}=0 \Rightarrow x= \pm \frac{1}{\sqrt{3}}[/tex]


Meaning that between 1 and 0 the curve passes through the x-axis, you would probably need to change from 1 to 0 to

[tex]\int ^{1} _{\frac{1}{\sqrt{3}}} \frac{3x^2-1}{\sqrt{x-x^3}} dx + \left | \int ^{\frac{1}{\sqrt{3}}} _{0} \frac{3x^2-1}{\sqrt{x-x^3}}dx \right |[/tex]
 
Last edited:
  • #6
33,646
5,313
You can change the limits of integration as you were thinking, or you can leave them unchanged, integrate using the substitution, and then undo the substitution, and finally evaluate. When I do it this way, I remind myself that these are x limits by writing x = <whatever> for the lower limit of integration.
 
  • #7
33,646
5,313
You should also note that this is an improper integral. The integrand is undefined at both limits of integration.
 

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