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Integrating Using Trigonometric Substitution

  • #1
105
0

Homework Statement



Use the method of trigonometric substitution to evaluate the following:


[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]


Homework Equations



The only relevant equation that I could think of for this one was the trig identity:

[itex]sin^{2}\vartheta + cos^{2}\vartheta = 1[/itex]

The Attempt at a Solution



This is what I got from trying this problem:

[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]

[itex]\sqrt{4-9x^{2}}[/itex] is a difference between two squares, so it's a leg of a right triangle. So I let [itex]4sin\vartheta = \sqrt{4-9x^{2}}[/itex] to satisfy the definition of sine (Opposite Side over Hypotenuse).

Then, I let [itex]x^{2} = \frac{4}{9}cos\vartheta[/itex] to satisfy the definition of cosine.

Differentiating this same function with respect to [itex]x[/itex], I got:

[itex]\frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}[/itex]

Substituting these into the original function:

[itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}d\vartheta}[/itex]

[itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\frac{2}{3}\sqrt{cos\vartheta}}d\vartheta}[/itex]

[itex]\int{\frac{1}{9} \times \frac{cos\vartheta}{sin\vartheta} \times \frac{-1}{3} \times \frac{sin\vartheta}{\sqrt{cos\vartheta}}d\vartheta}[/itex]

[itex]\frac{-1}{27}\int{\frac{cos\vartheta}{\sqrt{cos\vartheta}}d\vartheta}[/itex]

At this point, I got stuck and didn't know what to do. I tried using a modified version of the identity above and just went on to see if I could get anywhere but failed.

Thanks
 

Answers and Replies

  • #2
1,196
0
they hypotenuse is actually 2, remember c^2=a^2 + b^2, the other leg is actually 3x
 
Last edited:
  • #3
867
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[itex]\sqrt{4-9x^{2}}[/itex] is a difference between two squares, so it's a leg of a right triangle. So I let [itex]4sin\vartheta = \sqrt{4-9x^{2}}[/itex] to satisfy the definition of sine (Opposite Side over Hypotenuse).

Then, I let [itex]x^{2} = \frac{4}{9}cos\vartheta[/itex] to satisfy the definition of cosine.
I can't really see how you got these substitutions... A standard substitution for this problem would be [itex]x = \frac{2}{3}\cos\vartheta[/itex]
 
  • #4
SammyS
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Homework Statement



Use the method of trigonometric substitution to evaluate the following:


[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]


Homework Equations



The only relevant equation that I could think of for this one was the trig identity:

[itex]sin^{2}\vartheta + cos^{2}\vartheta = 1[/itex]

The Attempt at a Solution



This is what I got from trying this problem:

[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]

[itex]\sqrt{4-9x^{2}}[/itex] is a difference between two squares, so it's a leg of a right triangle. So I let [itex]4sin\vartheta = \sqrt{4-9x^{2}}[/itex] to satisfy the definition of sine (Opposite Side over Hypotenuse).

...
It may help you to indicate that x is the variable of integration, by including dx in your integral: [itex]\displaystyle \int\frac{x^{2}}{\sqrt{4-9x^{2}}}\,dx\,.[/itex]

To elaborate on GreenPrint's post:

If [itex]\sqrt{4-9x^{2}}[/itex] is a leg of a right triangle, then the hypotenuse would likely be 2 and the other leg 3x.

As you say, sine is opposite over hypotenuse, so either [itex]\displaystyle \sin(\theta)=\frac{3x}{2}[/itex] or [itex]\displaystyle \sin(\theta)=\frac{\sqrt{4-9x^{2}}}{2}\,.[/itex]

cos(θ) is the other.
 
  • #5
209
0
Just making info easier to understand.

[PLAIN]http://img856.imageshack.us/img856/4011/derpto.png [Broken]

Make sure you find dx to get dθ.
 
Last edited by a moderator:
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
955

Homework Statement



Use the method of trigonometric substitution to evaluate the following:


[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]


Homework Equations



The only relevant equation that I could think of for this one was the trig identity:

[itex]sin^{2}\vartheta + cos^{2}\vartheta = 1[/itex]

The Attempt at a Solution



This is what I got from trying this problem:

[itex]\int\frac{x^{2}}{\sqrt{4-9x^{2}}}[/itex]

[itex]\sqrt{4-9x^{2}}[/itex] is a difference between two squares, so it's a leg of a right triangle. So I let [itex]4sin\vartheta = \sqrt{4-9x^{2}}[/itex] to satisfy the definition of sine (Opposite Side over Hypotenuse).

Then, I let [itex]x^{2} = \frac{4}{9}cos\vartheta[/itex] to satisfy the definition of cosine.
No, you must have [itex]x= (2/9)cos(\vartheta)[/tex].

Differentiating this same function with respect to [itex]x[/itex], I got:

[itex]\frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}[/itex]


Substituting these into the original function:

[itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}d\vartheta}[/itex]

[itex]\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\frac{2}{3}\sqrt{cos\vartheta}}d\vartheta}[/itex]

[itex]\int{\frac{1}{9} \times \frac{cos\vartheta}{sin\vartheta} \times \frac{-1}{3} \times \frac{sin\vartheta}{\sqrt{cos\vartheta}}d\vartheta}[/itex]

[itex]\frac{-1}{27}\int{\frac{cos\vartheta}{\sqrt{cos\vartheta}}d\vartheta}[/itex]

At this point, I got stuck and didn't know what to do. I tried using a modified version of the identity above and just went on to see if I could get anywhere but failed.

Thanks
 
  • #7
105
0
Oh wow. Okay. Yeah I know what I did wrong now. Thanks everyone!

I'm still a little confused about choosing which leg to be the radical though. Does it matter? I was taught to just put the radical term on the opposite side of the reference angle. But are there instances in which this won't work?
 
  • #8
1,196
0
just think about it
c^2 = a^+b^2

in order to get something in the form sqrt(z^2 - y^2)
z can only equal one thing, the hypotenuse and y can be any leg
 
  • #9
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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Oh wow. Okay. Yeah I know what I did wrong now. Thanks everyone!

I'm still a little confused about choosing which leg to be the radical though. Does it matter? I was taught to just put the radical term on the opposite side of the reference angle. But are there instances in which this won't work?
No, it doesn't matter.
 

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