Trignometric Substiution Problem

  • Thread starter Archimedes II
  • Start date
In summary, the code is messed up, and you need to square the sin of the original equation to get an answer that is in the correct range.
  • #1
Archimedes II
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0

Homework Statement



[itex]\displaystyle \int _0 ^a 4x^2 \sqrt{a^2-x^2} dx[/itex]

Homework Equations



[itex]\cos^2 x =1 - \sin^2 x[/itex]

The Attempt at a Solution



Substuting

Substituting [itex]x=a \sin(\theta)[/itex]

[itex]dx=a\cos( \theta) d\theta[/itex]

x=0
asin[itex]\theta[/itex]=0
sin[itex]\theta[/itex]=0
[itex]\theta[/itex]=0

[itex]\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =[/itex]

[itex]\displaystyle \int_{0}^{\pi/2} 4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=[/itex]

[itex]\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta \sqrt{a^2-a^2sin^2\theta}a\cos\theta d\theta=[/itex]

[itex]\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a \sqrt{1-sin^2\theta}a\cos\theta d\theta=[/itex]

[itex]\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a cos\theta a\cos\theta d\theta=[/itex]

[itex]\displaystyle 4a^4\int_{0}^{\pi/2} sin \theta cos^2 \theta d\theta=[/itex]

U substitution

[itex]u=cos\theta[/itex]

[itex]-du= sin\theta d\theta[/itex]

[itex]\displaystyle 4a^4\int u^2 du=[/itex]
When I put in the bounds it yields from 1 to 0. What did I do wrong
 
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  • #2
So the code is messed up basically

Integral from 0 to a [4x^2 squareroot(a^2 - x^2)]dx =

4a^4 integral from 0 to pi/2 [sin theta cos^2 theta] dtheta

I'm stuck trying to change the bounds with a u substitution
 
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  • #3
Archimedes II said:

Homework Statement



[itex]\displaystyle \int _0 ^a 4x^2 \sqrt{a^2-x^2} dx[/itex]

Homework Equations



[itex]\cos^2 x =1 - \sin^2 x[/itex]

The Attempt at a Solution



Substituting [itex]x=a \sin(\theta)[/itex]

[itex]dx=a\cos( \theta) d\theta[/itex]

New Bounds

x=a
a sin[itex]\theta[/itex]=a
sin[itex]\theta[/itex]=1
[itex]\theta[/itex]=[itex]\pi[/itex]/2x=0
asin[itex]\theta[/itex]=0
sin[itex]\theta[/itex]=0
[itex]\theta[/itex]=0

[itex]\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =[/itex]

[itex]\displaystyle \int_{0}^{\pi/2} 4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2}\ \, a\cos\theta d\theta=[/itex]
I fixed your LaTeX code up to this point.

[itex]\int from 0 to \pi/2 4(a^2sin^2\theta)\sgrt{a^2-(a^2sin^2\theta)} acos\theta d\theta[/itex]=

[itex]\int from 0 to \pi/2 4(a^2sin^2\theta) a \sgrt{(1- sin\theta)} acos\theta d\theta[/itex]=

[itex]\int from 0 to \pi/2 4(a^2sin^2\theta) a sos\theta acos\theta d\theta[/itex]=

[itex]4a^4\int form 0 to \pi/2 sin \theta cos^2 \theta d\theta[/itex]

Let u= [itex]cos/theta[/itex]

thus -du=[itex]sin/theta d\theta[/itex]

My problems is when I set the new bounds I get from 1 to 0 not 0 to 1. What am I doing wrong?
Here is the Code for some of the lines I fixed:

Code:
[itex]\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =[/itex]

[itex]\displaystyle \int_{0}^{\pi/2}  4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=[/itex]

The "\displaystyle" modifier allows the integral symbol to be larger than otherwise.

I'll let you edit the Original Post .
 
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  • #4
Ok, I fixed it thank you very much. I didn't know how to put a bound into the intergral code.
 
  • #5
Archimedes II said:
...

[itex]\displaystyle 4a^4\int_{0}^{\pi/2} sin \theta cos^2 \theta d\theta=[/itex]

U substitution

[itex]u=cos\theta[/itex]

[itex]-du= sin\theta d\theta[/itex]

When I put in the bounds it yields from 1 to 0. What did I do wrong
So you get

[itex]\displaystyle -4a^4\int_{1}^{0} u^2\,du[/itex]

Right?
 
  • #6
SammyS said:
So you get

[itex]\displaystyle -4a^4\int_{1}^{0} u^2\,du[/itex]

Right?

Yes but the how can the bound be from a higher number to a smaller one.
 
  • #7
Archimedes II said:
Yes but the how can the bound be from a higher number to a smaller one.
What does switching the order do to the integral?
 
  • #8
SammyS said:
What does switching the order do to the integral?

I solved it I forgot to square the sin at the end. I did't need to use a u substitution.

Any regardless if you have that bound it yields a negative answer.

Thanks for helping though.

The answer is

[itex] \pi a^4/4 [/itex]
 
  • #9
Archimedes II said:
I solved it I forgot to square the sin at the end. I didn't need to use a u substitution.

Any regardless if you have that bound it yields a negative answer.

Thanks for helping though.

The answer is

[itex] \pi a^4/4 [/itex]
Yes. It works out just fine either way.

Switching the limits of integration switches the sign. That's good because it got rid of that leading negative sign.

... and you learned a bit more of LaTeX coding .
 

FAQ: Trignometric Substiution Problem

What is a Trignometric Substitution Problem?

A Trignometric Substitution Problem is a type of integration problem where trigonometric functions are used to substitute for other variables in order to simplify the integration process. It is often used when the integral involves square roots of quadratic expressions or rational functions.

Why is Trignometric Substitution used?

Trignometric Substitution is used to simplify integration problems that involve trigonometric functions. It allows for the use of basic trigonometric identities to solve the integral, making it easier and more efficient to solve.

What are the steps for solving a Trignometric Substitution Problem?

The steps for solving a Trignometric Substitution Problem are:1. Identify the type of substitution needed (based on the form of the integral).2. Substitute the given variable with a trigonometric function and rewrite the integral in terms of that function.3. Use trigonometric identities to simplify the integral.4. Solve the integral using standard integration techniques.5. Substitute back the original variable to get the final answer.

What are some common trigonometric substitutions used?

Some common trigonometric substitutions used in integration problems are:1. For integrals involving √(a^2 - x^2), use x = a sinθ.2. For integrals involving √(a^2 + x^2), use x = a tanθ.3. For integrals involving √(x^2 - a^2), use x = a secθ.4. For integrals involving √(x^2 + a^2), use x = a cotθ.

Are there any limitations or restrictions when using Trignometric Substitution?

Yes, there are some limitations and restrictions when using Trignometric Substitution. Some common restrictions include:1. The integral must involve a quadratic expression or a rational function.2. The limits of integration may need to be adjusted after the substitution.3. The substitution may introduce new trigonometric functions that need to be simplified using identities.4. The integral may not be solvable using trigonometric substitutions alone and may require additional techniques.

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