Integrating Using Trigonometric Substitution

[prosteve037]

Homework Statement

Use the method of trigonometric substitution to evaluate the following:


\int\frac{x^{2}}{\sqrt{4-9x^{2}}}

Homework Equations

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

sin^{2}\vartheta + cos^{2}\vartheta = 1

The Attempt at a Solution

This is what I got from trying this problem:

\int\frac{x^{2}}{\sqrt{4-9x^{2}}}

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

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

Differentiating this same function with respect to x, I got:

\frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}

Substituting these into the original function:

\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}d\vartheta}

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

\int{\frac{1}{9} \times \frac{cos\vartheta}{sin\vartheta} \times \frac{-1}{3} \times \frac{sin\vartheta}{\sqrt{cos\vartheta}}d\vartheta}

\frac{-1}{27}\int{\frac{cos\vartheta}{\sqrt{cos\vartheta}}d\vartheta}

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

 

[GreenPrint]

they hypotenuse is actually 2, remember c^2=a^2 + b^2, the other leg is actually 3x

 

[Bohrok]

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

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

I can't really see how you got these substitutions... A standard substitution for this problem would be x = \frac{2}{3}\cos\vartheta

 

[SammyS]

Homework Statement

Use the method of trigonometric substitution to evaluate the following:


\int\frac{x^{2}}{\sqrt{4-9x^{2}}}

Homework Equations

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

sin^{2}\vartheta + cos^{2}\vartheta = 1

The Attempt at a Solution

This is what I got from trying this problem:

\int\frac{x^{2}}{\sqrt{4-9x^{2}}}

\sqrt{4-9x^{2}} is a difference between two squares, so it's a leg of a right triangle. So I let 4sin\vartheta = \sqrt{4-9x^{2}} 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: \displaystyle \int\frac{x^{2}}{\sqrt{4-9x^{2}}}\,dx\,.

To elaborate on GreenPrint's post:

If \sqrt{4-9x^{2}} 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 \displaystyle \sin(\theta)=\frac{3x}{2} or \displaystyle \sin(\theta)=\frac{\sqrt{4-9x^{2}}}{2}\,.

cos(θ) is the other.

 

[Harrisonized]

Just making info easier to understand.

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

Make sure you find dx to get dθ.

 

[HallsofIvy]

Homework Statement

Use the method of trigonometric substitution to evaluate the following:


\int\frac{x^{2}}{\sqrt{4-9x^{2}}}

Homework Equations

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

sin^{2}\vartheta + cos^{2}\vartheta = 1

The Attempt at a Solution

This is what I got from trying this problem:

\int\frac{x^{2}}{\sqrt{4-9x^{2}}}

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

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

No, you must have x= (2/9)cos(\vartheta)[/tex].<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Differentiating this same function with respect to x, I got:<br /> <br /> <span style="font-size: 12px">\frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}<br /> <br /> <br /> Substituting these into the original function:<br /> <br /> <span style="font-size: 12px">\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\sqrt{\frac{4}{9}cos\vartheta}}d\vartheta}<br /> <br /> <span style="font-size: 12px">\int{\frac{\frac{4}{9}cos\vartheta}{4sin\vartheta} \times \frac{\frac{-2}{9}sin\vartheta}{\frac{2}{3}\sqrt{cos\vartheta}}d\vartheta}<br /> <br /> <span style="font-size: 12px">\int{\frac{1}{9} \times \frac{cos\vartheta}{sin\vartheta} \times \frac{-1}{3} \times \frac{sin\vartheta}{\sqrt{cos\vartheta}}d\vartheta}<br /> <br /> <span style="font-size: 12px">\frac{-1}{27}\int{\frac{cos\vartheta}{\sqrt{cos\vartheta}}d\vartheta}<br /> <br /> At this point, I got stuck and didn&#039;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.<br /> <br /> Thanks</span></span></span></span></span> </div> </div> </blockquote>

 

[prosteve037]

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?

 

[GreenPrint]

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

 

[SammyS]

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.