Solving the Square Root of a Difference: Integrating √(9-x^2)

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In summary: I on one side of the equation and simplifying);3. Write a general solution to this equation: I = ... (you have two of them and can pick one);4. Verify that the chosen solution is indeed correct by differentiating it and checking that it gives back the original function under the integral.I hope that I helped a bit by giving yet another method of solving this integral. :)In summary, there are multiple methods for solving the integral ∫√(9-x^2). One method is substitution, where x is replaced with 3sin(u) and dx is replaced with 3cos(u)du. Another method is using trigonometric identities to simplify the integral. Another approach is
  • #1
neshepard
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Homework Statement


∫√(9-x^2)


Homework Equations





The Attempt at a Solution


(9-x^2)^1/2
1/2(9x-(x^3/3))^3/2
(9x-(x^3/3))/3

But this is wrong and I can't see where or how.
 
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  • #2
How about if we let,

[tex] x = 3sin(u) [/tex]

and,

[tex] dx = 3cos(u)du[/tex]

Where,

[tex] u = sin^{-1}(\frac{x}{3}) [/tex]

Does that help?
 
  • #3
You cannot antidifferentiate stuff like this. That would be ok if the chain rule was something like " [f(g(x))]'=f'(g'(x))", but it is not. To conquer this integral, I suggest to substitute [tex]x=3\sin\theta[/tex] to simplify the radical.
Good luck!
 
  • #4
Where or how do you get sin and cos into this equation?
 
  • #5
Do you know integration by substitution?
 
  • #6
Where or how do you get sin and cos into this equation?

We are the ones that defined the variable u, so we can define them in any manner we wish. (In this case we chose to use sin's)

If you do enough of these problems you will begin to recognize what you should substitute and where.

Give the substitution I posted a shot(post your work), things should start to unfold.
 
  • #7
∫(9-3sin(u)^2)^1/2*3cos(u)du So I get I'm seeing substitution, though it's very different from anything we've done in our class. We never ever added in sin's and cos's, we would find the u and du and pull those out.

(1/2)[9-3sin(u)^2)^3/2]/(3/2)
3[(3x+cos(u)^2)^3/2]/3
[3x+cos(u)^2]^3/2
[3x+cos(sin^-1(x/3))^2]^3/2

And I'm quite sure this is wrong.
 
  • #8
Remember, it's (3sin(u))^2, not 3(sin(u))^2.
Unfortunetely, there is no simple way here to just "see" the answer, eg. find u and du. If that helps, the answer is
[tex]\frac{1}{2}\arcsin x+\frac{1}{2}x\sqrt{1-x^2}[/tex].
It can be done by substitution or by parts, but I'm afraid it can't be solved any simpler.
 
  • #9
∫(9-3sin(u)^2)^1/2*3cos(u)du

This isn't correct,

[tex] (3sin(u))^{2} \neq 3sin^{2}(u) [/tex]

You must also square the 3.

You should obtain,

[tex]3 \int \sqrt{9 - 9 sin^{2}(u)}cos(u)du [/tex]

You should be able to take it from there.

HINT: [tex]sin^{2}(t) + cos^{2}(t) = 1[/tex]
 
  • #10
Okay, I feel good about this one.

∫(9-(3sin(u))^2)^1/2*3cos(u)du
∫(9-9sin^2(u))^2)^1/2*3cos(u)du
3∫(1-sin^2(u))^1/2*cos(u)*du
3∫(cos^2(u))^1/2*cos(u)*du
2(sin^2(sin^-1(x/3)))^3/2
 
  • #11
∫(9-9sin^2(u))^2)^1/2*3cos(u)du

The part in bold should not be there.

3∫(1-sin^2(u))^1/2*cos(u)*du

Here you factored 9(hence 1-1sin^2(u)) and you should have taken the square root of 9 giving you 3, this multiplying the other 3 would result in a constant of 9 outside your integral.

So,

[tex]9\int \sqrt{1-sin^{2}(u)} cos(u)du[/tex]

Assuming you change the 3 outside the integral to a 9, this is also correct,

9∫(cos^2(u))^1/2*cos(u)*du

However you're last line after this is incorrect.

2(sin^2(sin^-1(x/3)))^3/2

Can you try to simplify the line below to make integrating it easier?

[tex] 9 \int \sqrt{cos^{2}(u)}cos(u)du[/tex]

Good luck! Keep trying you're almost there!
 
  • #12
The extra ^2 you hi-lited is not on my paper, just a typo.
As for the 3 and the 9 I thought the 3 is treated as a common factor, not a multiplier?
To simplify, use a lowering power rule? 1/2+cos(2u)/2?
 
  • #13
neshepard said:
Where or how do you get sin and cos into this equation?

http://www.5min.com/Video/Converting-Radicals-into-Trigonometric-Expressions-169056202

http://www.5min.com/Video/Converting-Radicals-into-Trigonometric-Expressions-169056202

http://www.5min.com/Video/Trigonometric-Substitutions-on-Rational-Powers-169056330

http://www.5min.com/Video/An-Overview-of-Trigonometric-Substitution-Strategy-169056433

http://www.5min.com/Video/Trigonometric-Substitution-Involving-a-Definite-Integral-Part-One-169056540 & part 2!

This is actually a very easy way to visualise these kinds of problems and,
although not foolproof, can be used to
evaluate really crazy looking integrals by smart substitutions.

To me, this is a lot more intuitive than randomly choosing u = sinx, or whatever.
Only if the method I've given you fails should you really use the other methods,
this one can be manipulated to extremes just by drawing right triangles &
you don't need to memorize anything!
You can use completing the square with this method to make it work as well.
Come back if you like this idea & have any questions on it, the example
you gave in your original post can be done in 20 seconds using this method :biggrin:
 
  • #14
Yes. Show your work next time. It should look something like this,


[tex] 9 \int \sqrt{cos^{2}(u)}cos(u)du[/tex]

[tex] 9 \int cos^{2}(u)du[/tex]

Since,

[tex] cos(2u) = cos^{2}(u) - sin^{2}(u) [/tex]

[tex] cos(2u) + sin^{2}(u) = cos^{2}(u) [/tex]

[tex] cos(2u) + (1 - cos^{2}(u)) = cos^{2}(u) [/tex]

[tex] cos(2u) + 1 = 2cos^{2}(u)[/tex]

[tex] \frac{cos(2u)+1}{2} = cos^{2}(u) [/tex]

So now,

[tex]9 \int \left(\frac{cos(2u) + 1}{2}\right)du[/tex]

Can you finish it off now?
 
  • #15
this is a lot more intuitive

Great video! Thanks for sharing sponsoredwalk!
 
  • #16
If you have trouble with trigonometric substitutions, perhaps you will find this roundabout way more convenient:

[tex]
I = \int{\sqrt{9 - x^{2}} \, dx} = \int{\frac{9 - x^{2}}{\sqrt{9 - x^{2}}} \, dx} = 9 \, \int{\frac{dx}{\sqrt{9 - x^{2}}}} - \int{\frac{x^{2}}{\sqrt{9 - x^{2}}} \, dx}
[/tex]

In the first of these integrals, you need to perform a simple substitution:

[tex]
x = a \, t
[/tex]

[tex]
dx = a \, dt
[/tex]

where we will chose the constant a suitably so that the resulting integral simplifies:

[tex]
9 \, \int{\frac{a \, dt}{\sqrt{9 - (a t)^{2}}}} = 9 \, a \, \int{\frac{dt}{\sqrt{9 - a^{2} \, t^{2}}}}
[/tex]

If we choose:

[tex]
9 = a^{2} \Rightarrow a = 3
[/tex]

we can take out the 9 as a common multiple in front of both terms from the expression under the square root. We have:

[tex]
9 \cdot 3 \, \int{\frac{dt}{\sqrt{9(1 - t^{2})}}} = \frac{9 \cdot 3}{3} \, \int{\frac{dt}{\sqrt{1 - t^{2}}}} = 9 \, \int{\frac{dt}{\sqrt{1 - t^{2}}}}
[/tex]

I claim that the obtained integral is a standard one and can be read from a table.

As for the second integral, one needs to perform integration by parts:

[tex]
\int{\frac{x^{2}}{\sqrt{9 - x^{2}}} \, dx} = \int{x \, \frac{x \, dx}{\sqrt{9 - x^{2}}}}
[/tex]

Take

[tex]
u = x \Rightarrow du = dx
[/tex]

[tex]
dv = \frac{x \, dx}{\sqrt{9 - x^{2}}}
[/tex]

To find v, we integrate:

[tex]
v = \int{\frac{x \, dx}{\sqrt{9 - x^{2}}}}
[/tex]

This integral can be solved by a very clever substitution:

[tex]
s = \sqrt{9 - x^{2}}
[/tex]

[tex]
ds = \frac{1}{2 \, \sqrt{9 - x^{2}}} \, (-2 x) \, dx = -\frac{x \, dx}{\sqrt{9 - x^{2}}}
[/tex]

so:

[tex]
v = -\int{ds} = -s = -\sqrt{9 - x^{2}}
[/tex]

where, in the last step, I substituted back the old variable x instead of s (a necessary step in the method of substitution). So, integration by parts gives:

[tex]
\int{\frac{x^{2}}{\sqrt{9 - x^{2}}} \, dx} = u \, v - \int{v \, du} = -x \, \sqrt{9 - x^{2}} + \int{\sqrt{9 - x^{2}} \, dx}
[/tex]

Notice that the remaining integral is equal to the integral we had started with and this is the beauty of this approach. Now, we can combine everything to have:

[tex]
I = 9 \, \int{\frac{dt}{\sqrt{1 - t^{2}}}} + x \, \sqrt{9 - x^{2}} - I
[/tex]

Notice the different sign in front of I on both sides of the equality. That is why they don't cancel.

To finish the solution, you need to:
1. Express the remaining integral from a table of standard integrals. Do not forget to go back from t to x by the substitution we used in the beginning;

2. Solve this equation for I (by moving all I on one side of the equation and dividing by the coefficient in front of it);

3. Add an arbitrary integrating constant in the end.

IF you can't follow the above steps, either you have not learned about the methods of integration using substitutions and integration by parts or, in case you didn't follow the intermediate steps, your knowledge in algebra and differentiation is so poor that you actually need to learn those things before you can go on to integration techniques.
 
  • #17
I've made another attempt using the intution given by sponsoredwalk's video

See figure.

First,

[tex] 3cos(\theta) = \sqrt{9-x^{2}}[/tex]

Now solving for dx,

[tex] \frac{d}{dx}\left(sin(\theta) = \frac{x}{3}\right)[/tex]

[tex] cos(\theta) \frac{d\theta}{dx} = \frac{1}{3} [/tex]

So,

[tex] dx = 3cos(\theta)d\theta [/tex]

Remember that,

[tex] 3cos(\theta) = \sqrt{9-x^{2}}[/tex]

So we now have,

[tex] \int (3cos(\theta))(3cos(\theta))d\theta[/tex]

After some simplification,

[tex] 9 \int cos^{2}(\theta)d\theta[/tex]

Same thing! Just another (more intuitive) way of getting there!
 

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  • #18
Your idea is correct :wink:

[PLAIN]http://img405.imageshack.us/img405/7181/integral.jpg [Broken]

[tex] - \ 9 \int sin^2(\gamma )\,d\gamma [/tex]

Then, use the identity
cos(2γ) = cos²(γ) - sin²(γ)

______= (1 - sin²(γ)) - sin²(γ)

______= 1 - sin²(γ) - sin²(γ)

______= 1 - 2sin²(γ)

cos(2γ) - 1 = - 2sin²(γ)

2sin²(γ) = 1 - cos(2γ)

sin²(γ) = ½(1 - cos(2γ))

Distribute, and integrate :wink:

No matter what way you do it, i.e. whether or not you end up with a cos²(γ), or a - sin²(γ),
it will all work out in the end by the + C, or the limits of integration.

When you see those scary fraction functions this is a good idea to try & be smart with
how you evaluate it. Also, don't forget about the completing the square method,
you can do this underneath a square root if you are smart about it :wink:
 
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  • #19
So I too watched the videos and found them to be extremely helpful. I appreciate all the help from this post. The only thing I can see as to how I was not getting the answer was both not understanding from the start where sin and cos came from, but the video helped. After multiple tries and a large pile of eraser dust on the table I solved it.

∫√(9-x^2) dx and instead of the u's I choose the θ, so x=3sinθ, dx=3cosθ*dθ and θ=arcsin(x/3)

∫[√(9-9sin^2θ)*3cosθ]*dθ
9∫[√(1-sin^2θ)*cosθ]*dθ
9∫[√(cos^2θ)*cosθ]*dθ
9∫[|cosθ|cosθ]*dθ <but since positive angle use positive cosθ>
9∫cos^2θ*dθ
9∫(1+cos2θ)/2*dθ
9/2∫(1+cos2θ)*dθ
9/2[θ+(sin2θ)/2]
9/2[θ+(2sinθcosθ)/2]
9/2[θ+sinθcosθ]

9/2[arcsin(x/3)+sin(arcsin(x/3)cos(arcsin(x/3)]
9/2[arcsin(x/3)+(x√(9-x^2))/3]+C

Sorry to have drawn this post out, but this is a problem on our final exam review sheet plus I am one of those people that needs to know to the base level where or how a thing works. Knowing where the sin and cos came from in the first posts as told to me in the videos was most helpful in solving because I could then cipher exactly what the next step was.

Thanks all.
 
  • #20
I'm a little confused about the last two lines,

9/2[arcsin(x/3)+sin(arcsin(x/3)cos(arcsin(x/3)]
9/2[arcsin(x/3)+(x√(9-x^2))/3]+C

How did you reduce,

[tex] sin(sin^{-1}(\frac{x}{3}))cos(sin^{-1}(\frac{x}{3}))[/tex]

to

[tex] x \frac{ \sqrt{9-x^{2}}}{3}[/tex]
 
  • #21
Inverse functions.
 
  • #22
Dickfore said:
Inverse functions.

Could you please demonstrate Dickfore?(Or anyone else for that matter)

I'm trying to figure out how to do so but I can't get it.

Thanks again!
 
  • #23
By, definition [itex]f[f^{-1}(x)] = x[/itex].
 
  • #24
So,

[tex] sin(sin^{-1}(\frac{x}{3})) = \frac{x}{3} [/tex]

But what about,

[tex]cos(sin^{-1}(\frac{x}{3})) = ? [/tex]
 
  • #25
You need to express the cosine in terms of the sine.
 
  • #26
Is this it by chance?

[tex] cos(x) = \sqrt{1-sin^{2}(x)}[/tex]

EDIT:

[tex]\sqrt{(1-sin^{2}(sin^{-1}(\frac{x}{3}))} [/tex]

[tex] \sqrt{1 - (\frac{x}{3})^{2}} = \sqrt{\frac{9-x^{2}}{9}}[/tex]

Still not there yet, any help?
 
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  • #27
[tex]
\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, \sqrt{9} = 3
[/tex]

The questions you are asking have nothing to do with Calculus, though. Are you sure you had a pre Calculus course?
 
  • #28
Well if,

[tex] sin(sin^{-1}(\frac{x}{3})) = \frac{x}{3} [/tex]

and

[tex]cos(sin^{-1}(\frac{x}{3})) = \frac{\sqrt{9-x^{2}}}{3}[/tex]

then,

[tex]\left(\frac{x}{3}\right)\left(\frac{\sqrt{9-x^{2}}}{3}\right) \neq x \frac{ \sqrt{9-x^{2}}}{3}[/tex]

This is why I'm confused.

The questions you are asking have nothing to do with Calculus, though. Are you sure you had a pre Calculus course?

If you're just going to be rude, please don't bother replying. I'm trying to learn, regardless of what courses I've taken in the past.
 
  • #29
Yes, he made an error with a factor of 1/3. Good thing you caught it. I am not trying to be rude, but you must know that we have written way more than we are allowed. In fact, you have the complete solution here, with some minor typos.
 
  • #30
Dickfore said:
Yes, he made an error with a factor of 1/3. Good thing you caught it.

Thanks for all your help Dickfore I really do appreciate. (No sarcasm intended)

Although, next time please spare me of the rude comments.
 
  • #31
[STRIKE]What makes you think next time you will have any comments?[/STRIKE]

EDIT:

I just noticed you were not the person looking for the solution, but actually checked through their solution. My bad.
 
  • #32
Jeges don't mind him just examine what he's said, as he clearly is lacking social skills! I have another way you can visualize how to find cos(sin^-1(x/3)) and it involves constructing your triangles similar to what you've already done. You know that for right triangles sin(x)=opp/hyp and cos(x)=adj/hyp. First construct your triangle with theta as your angle. You know that sin(theta)=x/3 This means that the leg of the triangle opposite theta is of length x and the hypotenuse is of length 3. You can use the Pythagorean theorem to find out what the adjacent length of triangle is and you know it is sqrt(9-x^2). Now you have constructed your triangle and you know that cos(theta)=adj/hyp=[sqrt(9-x^2)]/3 but since theta=sin^-1(x/3) then cos(theta)=cos(sin^-1(x/3))=[sqrt(9-x^2)]/3

And a little algebraic mistake was made above because you are right the expression above should read that sin (sin^-1(x/3))(cos(sin^-1(x/3))=(x/3)[sqrt(9-x^2)]/3=x[sqrt(9-x^2)]/9
 
  • #33
It seems like many of these explanations are somewhat convoluted. Physically speaking the integrand is the equation of a semicircle, therefore using circular functions like sine and cosine is warranted. What about this? x^2 + y^2=r^2, here r=3 recall the usual parameterization of the equation in terms of sines and cosines=> x=rcos(theta) y=rsin(theta) For this case x=3cos(theta) Then use the trigonometric identity that sin^2(theta)+cos^2(theta)=1 Transform integrand from dx to dtheta and proceed. I think the solution is easier to find that way. Then in the end you can use trig inverse functions to get explicit expression for theta in terms of x.
 
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  • #34
Wow did my review problem flow into a bad area. Sorry. As to my having pre-calc, yes. and I got a B in part 1 and 2. (As an aside, I had to teach myself part 1 during summer school because the teachers favorite expression is "I do not have to legally tell you this" as if we were in law school instead of community college. But I digress.

To tell you where my numbers came from, and please realize I could very well still be wrong, or stumbled onto the right answer blindly, or mistyped (not my best area).

Jegues, and all, you are correct in the multiplication error. I missed that on my paper and then typed it in wrong.

As to the sin(arcsin) and cos(arcsin), I know sin(arcsin(x/3))=x/3 and I cheated and looked up cos(arcsin(x))=√(1-x^2) from one of my formula memorization sheets from class. So cos(arcsin(x/3)=√(9-x^2)/3 but this is one of those I don't know why. This is why I need to get to the nitty gritty of a problem, just being told something is something doesn't help.

This whole problem is a definite integral and in working out the answer I got what my answer key told me. So somewhere in the definite integral I screwed up and lucked into the answer at the same time. lol
 
  • #35
neshepard said:
community college
Ah, that explains everything.
 
<h2>1. What is the purpose of solving the square root of a difference?</h2><p>The purpose of solving the square root of a difference is to find the value of the variable (x) that satisfies the equation √(9-x^2). This type of equation is commonly used in geometry and physics to find the length of a side or the magnitude of a vector.</p><h2>2. How do you solve the square root of a difference?</h2><p>To solve the square root of a difference, you can use the Pythagorean theorem or the trigonometric identities. You can also use a calculator or a computer program to find the numerical value of the square root.</p><h2>3. What is the domain of the square root of a difference?</h2><p>The domain of the square root of a difference is all real numbers that make the expression under the radical sign (9-x^2) equal to or greater than zero. This means that the domain is all values of x that make the equation solvable.</p><h2>4. Are there any special cases when solving the square root of a difference?</h2><p>Yes, there are two special cases when solving the square root of a difference: when the expression under the radical sign is equal to zero, and when it is less than zero. In the first case, the solution is simply x=3 since the square root of zero is zero. In the second case, there is no real solution, as the square root of a negative number is not a real number.</p><h2>5. How can solving the square root of a difference be applied in real life?</h2><p>Solving the square root of a difference can be applied in various real-life situations, such as calculating the length of a diagonal in a rectangle, finding the magnitude of a force in physics, or determining the distance between two points on a coordinate plane. It can also be used in engineering and construction to calculate the length of a diagonal support beam or the height of a building.</p>

1. What is the purpose of solving the square root of a difference?

The purpose of solving the square root of a difference is to find the value of the variable (x) that satisfies the equation √(9-x^2). This type of equation is commonly used in geometry and physics to find the length of a side or the magnitude of a vector.

2. How do you solve the square root of a difference?

To solve the square root of a difference, you can use the Pythagorean theorem or the trigonometric identities. You can also use a calculator or a computer program to find the numerical value of the square root.

3. What is the domain of the square root of a difference?

The domain of the square root of a difference is all real numbers that make the expression under the radical sign (9-x^2) equal to or greater than zero. This means that the domain is all values of x that make the equation solvable.

4. Are there any special cases when solving the square root of a difference?

Yes, there are two special cases when solving the square root of a difference: when the expression under the radical sign is equal to zero, and when it is less than zero. In the first case, the solution is simply x=3 since the square root of zero is zero. In the second case, there is no real solution, as the square root of a negative number is not a real number.

5. How can solving the square root of a difference be applied in real life?

Solving the square root of a difference can be applied in various real-life situations, such as calculating the length of a diagonal in a rectangle, finding the magnitude of a force in physics, or determining the distance between two points on a coordinate plane. It can also be used in engineering and construction to calculate the length of a diagonal support beam or the height of a building.

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