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

[neshepard]

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.

 

[jegues]

How about if we let,

x = 3sin(u)

and,

dx = 3cos(u)du

Where,

u = sin^{-1}(\frac{x}{3})

Does that help?

 

[losiu99]

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 x=3\sin\theta to simplify the radical.
Good luck!

 

[neshepard]

Where or how do you get sin and cos into this equation?

 

[losiu99]

Do you know integration by substitution?

 

[jegues]

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.

 

[neshepard]

∫(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.

 

[losiu99]

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
\frac{1}{2}\arcsin x+\frac{1}{2}x\sqrt{1-x^2}.
It can be done by substitution or by parts, but I'm afraid it can't be solved any simpler.

 

[jegues]

∫(9-3sin(u)^2)^1/2*3cos(u)du

This isn't correct,

(3sin(u))^{2} \neq 3sin^{2}(u)

You must also square the 3.

You should obtain,

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

You should be able to take it from there.

HINT: sin^{2}(t) + cos^{2}(t) = 1

 

[neshepard]

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

 

[jegues]

∫(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,

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

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?

9 \int \sqrt{cos^{2}(u)}cos(u)du

Good luck! Keep trying you're almost there!

 

[neshepard]

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?

 

[sponsoredwalk]

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

 

[jegues]

Yes. Show your work next time. It should look something like this,


9 \int \sqrt{cos^{2}(u)}cos(u)du

9 \int cos^{2}(u)du

Since,

cos(2u) = cos^{2}(u) - sin^{2}(u)

cos(2u) + sin^{2}(u) = cos^{2}(u)

cos(2u) + (1 - cos^{2}(u)) = cos^{2}(u)

cos(2u) + 1 = 2cos^{2}(u)

\frac{cos(2u)+1}{2} = cos^{2}(u)

So now,

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

Can you finish it off now?

 

[jegues]

this is a lot more intuitive

Great video! Thanks for sharing sponsoredwalk!

 

[Dickfore]

If you have trouble with trigonometric substitutions, perhaps you will find this roundabout way more convenient:

<br /> 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}<br />

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

<br /> x = a \, t<br />

<br /> dx = a \, dt<br />

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

<br /> 9 \, \int{\frac{a \, dt}{\sqrt{9 - (a t)^{2}}}} = 9 \, a \, \int{\frac{dt}{\sqrt{9 - a^{2} \, t^{2}}}}<br />

If we choose:

<br /> 9 = a^{2} \Rightarrow a = 3<br />

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

<br /> 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}}}}<br />

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:

<br /> \int{\frac{x^{2}}{\sqrt{9 - x^{2}}} \, dx} = \int{x \, \frac{x \, dx}{\sqrt{9 - x^{2}}}}<br />

Take

<br /> u = x \Rightarrow du = dx<br />

<br /> dv = \frac{x \, dx}{\sqrt{9 - x^{2}}}<br />

To find v, we integrate:

<br /> v = \int{\frac{x \, dx}{\sqrt{9 - x^{2}}}}<br />

This integral can be solved by a very clever substitution:

<br /> s = \sqrt{9 - x^{2}}<br />

<br /> ds = \frac{1}{2 \, \sqrt{9 - x^{2}}} \, (-2 x) \, dx = -\frac{x \, dx}{\sqrt{9 - x^{2}}}<br />

so:

<br /> v = -\int{ds} = -s = -\sqrt{9 - x^{2}}<br />

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:

<br /> \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}<br />

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:

<br /> I = 9 \, \int{\frac{dt}{\sqrt{1 - t^{2}}}} + x \, \sqrt{9 - x^{2}} - I<br />

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.

 

[jegues]

I've made another attempt using the intution given by sponsoredwalk's video

See figure.

First,

3cos(\theta) = \sqrt{9-x^{2}}

Now solving for dx,

\frac{d}{dx}\left(sin(\theta) = \frac{x}{3}\right)

cos(\theta) \frac{d\theta}{dx} = \frac{1}{3}

So,

dx = 3cos(\theta)d\theta

Remember that,

3cos(\theta) = \sqrt{9-x^{2}}

So we now have,

\int (3cos(\theta))(3cos(\theta))d\theta

After some simplification,

9 \int cos^{2}(\theta)d\theta

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

 

[sponsoredwalk]

Your idea is correct

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

- \ 9 \int sin^2(\gamma )\,d\gamma

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

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

 

[neshepard]

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.

 

[jegues]

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,

sin(sin^{-1}(\frac{x}{3}))cos(sin^{-1}(\frac{x}{3}))

to

x \frac{ \sqrt{9-x^{2}}}{3}

 

[Dickfore]

Inverse functions.

 

[jegues]

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!

 

[Dickfore]

By, definition f[f^{-1}(x)] = x.

 

[jegues]

So,

sin(sin^{-1}(\frac{x}{3})) = \frac{x}{3}

But what about,

cos(sin^{-1}(\frac{x}{3})) = ?

 

[Dickfore]

You need to express the cosine in terms of the sine.

 

[jegues]

Is this it by chance?

cos(x) = \sqrt{1-sin^{2}(x)}

EDIT:

\sqrt{(1-sin^{2}(sin^{-1}(\frac{x}{3}))}

\sqrt{1 - (\frac{x}{3})^{2}} = \sqrt{\frac{9-x^{2}}{9}}

Still not there yet, any help?

 

[Dickfore]

<br /> \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, \sqrt{9} = 3<br />

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

 

[jegues]

Well if,

sin(sin^{-1}(\frac{x}{3})) = \frac{x}{3}

and

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

then,

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

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.

 

[Dickfore]

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.

 

[jegues]

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.