Plus or minus in integrals? And figuring out new bounds

[flyingpig]

Homework Statement

Evaluate this integral using trigonometric substitution.

$$\int_{0}^{2} \frac{x^3}{\sqrt{4-x^2}}dx$$


Now I can do this the "textbook memorization" method like every calculus student does, but I want to go ahead an analyze this further. But I will show you the intermediate work and then ask my question

Let $$x = 2\cos(\theta)$$ and so, $$dx = -2\sin(\theta)d\theta$$

My indefinite integral then becomes

$$\int\frac{-16\cos^3(\theta)sin(\theta)\;d\theta}{\sqrt{4 - 4\cos^2(\theta)}}$$

This is a mess, so simplifying further, we get

$$\int\frac{-16\cos^3(\theta)sin(\theta)\;d\theta}{\sqrt{4}\sqrt{1 - \cos^2(\theta)}}$$

Now watch, closely

$$\int\frac{-16\cos^3(\theta)\;d\theta}{\sqrt{4}}$$

Now here is my first question for the square root of 4, how do we know that it is going to be +2 and not -2?

Also, if I continue solving using +2 I get

$$\int-8\cos^3(\theta)\;d\theta$$

Suppose now I want to evaluate the definite integral and I put back my limits of integration in terms of $$\theta$$ such that

$$x = 2\cos(\theta)$$

$$0 = 2\cos(\theta)$$ and $$2 = 2\cos(\theta)$$

Now back in old trig, we always know that theta could have coterminal angles, so how do we decide which is which when we compute the integral? I know it has to do something with the interval with which we are integrating, but I just couldn't come up with a good relation and answer.

So I really appreciate it if someone could find me a relation, thank you!

 

[SammyS]

In the case you have, when x = 2, θ = 0, and when x = 0, θ = π/2

As to your question about $$\sqrt{4}\ :$$

There is no reason to think it would be negative.

It comes from the denominator $$\sqrt{4-x^2}$$ which is non-negative.

 

[flyingpig]

Why couldn't θ = 3π/2 though?

 

[SammyS]

θ for which value of x?

 

[flyingpig]

x = 0 that is

 

[SammyS]

In the original integral, x takes on all values from 0 through 2 and no others.

If you have θ go from 3π/2 to 0, then x = 2cosθ will take on values from 0 through -2, back to zero and finally to 2. Will that give the same answer? I doubt it. Partly because of the following:

$$\text{When }\pi<\theta<2\pi\,,\ \text{ then, }\sin(\theta)<0\ \to\ \sqrt{1-\cos^2(\theta)}=\sqrt{\sin^2(\theta)}=-\sin(\theta)$$
.

 

[flyingpig]

Also, noticed how we used x = 2cosθ instead of x = 2sinθ?

In the form of $$\sqrt{a^2-x^2}$$

 

[SammyS]

Also, noticed how we used x = 2cosθ instead of x = 2sinθ?

In the form of $$\sqrt{a^2-x^2}$$

Yes, I noticed.

BTW: Just what is the question for this thread?

 

[flyingpig]

Figuring out limits of integration when it involves square roots. I was going to also ask what happens when we are evaluting the integral and there are square roots. How do we know to not take the negative roots?

 

[SammyS]

I thought this was answered in an earlier https://www.physicsforums.com/showthread.php?t=483500", but it looks like you didn't understand.

By definition the principal square root (or any even root for that matter) is positive for a positive radicand ! ! ! !

$$\sqrt{x}\ge0$$

 

[flyingpig]

Oh okay, tricky...

 

[SammyS]

Why couldn't θ = 3π/2 though?

I've been thinking about this integration problem, especially the above question ...

There's a more fundamental reason that you can't use the substitution x = 2cosθ and have both:

θ = 3π/2 correspond to x = 0​

 and

θ = 0 correspond to x = 2
​

The reason is that θ must be a function of x (emphasis on the word function). If x = 2 cosθ, and θ spans the interval from 0 to 3π/2, then the function defined by x = 2cosθ is not a one-to-one function of θ, so that this function does not have an inverse.

Of course, a way around this is to restrict the values of θ, as is done in defining the inverse trig functions. The standard way to do this is to restrict the domain of our function, x = 2cosθ, to [0, π]. Inverting this gives: θ = arccos(x/2). This will eliminate θ = 3π/2 .

Yes, you can change things around so that θ = 3π/2 can be one of the limits of integration, but then the other integration limit will be π or 2π .

 

[flyingpig]

How did you "restrict" the domain?

you mean x = 2cos(θ(x))

θ as a function of x? Implicit function!

 

[SammyS]

When you did the following step:

Let $$x = 2\cos(\theta) \,,$$

you where defining a trig substitution in which θ is an implicit function of x.

The above defines x explicitly as a function of θ. To use this as a substitution for evaluating an integral, θ must be a function of x. Yes, you can consider the above to define θ as an implicit function of x, but θ can only be a function of x, consistent with the above explicit definition, if the domain of that explicit function is restricted to an interval which makes that explicit function one-to-one.

 

[flyingpig]

Sammy, I tried this

http://www.wolframalpha.com/input/?i=polarplot{2cos%28t%29%2C{t%2C0%2C2pi}}

I tried to look at it in polar coordinates, but I still couldn't make it out...

I am probably not doing much helping lol

 

[SammyS]

No need for plotting in polar coordinates.

Plot http://www.wolframalpha.com/input/?i=plot+x=2cos(theta),+theta=-2pi..4pi". That's straight forward. This shows x as a function of θ.

Now, http://www.wolframalpha.com/input/?i=parametric+plot+(2*cos(theta),+theta),+theta=-2pi..4pi". I did this as a parametric plot.

Why is θ not a real function of x as shown by this graph?​

 

[flyingpig]

Wait which one is theta in the parametric curve?

 

[Mark44]

x = 2cos(θ)
y = θ
-2π <= θ <= 4π

 

[flyingpig]

No need for plotting in polar coordinates.

Plot http://www.wolframalpha.com/input/?i=plot+x=2cos(theta),+theta=-2pi..4pi". That's straight forward. This shows x as a function of θ.

Now, http://www.wolframalpha.com/input/?i=parametric+plot+(2*cos(theta),+theta),+theta=-2pi..4pi". I did this as a parametric plot.

Why is θ not a real function of x as shown by this graph?​

I still don't understand

 

[SammyS]

Please read and re-read this whole post , and try to understand it before responding. Be more complete and specific in your responses.

Maybe I didn't explain the following completely enough...

No need for plotting in polar coordinates.

Plot http://www.wolframalpha.com/input/?i=plot+x=2cos(theta),+theta=-2pi..4pi". That's straight forward. This shows x as a function of θ.

In the above graph: The vertical axis is the x-axis, the horizontal axis is the θ-axis.

But, in your original integral, x is the variable of integration, so to do a "legal" substitution, we need θ as a function of x, not x as a function of θ. To do this, we interchange the roles of x and θ. [ This process is that of finding the inverse of the function f, defined by: f (θ) = 2cos(θ). It is usually covered in all of the following courses: College Algebra, Trigonometry, Pre-Calculus, and Calculus I. It is usually done with the variables x & y and some function f, where y = f (x). (I thought of switching variables to x & y, but at this stage that would add extra confusion, so I'm sticking with θ and x.) ]

Graphically, this means that we graph the same relationship (between x and θ) with x as the dependent variable (horizontal axis) and θ as the dependent variable (vertical axis). This can also be done by reflecting the graph through the line, x = θ. This graph is shown in the following link -- in the following quote. (I did the graph as a parametric graph, because that's the easiest method I knew for obtaining the desired graph using WolframAlpha.)

Algebraically, this can be accomplished by solving the equation, x = 2cosθ, for θ. This gives:

θ = ± arccos(x/2) ± 2πn, for n an integer. ​

Now, http://www.wolframalpha.com/input/?i=parametric+plot+(2*cos(theta),+theta),+theta=-2pi..4pi". I did this as a parametric plot.

Why is θ not a real function of x as shown by this graph?​

Now, to repeat the last question I asked: "Why is θ not a real function of x as shown by the graph in the second quote above?"

 

[flyingpig]

I don't know why PF was down for me last night and it completely erased my message. Anyways.

θ is not a real function of x because it fails the vertical line test right? But if so, how does that help us eliminate the other "non-relevant" angles?

 

[SammyS]

You need to pick (restrict) a range for θ, so that the graph passes the vertical line test. This is equivalent to restricting the domain of the function x = 2cos(θ) so that this function is one to one. This is what is done to the trig functions so that the inverse trig functions can be defined.

 

[SammyS]

...
θ is not a real function of x because it fails the vertical line test right? But if so, how does that help us eliminate the other "non-relevant" angles?

One way to do this is to sketch a right triangle with hypotenuse of length 2 and a leg of length x. The other leg is √(2²-x²). Then choose θ so that x/2 = cos(θ).

Another way is to let θ = arccos(x/2), so 0 ≤ θ ≤ π, which is the range for the arccos function.

 

[flyingpig]

One way to do this is to sketch a right triangle with hypotenuse of length 2 and a leg of length x. The other leg is √(2²-x²). Then choose θ so that x/2 = cos(θ).

Another way is to let θ = arccos(x/2), so 0 ≤ θ ≤ π, which is the range for the arccos function.

That makes so much more sense lol

 

[Mark44]

One way to do this is to sketch a right triangle with hypotenuse of length 2 and a leg of length x. The other leg is √(2²-x²). Then choose θ so that x/2 = cos(θ).

Another way is to let θ = arccos(x/2), so 0 ≤ θ ≤ π, which is the range for the arccos function.

That makes so much more sense lol

Also see post #3 in this thread. It's pretty much the same as what SammyS said.