Integrating cos^3(x) and √cos^3(x): Step-by-Step Guide | Helpful Tips

[Player_13]

How do you integrate cos^3/2(x) or square root of cos^3(x), I know for the integration of cos^3(x) you break it apart to cos^2(x)cos(x). Then chnage that to (1 - sin^2(x))(cos(x) and do a u subsitution, with u set equal to sin. However, because of that square root it completely changes the problem, or at least in my eyes it does. Step by step help would be appreciated.

Thanks.

 

[HallsofIvy]

Do you have any reason to believe that this has a simple integral?

 

[Player_13]

It's from a practice AP booklet, so I am assuming you can do this by hand.

 

[cepheid]

If the problem is:

\int{\sqrt{\cos^3 x}dx}

Then I can't do it. Mathematica gave an answer involving "EllipticF" of something. I don't know what elliptic functions are.

=/

 

[Player_13]

Aren't eliptic funtions x^2/a + y^2/b = c, or something near that?

 

[cepheid]

No...that's just the equation of an ellipse in cartesian coordinates. That function, if you solve for y, is quite ordinary: y = square root of a bunch of stuff. Elliptic functions refer to something else entirely (not ordinary functions). I found it here: but don't really understand what it means: http://mathworld.wolfram.com/EllipticFunction.html

I doubt you are expected to either, so maybe there is some other way of solving this integral. I don't understand why they would give you one that isn't solvable at a high school level.

 

[X-43D]

Could it be that the question is \int \cos^3 x dx/2x

\cos^3 x = \cos x \cos^2 x = \cos x (1- \sin^2 x) = \cos x - \cos x \sin^2 x and now you have to get the derivative of 2x which is 2. So the answer should be: \frac{\cos x - \cos x \sin^2 x}{2}

Anyway, check out Integration of trigonometric functions

Elliptic functions are part of complex analysis and are more sophisticated than these ones. This is beyond the first two or three semesters of college calculus.

 

[HallsofIvy]

How does that help one integrate cos^3/2x?

 

[X-43D]

How does that help one integrate cos^3/2x?

I don't think you can integrate this...

 

[Data]

You can try the simple substitution u = \sqrt{\cos^3{x}}. I don't know how far it will get you (I got it to a reasonably friendly form, but it still required the EllipticF function for Maple to evaluate).

One of the main problems with finding this antiderivative is, of course, that the integrand is real sometimes and imaginary other times.

 

[Player_13]

I think it should be possible, it's a practice problem from the ap test for calculus ab. Maybe I don't need to show work for it? Anybody know how the calculus ap exam works?

That was only part of the problem, and I figured I could just do the rest after this step was accomplished. The whole problem is:

Consider the graphs of y=2/3x and y=cos^3/2(x). The two curves intersect at point P. Region R is bounded by the two curves and the y-axis. Region S is bounded by the two curves and the x-axis.

a) Find the slope of the tangent line to y=cos^3/2(x), or y=sqrt(cos^3(x)) at point P. (I already know how to do this.)

b) Find the area of R.
c) Find the area of S.

b) R = int(cos^3/2(x) - 2/3x)dx
c) S = int(2/3x)dx + int(cos^3/2(x))dx

Assuming I set those up right, you would need to do an itegration of cos^3/2(x) either way, which is where I get stuck. Maybe I'm just headed in the wrong direction? I can't think of any other way to find the area, besides the use of integrals.

 

[tony873004]

In the order of operations, doesn't cos^3/2(x) mean

\frac{cos^3}{2(x)}

instead of

cos^{3/2}x

since the parenthesis are around the x and not around the 3/2 ?

 

[cepheid]

You can't apply order of operations here because cos^3/2(x) is just the way he chose to try and write cos^3/2x on the forum. In fact, the reason why he put parentheses around the x was to try and isolate it, to show that it was the argument of cosine, and that it wasn't related to those number 3 and 2 in any way, for they make up the exponent.

By the way, cos³x is just a shorthand for (cos x)³. So

cos³ by itself, without an argument, is meaningless. That should have tipped you off that you were barking up the wrong tree.

 

[Player_13]

LOL, yeah I'm sorry, I don't know how to do your fancy javascript/math notation things. I found this forum by accident, searching google for an answer to my problem. I was just hoping that a large group of math/physics gurus would be able to answer this "simple" problem quick and easily. However, I guess I assumed incorrectly. Once again to clarify y=(cos x)^3/2.

 

[cepheid]

For what it's worth, I get:

Area of R:

\int_0^P {\left(\sqrt{\cos^3(x)} - \frac{2}{3}x\right)}dx

Area of S:

\int_0^{\pi/2} {[\sqrt{\cos^3(x)}]}dx \ \ - \ \ \text{Area of R}

 

[Player_13]

Yeah, I got the same. Just how are you suppose to integrate (cos x)^3/2...

 

[X-43D]

I think that in order to solve u = \sqrt{\cos^3{x}} you need to use this equation u = \int_0^x \frac{ds}{\sqrt{(1-s^2)(1-k^2 s^2)}}

 

[tony873004]

...That should have tipped you off that you were barking up the wrong tree.

It did, as soon as I hit Submit

 

[Player_13]

*shrug* X-43D, I've never seen that equation before, or I just don't remember it... I'm so lost on this problem, I just hope my teacher doesn't take a grade on this problem. I've asked two others in my class, and they can't seem to figure it out either.

 

[X-43D]

*shrug* X-43D, I've never seen that equation before, or I just don't remember it... I'm so lost on this problem, I just hope my teacher doesn't take a grade on this problem. I've asked two others in my class, and they can't seem to figure it out either.

I don't get it either but i think it's an elliptic integral.

 

[Player_13]

So, from what I've gathered, the answer is derived by a way that I haven't learned yet... Again, does anyone know how the AP grading system works? Showing work is or isn't required? I ask this, because maybe I don't even have to show work, but the answer is most likely I have to.

 

[Data]

No idea. I highly doubt they'll want you to write down the answer to a question as

\frac{2}{3}\left( \mbox{EllipticF}\left(\frac{x}{2}, \ 2\right) + \sqrt{\cos{x}}\sin{x} + C\right)

without qualification, though

 

[X-43D]

I think i found a way to solve this (at least in principle). It's not an elliptic integral. The basic idea here is the fundamental theorem of calculus. The real meaning of the fundamental theorem of calculus is that differentiation and integration are inverse processes. Even squaring and taking the root are inverse to each other if an additional condition is tacked on: only positive numbers are allowed.

We need to use F'(x) = f(x) or \frac{F (x + h) - F (x)}{h}.

See also http://ugrad.math.ubc.ca/coursedoc/math101/notes/integration/ftc.htm

 

[Data]

How will that help find the antiderivative of \cos^{3/2}{x} in terms of simple functions?

 

[Player_13]

*Pulls out hair* So frustrated with this problem. I've gone ahead and just set up the problem and used my calculator to get the answer. You guys can forget about this, or not, I'd still like to know if its possible to figure it out with simple rules. If my teacher explains it to the class tommorow, I'll be sure to tell you guys.

 

[Data]

If there is a way to do it simply, then tell us too!

 

[X-43D]

It gives me \frac{\sqrt \cos{x}}{0.5}.

 

[Player_13]

X-43D, I don't think that's correct. I plugged a value for each in my calculator, and the answers were different.

 

[whozum]

Have you guys kept in mind the bounds 0,P for your integrals? Perhaps maple and mathematica can't find indefinite integrals in simple terms due to the negative values of cos^3/2(x) being nonexistant. If you find P and use bounds, perhaps you will get a better answer.

 

[Data]

yes, but how can you evaluate that by hand, without finding the antiderivative?