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What are the units of the argument "x" for this cos(x) function integral?

  • Thread starter mech-eng
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753
11
Problem Statement
Show that the value of ##\int_0^1\sqrt(1-cosx)dx## is less than or equal to ##\sqrt2##
Relevant Equations
##1\ge \cos x\ge-1##
Show that the value of ##\int_0^1\sqrt(1-cosx)dx## is less than or equal to ##\sqrt2##

##1\ge cos x\ge-1##

The problem is a worked one but I am just confused by a simple thing. We integrate the function f ##\int_0^1\sqrt(1-cosx)dx in the interval [0,1] but I don't understand that what stands for x-axis? Radian, degree or nothing? I know this does not affect the maximum and minumum values of cos function but how should I think in such a situation?

Source: Thomas Calculus.
 
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SammyS

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Show that the value of ##\int_0^1\sqrt(1-cosx)dx## is less than or equal to ##\sqrt2##

##1\ge cosx\ge-1##

The problem is a worked one but I am just confused by a simple thing. We integrate the function f ##\int_0^1\sqrt{(1-cosx)}dx ## in the interval [0,1] but I don't understand that what stands for x-axis? Radian, degree or nothing? I know this does not affect the maximum and minimum values of cos function but how should I think in such a situation?

Source: Thomas Calculus.
You can depend on virtually all calculus related references to trig functions as being in radians. Derivative rules etc. for trig functions would be littered with the coefficient ##\dfrac{\pi}{180^\circ}##, if the arguments are in degrees.
 
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scottdave

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The argument for cosine or other trig function should be kept in radians for integration, and I'd say most math problems. If you are doing some type of engineering problem or navigation or surveying, then degrees are the norm.
 

SammyS

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@mech-eng
Thanks for using LaTeX.

Some LaTeX pointers:

To get all of ##1-\cos x ## under the radical, enclose 1-\cos x in braces: { } . Also, for most standard functions, precede the function name with a backslash, \ . Be sure to leave a space after the function name. LaTeX then uses a different font to render the function name.

So ##\sqrt{1-\cos x}## gets rendered as: ##\sqrt{1-\cos x}##
 

robphy

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In general, in [itex]\cos x [/itex], [itex]\exp x [/itex], [itex]\log x [/itex], [itex]\arctan x [/itex],etc...
[itex]x [/itex] is dimensionless.

For example,
since [itex]\exp x =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots[/itex], what would the right-hand side mean if [itex]x [/itex] had units of length?

Another example:
If [itex]E [/itex] has units of energy, then [itex]\log E [/itex] makes no sense--- [itex]\log \left(\frac{E}{\rm Joules}\right) [/itex] is more sensible.
I would even complain about "[itex]\log E_2-\log E_1 [/itex]" and would prefer either "[itex]\log\left(\frac{E_2}{\rm Joules}\right)-\log \left(\frac{E_1}{\rm Joules}\right) [/itex]" or [itex]\log\left(\frac{E_2}{E_1}\right)[/itex].
 
753
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@mech-eng
Thanks for using LaTeX.

Some LaTeX pointers:

To get all of ##1-\cos x ## under the radical, enclose 1-\cos x in braces: { } . Also, for most standard functions, precede the function name with a backslash, \ . Be sure to leave a space after the function name. LaTeX then uses a different font to render the function name.

So ##\sqrt{1-\cos x}## gets rendered as: ##\sqrt{1-\cos x}##
This is a relevant question to the above. I tried this but I cannot get all of the function inside the integration symbol. I use braces but something seems to me wrong.

##\int{1/ \sqrt x}##

Would you please explain how to get 1/ \sqrt x inside the integral symbol?
 

SammyS

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This is a relevant question to the above. I tried this but I cannot get all of the function inside the integration symbol. I use braces but something seems to me wrong.

##\int{1/ \sqrt x}##

Would you please explain how to get 1/ \sqrt x inside the integral symbol?
I'm not sure what you mean by "inside" the integral symbol, but here's a try.

Simply include dx to get ##\int {1/ \sqrt x } dx##, which doesn't look too good - the dx being so close to the square root. Use "\," for a small space, or either "~" or "\ " for a standard size space. (Yes, that is "\" followed by a space character. ##\ \int {1/ \sqrt x } \, dx##

To make a "stacked" fraction, use "\frac{numerator}{denominator}", giving : ##\int { \frac{1}{\sqrt x } } dx##
or use "\dfrac{numerator}{denominator}", giving : ##\int { \dfrac{1}{\sqrt x } } dx## .

Does any of that answer your question?

Be sure to consult the PF LaTeX Guide, if you haven't already done so.
 

WWGD

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I suggest ,if possible, learn to deal with these matters in the abstract too and wait until you find, if possible , a real-world interpretation. I was hung up on similar issues and this liberated me, made it easier to move forward.
 

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