Thread: Clarification on an identity of function

1. Clarification on an identity of function

Hi,
On my Notes of problem regarding surface area of revolution
There is a Step
Cosh2(x/c) integrating this

In next step it written as (c/2)sinh2(x/c)

The identity used is integ{cos2x} = sin2x/2

But I can't apply this one in solving that step.
I want to know how the "c" in (c/2) came on integrating

Please have a look at this !!

2. Re: Clarification on an identity of function

Hi,
On my Notes of problem regarding surface area of revolution
There is a Step
Cosh2(x/c) integrating this
In next step it written as (c/2)sinh2(x/c)
The identity used is integ{cos2x} = sin2x/2
But I can't apply this one in solving that step.
I want to know how the "c" in (c/2) came on integrating.
I see that this is your first post, welcome.
Now it is almost imperative that you post the entire original question.
What you have have posted is some incomplete step of a solution.
We have no way of knowing if that step is correct if we don't know the original question.

3. Re: Clarification on an identity of function

Thank You.

The question is about finding surface area by generated by revolving about the x axis

Q: Find area of the surface generated by revolving the arc of the catenary " y=ccosh(x/c) from x=0 to x=c about the x axis

My approach on Note : the surface area equation using integration

I think I can't write more of the answer steps . its very complex to text in keyboard

Thank you.

4. Re: Clarification on an identity of function

Sorry that the title of my thread is misleading one . no option to edit now. This problem mainly dealing with "integration step doubt"

5. Re: Clarification on an identity of function

Q: Find area of the surface generated by revolving the arc of the catenary " y=ccosh(x/c) from x=0 to x=c about the x axis
surface area of revolution ...

$\displaystyle S = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2} \, dx$

$f(x) = c \cdot \cosh\left(\dfrac{x}{c}\right)$

$f'(x) = c \cdot \sinh\left(\dfrac{x}{c}\right) \cdot \dfrac{1}{c}$

$[f'(x)]^2 = \sinh^2\left(\dfrac{x}{c}\right)$

$\displaystyle S = 2\pi \int_0^c c \cdot \cosh\left(\frac{x}{c}\right) \sqrt{1+ \sinh^2\left(\dfrac{x}{c}\right)} \, dx$

note $1+ \sinh^2\left(\dfrac{x}{c}\right) = \cosh^2\left(\dfrac{x}{c}\right)$

$\displaystyle S = 2\pi \cdot c \int_0^c \cosh^2\left(\frac{x}{c}\right) \, dx$

note that $\cosh^2(u) = \dfrac{1+\cosh(2u)}{2}$

can you finish from here?

6. Re: Clarification on an identity of function

Wow that really helped . i missed to take c as constant thank you so much