Calculus

swagata:

:roll: integration 0 to 1 cos inv x whole square

int 0 to 1 of [( cos ^-1x) ² ]

is that the question?



*

∫ (acos(x))^2 dx from 0 to 1

Definitely have to go to the integral tables for this one !!

Indefinite form :

-2 * (1 - x^2)^(1/2) * acos(x) - 2x + x * (acos(x))^2

Definite Output =

(π - 2) =

1.1459.......

Mmmm.......that's a really interesting result !!

Here's some Websites that might help with this kind of "sticky" integral !!


http://integrals.wolfram.com/index.jsp?expr=acos[x]^2&random=false

http://www.wolframalpha.com/input/?i=definite+integral+calculator&f1=%28acos[x]%29^2&f=DefiniteIntegralCalculator.integrand_%28acos[x]%29^2&f2=x&f=DefiniteIntegralCalculator.variable\u005fx&f3=0&f=DefiniteIntegralCalculator.rangestart\u005f0&f4=1&f=DefiniteIntegralCalculator.rangeend_1

http://en.wikipedia.org/wiki/List_of_integrals_of_inverse_trigonometric_functions#Arccosine_function_integration_formulas


The first one will calculate an indefinite integral, the second one a definite integral, and the third is Wikipedia's integral tables of inverse trig functions.

(Wolfram's method of inputting functions in the first two sites is a little tricky.......remember to use the brackets as specified!! )

Good luck !!

[quote="CPhill"]∫ (acos(x))^2 dx from 0 to 1

Definitely have to go to the integral tables for this one !!

Or use the Scientific calculator here:

Thanks for that tip, Alan.

I should have checked that first !!

CPhill:

Thanks for that tip, Alan.

I should have checked that first !!

Well, it's not always obvious how to use the calculator on this site! I've only just realised how to do the definite integral:

If you just want to know the result then using the scientific calculator is fine. If however you would like to know how the result is arrived at, that's different.
Here's a route through to the result, (and sorry, I don't know the best way of entering this maths stuff, maybe Melody could point me in the right direction ?)

int (cos ^-1x) ² dx = int (cos ^-1x) ².1.dx ,
(and integrating by parts,)
= x.(cos ^-1x) ² + int x.2cos ^-1x /sqrt(1 - x ²) dx
(Note that the negative sign from the parts formula is multiplied by a negative sign from the derivative of cos ^-1x .)

For this new integral, use the substitution u = cos ^-1x.
From that, du = -1/sqrt(1 - x ²) dx,
in which case the (new) integral becomes int (-1).2u.cos(u) du.

Integrating that by parts gets you (-2){u sin(u) - int sin(u) du} = (-2){u sin(u) + cos(u)},
(where again you have two negative signs giving you a positive sign.)

Now notice that if u = cos ^-1x, then cos(u) = x, so sin(u) = sqrt(1 - x ²).

So, substituting for u and putting this with the first term from the original integration by parts, we arrive at

x.(cos ^-1x) ² - 2sqrt(1 - x ²)cos ^-1x - 2x, (plus a constant of integration, or substitute in your limits).

unicorn in ur butt

Wow!! Nice job, Bertie!!!!

I definitely didn't see that one.

Of course, your way is superior.

Any time we don't have to lean on calculators or tables - then it's REAL math !!

Hi Bertie,

Considering that we don't have Latex here, I think that you have done a very good job of presenting this.
I do have a couple of little suggestions. I think that I would like to include other people in this so I will talk about it in my nightly wrap.
I'll do this either tonight or in the very near future.

Welcome to the forum Bertie I sent you a private message to welcome you ages ago, when you first joined up. I know that you haven't collected it yet.
I wrote a lot more here but I have decided to include it in tonight's wrap instead. Other new people need to know about the mail system as well.

Thanks you for this post. I am so glad that you decided to join in.
Melody.

edit.
The message information is not in tonight's wrap afterall - I'll probably include it tomorrow!