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yonski

Member

Registered: 2005-12-14

Posts: 67

Definite integral

Hi there,
what's the simplest way of finding the value of the folllowing definite integral, in terms of a (where a is a positive constant)?

I've tried expanding and rearranging the trig part into a form that I know how to integrate, but without success

Thanks.

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Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Definite integral

The constants a^2 and 2 can be pulled out of the integral to the outside.
Then I used "The Wolfram Integrator" online to obtain this answer.
Sort of cheating I know, but it's amazing.

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igloo myrtilles fourmis

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Definite integral

I forget what the rule is called, but you should be able to use this fact:

Where f(x) = g(x) = cos(x) + 1

Edit: Never mind, apparently that rule won't work if they are the same function.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

yonski

Member

Registered: 2005-12-14

Posts: 67

Re: Definite integral

The constants a^2 and 2 can be pulled out of the integral to the outside.
Then I used "The Wolfram Integrator" online to obtain this answer.
Sort of cheating I know, but it's amazing.

The answer is in the back of the book, or I can just use my calculator to get it Unfortunately I won't have that luxury in an exam lol

Edit: Never mind, apparently that rule won't work if they are the same function.

Yeah, I don't think the partial fractions thing will work.

This one's confusing because it comes amidst some relatively easy questions, so I wasn't sure if I was missing a simple trick.

Hmm, it's gonna bug me, but hopefully they won't ask anything this tough in an exam!

Thanks.

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Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

Oren

Guest

Re: Definite integral

Hello,

I tried transforming the integral into a double integral and used (1+Cosx)(1-Cosx) = SinxSinx

I reached the expression : integral = ( 1 / Sinx - 1 / tanx )( 1 / Sinx - 1 / tanx ) + C

Can anyone verify ?

Oren

Guest

Re: Definite integral

Ofcourse I kept the constants a.a and 1/2 out , but they can be added to get the right form of solution.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Definite integral

I tried transforming the integral into a double integral and used (1+Cosx)(1-Cosx) = SinxSinx

That rule does not apply unless it is the integral itself that is squared, not the function inside the integral.  I tried the same thing myself till realizing it.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Oren

Guest

Re: Definite integral

I just reviewed my work , you're right Ricky. I'm still learning about this. I shall rethink about it.

Oren

Guest

Re: Definite integral

I think I got it

take x = 1 / 1+cos

I = ∫ t / √( 2t - 1 ) dt

take u = 2t - 1

I = 1/2 ∫  [ √u - 1/√u ] du

Oren

Guest

Re: Definite integral

EDIT:

I think I got it

take t = 1 / 1+cosx

I = ∫ t / √( 2t - 1 ) dt

take u = 2t - 1

I = 1/2 ∫  [ √u - 1/√u ] du

Drew

Member

Registered: 2007-07-21

Posts: 2

Re: Definite integral

I = ∫ [ 1 / ( 1 + cosx ) ( 1 + cosx) ] dx

t = 1 / 1 + cosx

dt = [ sinx / ( 1 + cosx)( 1 + cosx) ] dx

Cosx = 1 / t - 1 so Sinx = sqrt [( 2t - 1 / t )] / t

I = ∫ [ t / sqrt( 2t -1 ) ] dt  = 1/2 ∫ [ ( 2t +1 - 1 ) / sqrt(2t - 1 ) ] dt

I = 1 / 2  ∫ [ sqrt ( 2t -1 ) - 1 / sqrt ( 2t - 1 ) ] dt

I = 1 /2 sqrt(2t -1)  [ ( 2t -1 ) / 3 - 1 ]

If im not mistaken the final expression should be something like this :

I = 1 /2 sqrt ( 1 - Cosx / 1 + Cosx ) [ (1 /3) (1 - Cosx) / ( 1 + Cosx ) - 1 ] + C

The definite integral can then be evaluated. Sorry for this unclearness as I still havent learned about LAtex

Last edited by Drew (2007-07-21 09:37:51)

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

First, using trig formula cos(2x)=cos²(x)-sin²(x)

The original problem equals to

Last edited by George,Y (2007-07-24 02:37:30)

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

Next, Define S(θ)

Then

So

Last edited by George,Y (2007-07-24 14:09:20)

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

when x=0, x/2=0; when x=π/2, x/2=π/4

Thus the result is

Last edited by George,Y (2007-07-24 14:12:37)

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X'(y-Xβ)=0

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Definite integral

George, your first step seems to be saying that:

Which doesn't seem to be the case.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

George, your first step seems to be saying that:

Which does seem to be the case.

Uhh... you are right! I'm correcting it.

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

This is correct instead:

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

My answer is 0. Can anyone evaluate it by software?

---my new answer is a²/3

Last edited by George,Y (2007-07-24 14:16:47)

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X'(y-Xβ)=0

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Definite integral

I get:

But I don't have time to check through your steps right now.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Definite integral

I think what you mentioned earlier, Ricky, was using partial fractions. But i don't think what you said is valid.

You can seperate two linear factors in the denominator provided the degree of the denominator is greater than that of the numerator.

1/ [(ax + b)*(mx + c)] = A/(ax+b) + B/(mx + c) but if a term is an irreducible quadradic factor, the numerator becomes of the form Ax + b. If a binomial in the denominator is raised to a power n > 1 then that alone becomes n separate terms when you separate it.
The rules are kind of strict so i don't think it applies to just f(x) and g(x) in the denomintor. Correct me if I'm wrong.

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Definite integral

You are correct, it will apply to all rational polynomials p(x)/q(x) given that the degree of q(x) is higher than that of p(x).  However, there are other cases where it will apply as well.

Although when you go through the steps it becomes very easily seen that this is a special case.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

Uhh, +/- mistake corrected. Now my result is not 0 but a²/3, consistent with your result in Post 19.

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

So the answer is:

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Definite integral

Ricky's result in Post 9 is 2a² times as large as my result, because what he evaluates is 2a² times as large as the original question, too.

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X'(y-Xβ)=0