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#1 2005-12-12 10:38:14

Chemist
Member
Registered: 2005-12-12
Posts: 35

Integration

Greetings mates,

If anyone can help me solve the following integral using the change of variable ჯ = cosმ

The ^ means to the power, so 2Cos^2მ is 2 multiplied by the squared of Cosმ ( just to avoid confusion ).

The given integral is:
I = ∫[(1+ჯ)/(1-ჯ)]^½dჯ

I already worked it out using other methods like taking t^2 = (1+ჯ)/(1-ჯ) and then solving in terms of t. I found out the solution. However, the method of change of variable using Cosმ is not working. For starters,

I = ∫[(1+cosმ)/(1-cosმ)]^½dcosმ = - ∫[(2Cos^2B)/(2Sin^2B)]^½ sinმdმ where B = მ/2

This implies, I = - ∫[1/|Tan^2B|] sinმdმ
--> I = - ∫ (sinმdმ / Tan^2B )

How to cotinue from here? I've tried many methods using integrations by parts, by nothing worked out.

Thanks in advance.

Last edited by Chemist (2005-12-12 10:39:18)


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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#2 2005-12-12 11:11:23

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Integration

You're making things more complicated than they need to be.

Inside the integral, the expression is (1 + cosθ)(1 - cosθ).

Multiplying out gives 1 - cos²θ, which is identical to sin²θ.

So now you have ∫ √[sin²θ] dcosθ and the powers cancel out to give ∫ sinθ dcosθ. Usually, when you are told to integrate by substitution, the reason is that the expression can be broken down into something nice and simple.


Why did the vector cross the road?
It wanted to be normal.

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#3 2005-12-12 12:04:13

Chemist
Member
Registered: 2005-12-12
Posts: 35

Re: Integration

Hello mathsyperson ,

thank you for your help, but you obviously didnt notice the / ( division ). The exp is (1 + cosθ) / (1 - cosθ). This is why I changed it to TanB/2


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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#4 2005-12-12 12:10:22

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Integration

Oh, I hate it when I do stupid things like that. It is past midnight, though, so I have a bit of an excuse.

In that case, I'd turn it into 1 + (2cosθ) / (1-cosθ), because I think there's an identity that can simplify that. Maybe there isn't, but it certainly looks familiar from somewhere.


Why did the vector cross the road?
It wanted to be normal.

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#5 2005-12-12 12:14:13

Chemist
Member
Registered: 2005-12-12
Posts: 35

Re: Integration

No problem mate, I'll try out different methods. Thanks anyway.


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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#6 2005-12-12 19:37:59

Flowers4Carlos
Member
Registered: 2005-08-25
Posts: 106

Re: Integration

hi yaz chemist!!

  (1+x)^(1/2)
∫---------------dx          and you are trying to solve this integral using x=cosθ ⇒ dx=-sinθdθ
  (1-x)^(1/2)

  (-sinθ)(1+cosθ)^(1/2)
∫-------------------------dθ     multiply everything by (1+cosθ)^(1/2)
  (1-cosθ)^(1/2)

  (-sinθ)(1+cosθ)^(1/2)*(1+cosθ)^(1/2)
∫--------------------------------------------dθ
  (1-cosθ)^(1/2)*(1+cosθ)^(1/2)

  (-sinθ)(1+cosθ)
∫--------------------------dθ
  (1-cos²θ)^(1/2)

  (-sinθ)(1+cosθ)
∫------------------dθ
        sinθ

i guess you can take it from here!!  i may have done something wrong so please √√

Last edited by Flowers4Carlos (2005-12-12 19:49:56)

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#7 2005-12-12 22:31:44

Chemist
Member
Registered: 2005-12-12
Posts: 35

Re: Integration

Greetings Flowers4Carlos,
Thank you for your help. I believe there's no mistake in your calculation, which is pretty good and simple. I shall double check the answer anyhow, thanks again.


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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#8 2005-12-12 22:43:50

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,551

Re: Integration

Greetings to you, Chemist, and welcome to the forum.

If you feel so inclined, you can tell us more about yourself in "Introductions" smile


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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