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MathsIsFun
2005-12-13 21:43:50

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

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

Chemist
2005-12-13 21:31:44

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.

Flowers4Carlos
2005-12-13 18:37:59

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 √√

Chemist
2005-12-13 11:14:13

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

mathsyperson
2005-12-13 11:10:22

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.

Chemist
2005-12-13 11:04:13

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

mathsyperson
2005-12-13 10:11:23

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.

Chemist
2005-12-13 09:38:14

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.