Integral: 1/(1+cos(x)+sin(x)) help

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Homework Help Overview

The discussion revolves around the integral of the function 1/(1+cos(x)+sin(x)), focusing on trigonometric substitutions and transformations to simplify the expression.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of half-angle formulas and the substitution of tan(x/2) to transform the integral. Questions arise regarding the correctness of the transformations and the necessity of further simplifications.

Discussion Status

Some participants have provided guidance on continuing the solution process, while others express uncertainty about the need for additional steps beyond a certain point. There is acknowledgment of a potential mistake in interpreting the final answer.

Contextual Notes

Participants are navigating through trigonometric identities and substitutions, with some expressing confusion about the correctness of their approaches and the validity of their final expressions.

Alexx1
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Can someone help me with this integral?

Integral: 1/(1+cos(x)+sin(x))
 
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Solving this will take some adroit trig substitutions. Half angle formulae, particularly that for tan(x/2), will come in handy here.
 


D H said:
Solving this will take some adroit trig substitutions. Half angle formulae, particularly that for tan(x/2), will come in handy here.

I've tried it, but my answer isn't correct..

cos(x) = (1 - (tan(x/2))^2) / (1 + (tan(x/2))^2)

sin(x) = (2*tan(x/2)) / (1 + (tan(x/2))^2)

And t = tan(x/2) ==> x = 2 bgtan(t) ==> dx = 2/(1+x^2)

If I do it like that i become this integral:

1/(t+1) dt

But that's not correct. Have I done something wrong?
 


That is exactly right, but you haven't finished yet. Continue on!
 


D H said:
That is exactly right, but you haven't finished yet. Continue on!

Than I get

ln (t+1)

= ln (tan(x/2) +1)

= ln ((sin(x/2) / cos(x/2)) +1)

= ln (sin(x/2) +1) - ln(cos(x/2)+1)

And what now?
 


D H said:
That is exactly right, but you haven't finished yet. Continue on!

Than I get

ln (t+1)

= ln (tan(x/2) +1)

= ln ((sin(x/2) / cos(x/2)) +1)

= ln (sin(x/2) +1) - ln(cos(x/2)+1)

And what now?
 


Alexx1 said:
Than I get

ln (t+1)

= ln (tan(x/2) +1)
Good so far.

= ln ((sin(x/2) / cos(x/2)) +1)

= ln (sin(x/2) +1) - ln(cos(x/2)+1)
Whoa! What's this last step?

For that matter, why do you need to go beyond ln(tan(x/2)+1) ? That is a perfectly good answer in and of itself.
 


D H said:
Good so far.


Whoa! What's this last step?

For that matter, why do you need to go beyond ln(tan(x/2)+1) ? That is a perfectly good answer in and of itself.

Ow, I'm verry sorry.. I made a mistake, I read the wrong answer in my book.. ln (tan(x/2) +1) is the correct answer..
 

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