- #1

Alexx1

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- 0

Can someone help me with this integral?

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

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

Last edited:

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- Thread starter Alexx1
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- #1

Alexx1

- 86

- 0

Can someone help me with this integral?

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

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

Last edited:

- #2

- 15,450

- 688

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

- #3

Alexx1

- 86

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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?

- #4

- 15,450

- 688

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

- #5

Alexx1

- 86

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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?

- #6

Alexx1

- 86

- 0

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?

- #7

- 15,450

- 688

Good so far.Than I get

ln (t+1)

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

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

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

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.

- #8

Alexx1

- 86

- 0

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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