Integrals

[Alexx1]

Can someone help me with this integral?

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

[D H]

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

[Alexx1]

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?

[D H]

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

[Alexx1]

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]

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]

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.

[Alexx1]

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..