1/(1+cos(x))

i swear iv gone brain dead in the last 4 hours

**0**

# how do you integrate...

Started by paddyb67, Nov 14 2007 09:48 PM

3 replies to this topic

### #1

Posted 14 November 2007 - 09:48 PM

### #2

Posted 27 November 2007 - 06:01 PM

It isn’t too bad.

Let rearrange it first

1/(1+cosx) = 1/( 1 + 2cos2(x/2)-1) cosx = 2cos2 (x/2)-1

= 1/ 2 cos2 (x/2)

=1/2 sec2 (x/2)

From now, if I weren’t wrong, integrate sec2 (x/2), we get tan (x/2) + C !!!

So the answer is 1/2 tan (x/2) + C

Let rearrange it first

1/(1+cosx) = 1/( 1 + 2cos2(x/2)-1) cosx = 2cos2 (x/2)-1

= 1/ 2 cos2 (x/2)

=1/2 sec2 (x/2)

From now, if I weren’t wrong, integrate sec2 (x/2), we get tan (x/2) + C !!!

So the answer is 1/2 tan (x/2) + C

### #3

Posted 27 November 2007 - 09:30 PM

That's nice I'll confess that I had a go at this question a while ago and gave up!

Just to clarify your solution, it relies on the identity , i.e. . So in the denominator, we have the LHS with . So we can replace it using the identity:

Note that the answer isn't - you can see this just by differentiating, as the chain rule comes in to play.

Another possible method is to use the " substitution". I vaguely recalled this from 1st year at uni (!) but didn't have time to hunt out my notes. There's a relevant bit on Wikipedia.

Basically, we can let and do a substitution. The hardest bit is getting that , and ; then you get:

Notice that I've left out some lines of working there; it's just manipulating the fractions to get a lot of cancellation.

Just to clarify your solution, it relies on the identity , i.e. . So in the denominator, we have the LHS with . So we can replace it using the identity:

Note that the answer isn't - you can see this just by differentiating, as the chain rule comes in to play.

Another possible method is to use the " substitution". I vaguely recalled this from 1st year at uni (!) but didn't have time to hunt out my notes. There's a relevant bit on Wikipedia.

Basically, we can let and do a substitution. The hardest bit is getting that , and ; then you get:

Notice that I've left out some lines of working there; it's just manipulating the fractions to get a lot of cancellation.

### #4

Posted 28 November 2007 - 01:01 PM

Oh thanks, silly me, forgot the chain rule, so careless !!!

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