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#1 paddyb67

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Posted 14 November 2007 - 09:48 PM

1/(1+cos(x))



i swear iv gone brain dead in the last 4 hours

#2 cK_hoangan

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Posted 27 November 2007 - 06:01 PM

It isnít 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 werenít wrong, integrate sec2 (x/2), we get tan (x/2) + C !!!

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



#3 George

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Posted 27 November 2007 - 09:30 PM

That's nice smile.gif 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 \cos 2\theta=2\cos^2\theta-1, i.e. 1+\cos 2\theta=2\cos^2\theta. So in the denominator, we have the LHS with \theta=\frac{x}2. So we can replace it using the identity:



\begin{align*}\int \frac{1}{1+\cos x} \;dx &= \int \frac{1}{2\cos^2{\frac{x}2}}\;dx \\
&= \int \frac12 \sec^2\frac{x}2\;dx\\
&=\tan \frac{x}2 +c
\end{align*}

Note that the answer isn't \frac12\tan \frac{x}2 +c - you can see this just by differentiating, as the chain rule comes in to play.


Another possible method is to use the "\tan\frac\theta2 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 t=\tan\frac{x}2 and do a substitution. The hardest bit is getting that dx=\dfrac{2}{1+t^2}dt, and \cos x = \frac{1-t^2}{1+t^2}; then you get:



\begin{align*}\int \frac{1}{1+\cos x} \;dx &= \int \frac{1}{1+\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt\;dx \\
&= \ldots \\
&= \int 1 \;dx \\
&= t + c\\
&=\tan \frac{x}2 +c
\end{align*}

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

#4 cK_hoangan

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Posted 28 November 2007 - 01:01 PM

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





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