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

Mar 2010
15
0
\(\displaystyle \int{\frac{x+sin(x)}{1+cos(x)}}dx\)


My attempt so far

\(\displaystyle \int{\frac{(x+sin(x))(1-cos(x))}{(1+cos(x))(1-cos(x))}}dx\) =

\(\displaystyle \int{\frac{x-x*cos(x)+sin(x)-cos(x)sin(x)}{sin^2(x)}}dx\) =

\(\displaystyle -x*cotan(x)+ln(sin(x))-x*cosec(x)-\frac{cos(x)}{sin^2(x)} + \int{\frac{1}{sin(x)}}dx-ln(sin(x))\) =

\(\displaystyle -x*cotan(x)-x*cosec(x)-\frac{cos(x)}{sin^2(x)} + \int{\frac{1}{sin(x)}}dx\)

I should get \(\displaystyle \frac{x(1-cos(x))}{sin(x)} = \frac{x}{sin(x)}-x*cotan(x)\) pretending that I don't know how to solve \(\displaystyle \int\frac{1}{sin(x)}dx\) so apparantly somehow I should be able to paritally integrate \(\displaystyle \int\frac{x*cos(x)}{sin^2(x)}dx\) in such a way that I get \(\displaystyle \int\frac{-1}{sin(x)}dx\) and so get rid of that integral.
 
Last edited:
Jun 2009
806
275


= \(\displaystyle \int{\frac{x}{1+cosx}}dx + \int{\frac{sinx}{1+cosx}}dx\)

= \(\displaystyle \int{\frac{x}{2cos^2{x/2}}}dx + \int{\frac{2sin(x/2)cos(x/2)}{2cos^2(x/2)}}dx\)

= \(\displaystyle \int{\frac{(x)(sec^2(x/2)}{2}}dx + \int{\tan(x/2)}dx\)

Now try to integrate.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
Try to only integrate the first one , there is a trick : we don't need to integrate the second one , it will be deleted finally !!


\(\displaystyle \frac{1}{2} \int x \sec^2(\frac{x}{2})~dx + \int \tan(\frac{x}{2})~dx \)

Using integration by parts ,

\(\displaystyle = x\tan(\frac{x}{2}) - \int \tan(\frac{x}{2})~dx + \int \tan(\frac{x}{2})~dx \)

we can see the second integral can be deleted but note that before writing this

\(\displaystyle = x\tan(\frac{x}{2}) + C \)

we should add one more line , it is

\(\displaystyle = x\tan(\frac{x}{2}) + \int 0~dx \)

so followed by

\(\displaystyle = x\tan(\frac{x}{2}) + C \)