Expert: Paul Klarreich - 4/29/2008

Question
Integration of exp(x)(1+Sinx)/(1+Cosx)....
Please sir i am not able to do this very hard for me to do this.......I tried some trig formulaes but again not able to do.

Answer
Questioner:   Suzan Thomas
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  Integration of exp(x)(1+Sinx)/(1+Cosx)....
Please sir i am not able to do this very hard for me to do this.......I tried some trig formulaes but again not able to do.
........................................
Hi, suzan,

I am sorry, but I cannot find a clever way to do it.  All I have to offer you is plenty of sweat and struggle.  Your integral was:

{ exp(x)(1 + sin x)
| ----------------- dx
}   (1 + cos x)

If you go to this web site, called THE INTEGRATOR:

integrals.wolfram.com/index.jsp

and enter your integral, you get this answer:

exp(x) tan(x/2)

If you go to THIS site:

en.wikipedia.org/wiki/Trigonometric_identity#Double-.2C_triple-.2C_and_half-angle_formulae

you find this identity:

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

So the Integrator's answer was:

 exp(x) sin x
---------------
  (1 + cos x)

which you can now use as a 'hint' as to how to do your integral.  Start by computing the derivative, as a 'check'.

Use the quotient rule:  [I will write e = exp x, s = sin x,  c = cos x]

(1 + c)(e s + e c) - (e s)(- s)
-------------------------------- << quotient and product rules
(1 + c)^2

e(1 + c)(s + c) - (s )(-s)]
-----------------------------  << factor out  e
(1 + cos x)^2

e [s + c + c s + c^2 + s^2]
----------------------------- << multiply out
(1 + cos x)^2

e [s + c + c s + 1]
-------------------- << identity
(1 + cos x)^2

e [s(1 + c) + 1 + c]
----------------------------  << factor out an s
(1 + cos x)^2

e [(s + 1)(1 + c)]
--------------------  << factor some more
(1 + cos x)^2

exp x [sin x + 1]
-----------------  << cancel, and you have your integral.
(1 + cos x)

...............................................
So it's the right answer (big deal). Now let's do it.  Here is your integral:

{ exp(x)(1 + sin x)
| ----------------- dx = J, just to give it a name,
}   (1 + cos x)

The nice thing about exp(x) is that it is VERY easy to integrate, which makes it great for being  dv in an Integration by Parts:

             1 + s
Try IBP:  u = ------  and dv = exp(x),  v = exp(x)
             1 + c

    (1 + c)(c) - (1 + s)(-s)
du = ------------------------ dx
        (1 + c)^2

    c + c^2 + s + s^2
du = ------------------ dx
        (1 + c)^2

    c + s + 1
du = ---------- dx
    (1 + c)^2

Your integral is  uv - INT( v du ). So J is:

exp(x)(1 + s)         exp(x)(c + s + 1)
------------- - INT ( ----------------- dx)
 1 + c                   (1 + c)^2

exp(x)(1 + s)         exp(x)[(c + 1) + s]
------------- - INT ( ------------------- dx)
 1 + c                   (1 + c)^2

Now split the second term into two parts, both minus, of course:

     exp(x)(c + 1)
INT ( -------------- dx)  [Int A]
       (1 + c)^2

     exp(x)(s)
INT ( ----------- dx)  [Int B]
      (1 + c)^2

Do A:

      exp(x)
INT ( -------- dx)
     (1 + c)

u = (1 + c)^-1,  du = -1(1 + c)^2(-s) dx
       s dx
du = ----------
    (1 + c)^2

dv = exp(x) dx,  v = exp(x)
exp(x)        exp(x) s dx
------- - INT (----------- )  << Wow! That is integral B, so
(1 + c)          (1+c)^2

    exp(x)    
A = ------- - B
   (1 + c)    

So we have:

       exp(x)
A + B = -------
       (1 + C)

AND FINALLY, OUR J IS:

exp(x)(1 + s)   
------------- - (A + B)
 1 + c         

exp(x)(1 + s)   exp(x)
------------- - -------
 1 + c         (1 + c)

exp(x)(1 + s - 1)
-----------------
 1 + c       

exp(x)(s)
---------
 1 + c

which is, in fact, exp(x) tan(x/2)