Example 4: Find the antiderivative of e^xcos(x).

Solution: With this one we can choose either

1. With f(x)=cos(x) and g(x)=e^x, we have fg¢=e^x cos(x), or
   
   
2. with f(x)=e^x and g(x)=sin(x), we have again fg¢=e^x cos(x).
   
   

We will follow through both of these.

1. 
   

   	ó õ ex cos(x) dx	=	ex cos(x) - ó õ ex ·(-sin(x)) dx
   		=	excos(x) + ó õ ex sin(x) dx.

   It is not clear that this is an improvement, as we are left with an integral of the same type. We now perform another integration by parts, using f(x)=sin(x) and g(x)=e^x. This gives fg¢ = e^x sin(x). Thus
   

   ó õ ex sin(x) dx

   =

   ex sin(x) - ó õ ex cos(x),

   so that our original antiderivative is
   

   	ó õ ex cos(x) dx	=	ex cos(x) + ó õ ex sin(x)
   		=	ex cos(x) +(ex sin(x) - ó õ ex cos(x))
   		=	ex cos(x) +exsin(x) - ó õ ex cos(x).

   It is still not clear that this is an improvement. However, a little algebra helps out here. Add òe^x cos(x) dx to both sides of the equation to get
   

   	2 ó õ excos(x) dx	=	excos(x) +exsin(x) +C
   		=	ex (cos(x) +sin(x)) +C,

   so that
   

   ó õ excos(x) dx =  ex 2 (cos(x) +sin(x)) +C.

   Again, one can check the verity of this by differentiating.
   
   
2. Similar is
   

   ó õ excos(x) dx

   =

   exsin(x) - ó õ exsin(x) dx.

   We have already seen the trick of integrating by parts twice, and we perform this again. Using f(x)=exp(x) and g(x)=-cos(x), we have fg¢=e^xsin(x). Thus
   

   ó õ exsin(x) dx

   =

   -excos(x) - ó õ ex(-cos(x)) dx,

   so that our original antiderivative is
   

   	ó õ excos(x) dx	=	exsin(x) - ó õ exsin(x) dx
   		=	exsin(x) -(-excos(x) - ó õ ex(-cos(x)))
   		=	exsin(x) +excos(x) - ó õ excos(x) dx.

   Again, add òe^xcos(x) dx to both sides of the equation to get
   

   	2 ó õ excos(x) dx	=	exsin(x) +excos(x) +C
   		=	ex(sin(x) +cos(x)) +C,

   so that
   

   ó õ excos(x) dx =  ex 2 (sin(x) +cos(x)) +C.

   This is the same as above.
   
   

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File translated from T_EX by T_TH, version 3.33.
On 09 Jan 2004, 02:52.