SQL Server 2005 Get First and Last date for any Month in any Year

First day of the month: SELECT DATEADD(mm, DATEDIFF(mm, 0, GETDATE()), 0)

Last day of the month: SELECT DATEADD(ms, -3, DATEADD(mm, DATEDIFF(m, 0, GETDATE()) + 1, 0))

Substitute a DateTime variable value for GETDATE().

I got that long ago from this very handy page which has a whole bunch of other date calculations, such as "Monday of the current week" and "first Monday of the month".

DECLARE @Month INTEGER
DECLARE @Year INTEGER
SET @Month = 10
SET @Year = 2010

DECLARE @FirstDayOfMonth DATETIME
DECLARE @LastDayOfMonth DATETIME

SET @FirstDayOfMonth = Str(@Year) + RIGHT('0' + Str(@Month), 2) + '01'
SET @LastDayOfMonth = DATEADD(dd, -1, DATEADD(mm, 1, @FirstDayOfMOnth))
SELECT @FirstDayOfMonth, @LastDayOfMonth

DECLARE @Month int
DECLARE @Year int

set @Month = 2
set @Year = 2004

select DATEADD(month,@Month-1,DATEADD(year,@Year-1900,0)) /*First*/

select DATEADD(day,-1,DATEADD(month,@Month,DATEADD(year,@Year-1900,0))) /*Last*/

But what do you need as time component for last day of the month? If your datetimes have time components other than midnight you may well be better off just doing something like

WHERE COL >= DATEADD(month,@Month-1,DATEADD(year,@Year-1900,0)) 
     AND COL < DATEADD(month,@Month,DATEADD(year,@Year-1900,0)) 

In this way your code will continue to work if you eventually migrate to SQL Server 2008 and the greater precision datetime datatypes.