Here I show TWO Ways to Do This! Can You Think of a Third??? ;-)

   The Integral of Sine Squared:

   (    2                          (      2                          (    2
   | sin(x) dx = - sin(x) cos(x) - | - cos(x) dx = - sin(x) cos(x) + | cos(x) dx
   )                               )                                 )

    +-   u = sin(x)    dv = sin(x) dx   (             (       -+
    |   du = cos(x) dx   v = - cos(x)   | u dv = uv - | v du   |
    +-  * Integration By Parts *        )             )       -+

                       (        2                          (      (    2
   = - sin(x) cos(x) + | 1 - sin(x) dx = - sin(x) cos(x) + | dx - | sin(x) dx
                       )                                   )      )

                            (    2
   = - sin(x) cos(x) +  x - | sin(x) dx      And Next, Collect Like Terms,
                            )

        (    2                                    (    2         - sin(x) cos(x)     x
  So: 2 | sin(x) dx = - sin(x) cos(x) +  x  And:  | sin(x) dx = ----------------- + ---
        )                                         )                     2            2

             ( pi  2        +-  - sin(x) cos(x)      x  -+ pi     pi
  Therefore: |  sin(x) dx = |  -----------------  + ---  |     = ----
             ) 0            +-         2             2  -+ 0      2

  And the Root of the Mean of the Square is calculated thusly:

                        +-  ( pi  2       -+1/2                   _
                        |   |  sin(x) dx   |    +-   pi  -+1/2   V2
  The RMS of Sine(x) =  |   ) 0            |  = |  -----  |    = --- = .707
                        |  --------------  |    +-  2pi  -+       2
                        +-       pi       -+

  And Another Way To Derive This:
  +-------------------------------------------------------------------------------+
  |  Some Trig Identities:                                                        |
  |                                                                               |
  |              2         2   |                                                  |
  |  Now: 1 = sin(x) + cos(x)  | Now: cos(x + y) = cos(x) cos(y) - sin(x) sin(y)  |
  |                            |  Then: cos(2x) = cos(x) cos(x) - sin(x) sin(x)   |
  |         2            2     |                                                  |
  |  So: cos(x) = 1 - sin(x)   |                          2        2              |
  |                            |    So Then: cos(2x) = cos(x) - sin(x)            |
  |                                                                               |
  |  So:            2        2              2        2                2           |
  |    cos(2x) = cos(x) - sin(x)  =  1 - sin(x) - sin(x)  =  1 - 2 sin(x)         |
  |                                                                               |
  |  And:                     +-         2   -+          2                        |
  |        1 - cos(2x) =  1 - | 1 - 2 sin(x)  |  =  2 sin(x)                      |
  |                           +-             -+                                   |
  |  Therefore:                                                                   |
  |         2      1 - cos(2x)     +-                2      1 + cos(2x) -+        |
  |      sin(x) = -------------    |  Likewise! : cos(x) = ------------  |        |
  |                    2           +-                           2       -+        |
  +-------------------------------------------------------------------------------+
  And Now Integrate:

    ( pi  2         1  ( pi                  1  ( pi      1  ( pi
    |  sin(x) dx = --- |   1 - cos(2x) dx = --- |  dx  - --- |  cos(2x) dx
    ) 0             2  ) 0                   2  ) 0       2  ) 0
                                                                 u = 2x
         Now: By Substitution, Inverse of the Chain Rule:       du = 2 dx
                                                                dx = du/2
       pi    1  ( x=pi     du     pi    1  ( x=pi  1
    = --- - --- |  cos(u) ---- = --- - --- |      --- cos(u) du 
       2     2  ) x=0      2      2     2  ) x=0   2

       pi    1         |x=pi    pi     1          |pi   pi     1  +-     -+    pi
    = --- - --- sin(u) |     = ---- - --- sin(2x) |  = ---- - --- | 0 - 0 | = ----
       2     4         |x=0     2      4          |0    2      4  +-     -+    2

      ( pi   2         pi
  So: |   sin(x) dx = ----  and the same RMS for Sine Squared Follows...
      ) 0              2

  Copyright (c) 2000  for the Public Domain by R. Steve Walz

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