integrals of trigametrics?

2 Answers

• 1) Cos(2x) = 1 - 2sin^2(x)

  2) Using this, sin^2(x) = [1 - cos(2x)]/(2)

  3) Now integrating this, we get, = (1/2)*[x - sin(2x)/2]

  Hence answer is = (1/4)*[2x - sin(2x)] + C

  Source(s): Self knowledge

• > int(sin(x)^2, x) =

                            1                        1  
  
                         - --- sin(x)*cos(x) + ----x + C

  2 2

  Source(s): Maple 13.01