Evaluation of Integrals involving Trig Functions

 

 

In a Nut Shell:  The evaluation of integrals involving trigonometric functions is typically

a multi-step process.  In many cases you may need to use substitution, trig identities,

grouping, and integration by parts. 

 

There are a few general guidelines that help but you need to work many different examples

to become proficient.  Let’s start with one simple example.

 

 

 

Example 1:      I1  =   ∫ sin9/19 x cos x dx 

 

Substitution:    u  = sin x          du  =  cos x  dx

 

   So   I1   =    ∫ u9/19  du    which is a standard integral.  You need to express the result

In terms of the original independent variable, x.

 

Similar reasoning for the integral:   I1a  =   ∫ cos21/19 x sin x dx

 

In this case   u  = cos x          du  =  - sin x  dx     and

 

    I1a  =  - ∫ u21/19 du      which again is a standard integral.

 

 

 

  For integrals of the type:  ∫ sinm x cosn x  dx 

     with  m an even integer, n an odd integer or vice-versa,

 

 

Example 2:   I2  = ∫ sin8 x cos7 x dx  =  ∫ sin8 x cos6 x cos x dx 

 

   Now let  cos2 x  = 1 – sin2x,       cos6 x  =  ( 1 – sin2x)3 

 

  Then   I2 =  ∫ sin8 x ( 1 – sin2x)3 cos x dx 

 

  Next let    u  =  sin x,  du  =  cos x dx,  so  I2 =  ∫ u8 (1 – u2)3  du

 

 

 Use a similar strategy for

 

         I2a    =  ∫ sin3 x cos6 x dx = ∫ cos6 x sin2 x sin x dx

 

 Now with  u  =  cos x,  du  =  -sin x dx

 

 And using    sin2x =  1 - cos2 x

 



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