Integrating the product of the power of sine by the power of cosine

Integrating a function of the product of sine by cosine: f(x) = sin^mxcos^nx ( read sine  exponent m x by cos exponent n x)

Method

1. If m is odd let u = cosx
2. If n is odd let u = sinx
3. If m and n are even use identities to reduce the power of sine and cosine

Example I

Solve ∫sin³xcos⁴xdx

 Since m is odd let u = cox then du =- sinxdx  dx = -du/sinx

Let's substitute cox and dx in the integral

∫sin³xcos⁴xdx = ∫sin³xu⁴.-du/sinx

Let's simplify by sinx:

∫sin³xcos⁴xdx = - ∫sin²xu⁴du

In order to have the integral as a function of u let's express sin²x as an expression of cosx

sin²x = 1-cos²x = 1-u²

Let's substitute sin²x

∫sin³xcos⁴xdx = -∫(1-u²)u⁴du

                     = -∫(u⁴-u⁶)du

                     = -(u⁵/5-u⁷/7 +C

                     = -u⁵/5-u⁷/7 +C

                     =  -cos⁵x/5-cos⁷x/7 +C ( by substituting u by cosx)

Example II

Solve ∫sin²xcos²xdx.

We have m and n even. We use identities to reduce power.

 sin²xcos²xdx. = (1 - cos2x)/2.(1 + cos2x)/2 = 1 - cos²2x/4 = sin²2x/4

Let's reduce the power of sin²2x

 sin²2x = (1 - cos4x)/2

 sin²2x/4 = (1 - cos4x)/8 = 1/8 - 1/8cos4x

 ∫sin²xcos²xdx = ∫(1/8 - 1/8cos4x)dx

                      =1/8 ∫dx - 1/8 ∫cos4xdx

                      = 1/8x - 1/8sin4x + C

Practice

Solve

1) ∫sin⁴xcos³x
2) ∫sin⁴xcos⁴x
.

Integration of trigonometric functions involving the power of sine or the power of cosine

Objective:

To compute the integral of a function involving the power of sine or the power of cosine

Method

In order to solve the integral of the power of sine or the power of cosine we have to reduce their power. We use the following formulas:

sin²x  = 1/2 (1-cos2x )
cos²x = 1/2 ( 1 + cos2x )

Example I

Solve   ∫ sin²xdx

Let's substitute sin²x by  1/2 ( 1-cos2x )

∫sin²xdx = ∫ [1/2 ( 1-cos2x )] dx
         
            = 1/2 ∫ ( 1-cos2x ) dx
         
            = 1/2 [ ∫ dx - ∫ cos2x dx ]

 Let's solve ∫ cos2x dx

Let's use the substitution method by writing u = 2x

Then du = 2xdx dx = du/2

∫ cos2x dx =  ∫ cosu.du/2
                = 1/2 ∫ cosudu

                = 1/2 sinu + C

                =  1/2 sin2x + C

Let's solve the expression between brackets:

 ∫sin²xdx = 1/2 ( x - 1/2 sin2x ) + C

             =  1/2 x - 1/4 sin2x + C
  
Example II   Evaluate ∫ cos⁴x

The technique used to solve  the integral of the power of cosine and the power of sine consists in reducing the power of these functions.

Let's first reduce the power of cos⁴x
cos⁴x = ( cos²x )²

Let's reduce the power of cos²x by using the formula cos²x = 1/2 ( 1 + cos2x )

cos⁴x = [ 1/2 ( 1 + cos2x ) ]²

         = 1/4 ( 1 + 2cos2x + cos²2x )

         = 1/4 + 1/2cos2x + 1/4cos²2x

Let's reduce the power of cos²2x:

cos²x = 1/2 ( 1 + cos2x ). By analogy  cos²2x = 1/2 ( 1 + cos4x ). Then 1/4cos²2x = 1/8 + 1/8cos4x

Therefore cos⁴x = 1/4 + 1/2cos2x + 1/8 + 1/8cos4x

                 cos⁴x = 3/8 + 1/2cos2x + 1/8cos4x
               
∫cos⁴xdx = ∫( 3/8 + 1/2cos2x + 1/8cos4x )dx
             
              = 3/8∫dx + 1/2∫ cos2xdx + 1/8∫ cos4xdx
           
In the example above we found ∫cos2xdx = 1/2 sin2x + C. By analogy ∫cos4xdx = 1/4 sin4x + C

  ∫cos⁴xdx = 3/8x + 1/2 ( 1/2 sin2x ) + 1/8 ( 1/4 sin4x ) + C 
             
                 = 3/8x + 1/4 sin2x  + 1/32 sin4x  + C

Practice 

Evaluate:

1) ∫cos²xdx
2) ∫sin⁴xdx

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