Solution to Problem 11.8-3


© 2001 by Karl Hahn 		KCT logo

3)The problem is to integrate 


   



 tan6(2x) dx

Step 1: Use the tangent substitution trick. This has a new wrinkle in that we have 2x inside the parenthesis instead of just x. So the substitution is just a little different, but not much different. See if you can handle it. Make the substitution, then click here to continue.

Your substitution is 



   u  =  tan(2x)
and 

1
 
2


   du
         =  dx
 u2 + 1

so your substituted integral is 


   



 tan6(2x) dx  =



  1
   
  2


 


   u6
        du
 u2 + 1

Step 2: You have a quotient of polynomials now, but highest power in the numerator is too big. Recall that the highest power in the numerator must be less than the highest power in the denominator. To fix that you have to use polynomial long division and apply the remainder rule. Do all that and then click here to continue.

The polynomial long division is 

               u4          -   u²          +   1              
   u2  +  1  ) u6  +  0u5  +  0u4  +  0u3  +  0u2  +  0u  +  0
               u6  +  0u5  +   u4
                                 
                              -u4  +  0u3  +  0u2
                              -u4  +  0u3  -   u2
                                                 
                                               u2  +  0u  +  0
                                               u2  +  0u  +  1
                                                              
                                                            -1
So the quotient is  u4 - u2 + 1,  and the remainder is -1. When you apply the remainder rule to that the integral becomes 


   



 tan6(2x) dx  =



  1
   
  2


 



 (u4 - u2 + 1) du  -



  1
   
  2


 


   du
       
 u2 + 1

Step 3: The right-hand integral needs trig substitution. Recall the substitution for this form of integral and apply it. Then click here to continue.

The substitution is 


  tan(v)  =  u
and 

  (tan2(v)) dv  =  du

and the integral becomes 


   



 tan6(2x) dx  =



  1
   
  2


 



 (u4 - u2 + 1) du  -



  1
   
  2


 



 dv

Step 4: Integrate. Everything is in easy form now. Do it and then click here to continue.

You should have 


   



 tan6(2x) dx  =



  1
   
  2


 


u5   u3
   -    + u
 4    3





     v
  -     +  C
     2

Step 5: Substitute back. First substitute the v back to u, then substitute u back to x. When you're done, click here to see the final answer.

When you substitute for the bare v with  tan(v) = u,  you have to use the arctangent. You get 


   



 tan6(2x) dx  =



  1
   
  2


 


u5   u3
   -    + u
 4    3





     1
  -    arctan(u)  +  C
     2

Then when you substitute back  u = tan(2x),  you get 


   



 tan6(2x) dx  =



  1
   
  2


 


tan5(2x)   tan3(2x)
         -          + tan(2x)
    5          3






  -  x  +  C

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