Solution to Exercise 11.5-1


© 2001 by Karl Hahn 		KCT logo

The problem was to use trig substitution to find the indefinite integral: 

   

  _______
 √25 - x2 dx
Step 1: Factor out the 25. Pull it all the way out of the radical, hence it will have to get square-rooted out there 


   5


 

 




1  -


  x2
    
  25




 dx
or equivalently 


   5


 

 




1  -


  

x
 
5



2






 dx
Step 2: Make the substitution. For this example the substitution (and taking its derivative implicitly) is: 

              x
   sin(u)  =   
              5
and 
       du     1
cos(u)     =   
       dx     5
or 

5 cos(u) du  =  dx
When you put that in for x and dx, the integral becomes 

   

  _______
 √25 - x2 dx



  =  25


 

  ___________
 √1 - sin2(u) cos(u) du  =


  25

 


 cos2(u) du
Step 3: Integrate. We already did the integral of cosine squared in the main text, so click here if you don't remember the trick. You will find that 

   25

 


 cos2(u) du  =  25


 

1       1
  u  +    sin(u)cos(u)
2       2




 + C
Step 4: Substitute back. You have to take the inverse function on the original substitution to substitute back the bare u, so it becomes  arcsin(x/5).  The  sin(u) becomes  x/5.  And the  cos(u)  is the same as 

    ___________
   √1 - sin2(u)
so it becomes 

   



      x2
1  −    
      25
Putting all the pieces together and multiplying the 25 through both the terms, you get 

   

  _______
 √25 - x2 dx


     25
  =     arcsin
      2



x
 
5



     5
  +    x
     2

 



      x2
1  −    
      25




  +  C  =



                                   25
                                      arcsin
                                    2



x
 
5



     1
  +    x
     2

  _______
 √25 - x2



  +  C
Now very carefully (paying close attention to what all the constant factors do) take the derivative above and confirm that you get back the original integrand.

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