Solution to Exercise 11.5-4


© 2001 by Karl Hahn 		KCT logo

The problem is to use completing the square and trigonometric substitution to find the indefinite integral: 

   


 x2


  ____________
 √15 + 2x - x2 dx

Step 1: Complete the square. When you have your v substitution, remember to apply it also to the x2 on the left of this integrand. When you have done that, click here to continue.

To complete this square, you start with the quadratic under the radical,  q(x) (and since the x2 term you find there is negative, you designate what you find under the radical as  -q(x)):  


   15 + 2x - x2  =  -q(x)
hence 

x2 - 2x - 15  =  q(x)

Now add half the middle coefficient to x to form v. Here the middle coefficient is -2, so 


   v  =  x - 1
and 

v2  =  x2 - 2x + 1

What do you have to add to v2 to make it equal to  q(x)?  

   v2 - 16  =  x2 - 2x - 15  =  q(x)

Step 2: Make the completed-square substitution. The easy part of this is substituting for the  -q(x)  that's under the radical. But think about what you have to do to substitute the x2 that's out there by its lonesome. And what about substituting the dx? When you have all that figured out, click here to continue.

If  v = x - 1,  then  v + 1 = x,  and  v2 + 2v + 1 = x2.  Also when you take the derivative of  v = x - 1,  you find that  dv = dx.  Now you are ready to substitute everything in this integral to turn it from an integral in x to an integral in v. 

   


 x2


  ____________
 √15 + 2x - x2 dx  =


  


 (v2 + 2v + 1)


  _______
 √16 - v2 dv

Step 3: Factor out the 16. Remember that when a factor gets pulled out from under the radical, it gets square-rooted. Click here when you're ready to continue.

This was a pretty easy step. Your integral should now look like 

   4

 


 (v2 + 2v + 1)


  __________
 √1 - (v/4)2 dv

Step 4: Multiply the stuff in the parentheses through by the radical so that your integrand is now the sum of three terms. Then use equation 11.2-13b to break it up into the sum of three separate integrals. Click here when you're ready to continue. (See if you can do the further substitutions that arise from those three integrals as well before you click).

Now you should have 

   4

 


 v2


  __________
 √1 - (v/4)2 dv  +


  8

 

    __________
 v √1 - (v/4)2 dv  +


  4

 

  __________
 √1 - (v/4)2 dv

Step 5: Make your final substitutions. The middle integral is susceptible to simple substitution. Its substitution is 



   s  =  1 -



 


v
 
4





2
and 


ds  =  -



 v dv
     
   8
or 


 -8 ds  =  v dv

For the first and the third integral your substitution is 


              v
   sin(u)  =   
              4
and 

              dv
cos(u) du  =    
               4
or 


4 cos(u) du  =  dv

Make all those substitutions, do a little simplifying, and then click here to continue.

Here's what it should look like all substituted: 


   256


 


 sin2(u)cos2(u) du  -



  64


 

  _
 √s ds  +



  16


 


 cos2(u) du

Step 6: Integrate. The integral on the right we have already done a bunch of times before, but click here if you need to see it again. The middle integral you should know by now, but if you need to look, it's on the table. For the one on the left, you might be tempted just now to have a panic attack over, but try to resist that temptation. You only have to apply some trig identities to it. This will take two trig-id steps. First look at equation 7.1b-6b, and see where you can go with that. Then look at equation 7.1b-9a and see if you can use that to get that left-hand integrand to something you can actually integrate. When you've done all that, click here.

I shall assume that you had no problem with the two integrals on the right. As for the one on the left, squaring equation 7.1b-6b tells you that 

   1
    
   4


 sin2(2u)  =  (sin(u)cos(u))2  =  sin2(u)cos2(u)

and equation 7.1b-9a should have told you that 

   1
    
   4


 sin2(2u)  =


  1
   
  8


 (1 - cos(4u))

That means the integrals we have to take are (including applying the trig identity, equation 7.1b-9b, to the integral on the right as well) 


   32


 


 (1 - cos(4u)) du  -



  64


 

  _
 √s ds  +



  8


 


 (1 + cos(2u)) du

and when you do them you get 


   32u  -  8 sin(4u)  -


  128
     
   3


 s3/2  +



  8u  +  4 sin(2u)  +  C

Step 7: Substitute back to v. The s term is easy -- just put back the substitution of  s = 1 - (v/4)2.  But to do the sine and cosine terms, you will have to apply trig identities again. In particular, apply equations 7.1b-6b and then 7.1b-6a in succession to the  sin(4u)  term. Then substitute back, remembering that 

   cos2(u)  =  1 - sin2(u)
and that 


   u  =  arcsin




v
 
4

Take your best shot at all that, then click here.

When you applied the trig identities you should have gotten 


   32u  -  32 (sin(u)cos(u))(cos2(u) - sin2(u))  -


                          128
                             
                           3


 (1 - (v/4)2)3/2  +



                          8u  +  8 sin(u)cos(u)  +  C

Of course the 32u and the 8u combined to make 40u. Now substitute all the sines and cosines (and  arcsin(v/4)  for the bare occurrence of the u): 


   40 arcsin




v
 
4




  +  32


 

v
 
4



  __________
 √1 - (v/4)2


 

    v2
1 −   
     8



  -

                          128
                             
                           3

 


1 -


 

v
 
4



2





3/2
     +



                          8


 

v
 
4



  __________
 √1 - (v/4)2  +  C

And with a little dividing out factors and gathering terms you can get it to this: 


   40 arcsin




v
 
4




  -


  

3     1
  v −  
2     4


 v3




  _______
 √16 - v2



  -


  2
   
  3


 (16 - v2)3/2  +  C

Step 8: Substitute back to x. Remember that  v = x - 1,  so plug that in everywhere you see the v, then simplify as best you can, and you will be done. Click here when you are through it all.

Without bothering to multiply out that cubed binomial (which you shouldn't have to do unless your instructor is a stickler for such things), you get: 


   40 arcsin




x - 1
     
  4




  -


  

3
  (x - 1)  -
2

  1
   
  4


 (x - 1)3




  ____________
 √15 + 2x - x2



  -


                                        2
                                         
                                        3


 (15 + 2x - x2)3/2  +  C

About checking this -- I can't tell you with a straight face that this thing isn't just the devil to take the derivative of and not make any mistakes doing so. So let me suggest a method that lets you do your checking in smaller increments. First observe that the last step of doing this integral -- that is substituting  x - 1  for v -- was pretty easy, and the likelihood that you made a mistake there is small. Remember that you originally made the v substitution and then did an integral in v to get your v solution. So you could check that -- that is take the derivative of the v solution and see if you can get it back to the v integral. This is still rather messy, but it is not as hard as taking the derivative of the above x expression. It also does not check whether or not you completed the square correctly, so you would have to go back and check that separately.

You can also break the checking of your v solution into even smaller pieces by looking at the sum of integrals you got when you were able to break up the v integral. In the development here I kept the intermediate results from each of the three summands separate until the end. If you look at this version of the v solution, you can see that the only terms we have combined so far are the 32u with the 8u to get  40 arcsin(v/4).  It's pretty easy to separate those two out again and move the  8 arcsin(v/4)  to the third line (look at the expression in u that we got just prior to that). That way each line corresponds to just one of the three integrals in the sum. So you can take the derivatives of each of the lines separately and try to work them back to their respective summands. If you find a discrepancy this way, you will know where to go looking for the mistake.

And remember one more thing. Even if you did take the derivative of the x expression and did all the exacting work it takes to munge it back to the original integrand, that is still considerably less work than you have already done integrating this monster in the first place. So even if you checked it that way, your time would be well spent.

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