Section 11: Methods of Integration


© 2001 by Karl Hahn 		KCT logo

11.5 Trigonometric Substitution

In the days before telephone service became widespread, a gentleman named Mr. Nakamura emigrated from Japan to San Francisco, bringing his aging mother along with him. They shared an apartment together, and Mr. Nakamura, who was an adept businessman, soon prospered in his new home. It wasn't long before he was expanding his business to take advantage of opportunities in Los Angeles as well. But to secure those opportunities Mr. Nakamura had to make a trip to Los Angeles, leaving his poor mother home alone. The thought of his mother fending for herself in a land that still seemed alien to her troubled Mr. Nakamura more and more each day he was away. On the fifth day he decided he needed to send his mother a message of reassurance that he would be returning soon. So he went to the Western Union office to send her a telegram.

Mr. Nakamura had written his message to his mother out in Japanese characters. But the telegraph operator had no idea how to encode Japanese characters into dots and dashes. So he suggested to Mr. Nakamura that he translate the message into English.

"Oh no," said Mr. Nakamura. "That would never do. My mother reads only Japanese such as you see here. A message in English might as well be in Persian as far as she's concerned."

"We could have somebody translate it back into Japanese at the other end," suggested the Western Union man.

Mr. Nakamura thought for a moment, then said, "No that won't do either. The words wouldn't be the same after two translations. My mother needs to read my exact words, for I have written my message to her in Haiku, and only the exact words will preserve the poetic form."

Mr. Nakamura stood silent for several minutes thinking. Then he said, "Instead, I shall spell out each of the Japanese words in my message phonetically, using your alphabet. You send that over the wire. Your Japanese translator at the other end will certainly recognize the words by their phonetic spellings, so he shall be able to transcribe them back into Japanese characters that my mother can read. Yes, and she will read my exact words and know how deeply her only son is missing her."

Imagine that the methods you've learned so far for finding the integral of a function are like passing a message over the telegraph wires. Now imagine that you have an integral on which none of the tools you've learned so far work. It is like Mr. Nakamura's message. It is spelled in an alphabet that renders your methods useless. The solution to the problem could very well be to transcribe your function into an alphabet that your methods do work on. You could then apply your method, then translate it back into its original alphabet when you're done.

Here is an example of doing just that: 

   

    dx
  ______                                                      eq. 11.5-1
 √1 - x2
Your first instinct might be to substitute  u = 1 - x2.  But if you tried that you would soon find that it didn't work. When you took the derivative of u, you'd find  du = -2x dx,  and there is no place to make that substitution in this integral.

In the section on simple substitution you learned how to make substitutions based upon finding some function,  u(x),  that you could slip in to simplify your integral. What you haven't tried yet is turning this method on its head. That is, we imagine that our independent variable, x, is actually a function of the substitution variable, u. Suppose we tried that substitution blindly in the example above. Then we'd have 

  x  =  f(u)                                                      eq. 11.5-2a
for some mysterious and as yet unknown function, f. Applying implicit differentiation to this we'd also have 
               du
   1  =  f'(u)                                                    eq. 11.5-2b
               dx
or more appropriately for the purpose of substitution 
  dx  =  f'(u) du                                                 eq. 11.5-2c
Now we put all that into the integral, substituting  f(u)  for x and  f'(u) du  for dx: 

   

  f'(u) du
  _________                                                   eq. 11.5-3
 √1 - f2(u)
If only we could find a function,  f(u),  such that 

              _________
   f'(u)  =  √1 - f2(u)                                           eq. 11.5-4
it would make life wonderful, because you could then substitute the right-hand side of equation 11.5-4 for  f'(u)  into the integral in 11.5-3, and all the hard stuff would cancel. You'd be left having only to integrate 

   


 du                                                           eq. 11.5-5
which is a piece of cake. Well I just happen to know of such a function. Remember from trigonometry that 

   sin2(u) + cos2(u)  =  1                                        eq. 11.5-6a
or equivalently 
               ___________
   cos(u)  =  √1 - sin2(u)                                        eq. 11.5-6b
But we already know that cosine is the derivative of the sine. So the function,  f(u) = sin(u),  exactly satisfies our fondest desires for  f(u)  as expressed in equation 11.5-4. Now let's do the whole substitution exercise over again using 

   x  =  sin(u)

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


     eq. 11.5-7

So using those substitutions for x and dx, the integral becomes 

   


    dx
  ______  =
 √1 - x2


  

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


  



 du  =  u + C            eq. 11.5-8
At this point we have, in a manner of speaking, transcribed the Japanese message into our English alphabet, sent it over the wire, and all we have left to do is transcribe it back to Japanese characters so that Mr. Nakamura's mother can read it. The original substitution was  x = sin(u).  So what do we put in for u? We have to use inverse sine or 

   arcsin(x)  =  u                                                eq. 11.5-9
and we find that when we substitute back we get 

   

    dx
  ______  =  arcsin(x) + C                                    eq. 11.5-10
 √1 - x2

You might have guessed this integral from our table of common derivatives. You will see the relationship between arcsin and our integrand near the bottom of that table. All we did was a fancy parlor trick to solve an integral that we already knew. But this same attack can integrate functions that are not so obvious. For example, let's integrate 

   

  ______
 √1 - x2 dx                                                   eq. 11.5-11

We'll try the same substitution as before of 

   x  =  sin(u)

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


   

  ______
 √1 - x2 dx  =


  

  ___________
 √1 - sin2(u)


  ___________
 √1 - sin2(u) du  =


                           


 (1 - sin2(u)) du  =


  


 cos2(u) du    eq. 11.5-12







And now you're left having to integrate the square of the cosine. But this
one isn't so bad once you apply a
trig identity
to it (which illustrates why you should sharpen up your command of trig identities
for this unit. You will need them time and again).
The integral becomes a sum that you can break up:








   




 cos2(u) du  =




  1
   
  2



 




 (1 + cos(2u)) du  =









                       1
                        
                       2



 








 du  +




  




 cos(2u) du









                 eq. 11.5-13a








Both of these are integrals you know how to do (and if you need
to you can still use the table
and equation 11.2-9).






   1
    
   2



 








 du  +




  




 cos(2u) du








     1       1
  =    u  +    sin(2u)  +  C         eq. 11.5-13b
     2       4





To translate back to our original variable, x, you will have to use
another trig identity.






   1
    
   2



 








 du  +




  




 cos(2u) du








     1       1
  =    u  +    sin(u)cos(u)  +  C    eq. 11.5-13c
     2       2





Now we can substitute back. The original substitution was
 sin(u) = x.  So running that in reverse on the bare
u in equation 11.5-13c, it becomes  arcsin(x). 
The  sin(u)  becomes x.
And we can use this to put  cos(u) 
in terms of  sin(u)  to get an x expression
for it as well. Hence








   



  ______
 √1 - x2 dx  =




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








                    1
                      arcsin(x)  +
                    2



  1    ______
    x √1 - x2  +  C             eq. 11.5-13d
  2







I'd like you to take the time right now to take the derivative of this result
so that you can see how intimately the terms above conspire to give you back
the original integrand (which is a nice way of saying that you'll have to do
some algebraic manipulations on the immediate derivative
to get all the way back to the original integrand).





As nicely as that works out, we have applied this method only to a very narrow
collection of integrands. Supposing you have a variation on the theme. How would
you integrate






   



    dx
  ______                                                      eq. 11.5-14a
 √9 - x2





The plan here is to do a little algebraic munging to turn this into something that
we can apply a substitution similar to the one we used on the first example.
First, factor 9 out of the expression under the radical. Remember
that when you pull that factor out of the radical, it gets square-rooted.






   1
    
   3



 



     dx
  __________                                                eq. 11.5-14b
 √1 - (x/3)2





The substitution is almost the same, but we have to account for the
factor of 1/3 that now goes with the x





Alternative method to integrate






   



    dx
  ______
 √9 - x2




Skip factoring out the 1/3.

Since  3 = √9,  substitute


   3 sin(u)  =  x



and you still get


   3 cos(u) du  =  dx


Since  3 cos(u) = √9 - x², 
you will find that you end up with the same expression to integrate as if you had
done it the way the main text suggests. Try it (it's really the same method).






              x
   sin(u)  =   
              3



   and   


              dx
cos(u) du  =    
               3





And that means that  3 cos(u) du = dx. 
I'll let you finish this one off. In the end make sure you get







   



    dx
  ______  =  arcsin(x/3) + C
 √9 - x2





                               eq 11.5-15







Let's try to apply this method to an integral that has an additional
complication.







   



      dx
  ___________                                                 eq. 11.5-16
 √3 - 2x - x2





When you see a quadratic under the radical, the first step always is
to complete the square. If there is no coefficient on the x2
term and the x2 is negative, then call the quadratic you have
under the radical,  -q(x).  That makes
 q(x) = x2 + 2x - 3. 

simply take half the middle coefficient of  q(x)  and
add it to x. That will be your first substitution. The middle coefficient for
this  q(x) is 2, hence





   v  =  x + 1



   and   


dv  =  dx



                                 eq. 11.17a





But look what happens when you square v


   v2  =  x2 + 2x + 1                                              eq. 11.5-17b



and then add or subtract whatever it takes from both sides to
make the right-hand side turn into the negative of what's
under the radical:


   v2 - 4  =  x2 + 2x - 3
                                                                  eq. 11.5-17c
   4 - v2  =  3 - 2x - x2



With this substitution, the integrand is now in a form we know how to integrate:






   



      dx
  ___________  =
 √3 - 2x - x2



 



    dv
  ______  =
 √4 - v2



  1
   
  2



 



      dv
  __________
 √1 - (v/2)2                eq. 11.5-18a





Now it is just like the last one we did. You can substitute





              v
   sin(u)  =   
              2



   and   


              dv
cos(u) du  =                             eq. 11.5-18b
               2





to find that






   



    dv
  ______  =
 √4 - v2




  arcsin(v/2) + C                                  eq. 11.5-18c






But you have one more step to do, and that is substitute back to an expression
in x. Remember that the first substitution we made was
 v = x + 1.  Wherever there's a v,
replace it with  x + 1. 








   



      dx
  ___________  =
 √3 - 2x - x2



   



    dv
  ______  =
 √4 - v2









                arcsin(v/2) + C  =  arcsin((x+1)/2) + C           eq. 11.5-19








Exercises



1) Use trigonometric substitution to find the indefinite
integral:






   



  _______
 √25 - x2 dx





See solution to exercise 1






2) Use completing the square followed
by trigonometric substitution to find the indefinite integral:






   



  ____________
 √35 + 2x - x2 dx





Check back here if you need to review how to
integrate the square of the cosine on either of the above two problems.



See solution to exercise 2




3) Use trigonometric substitution to find the indefinite integral:






   



    dx
  _______
 √1 - 9x2






See solution to exercise 3



Scroll down for more on Trigonometric Substitution









More Trig Substitution



© 2001 by Karl Hahn


KCT logo







Trigonometric substitution is a big topic because there are lots of
variations. Only with practice will you gain the skill of knowing
what variation to use on each integrand. Here is an integrand that
you've seen before because it's on the table,
but it is useful to see how we can use a trigonometric substitution on
it.






   



   dx
                                                              eq. 11.5-20
 1 + x2





Here the temptation is to try simple substitution using
 u = 1 + x2, 
but you will find that again you have no place to substitute
the  du = 2x dx  that arises as a result.




So let's ask ourselves, what would the ideal substitution function,
 f(u) = x  have to do in order to
be ideal for this integrand? Taking the derivative implicitly you
have




         du
   f'(u)     =  1
         dx



or



  f'(u) du  =  dx                       eq. 11.5-21






To make all the hard stuff cancel, we want an  f(u)  that
has the following property:


   f'(u)  =  1 + f2(u)                                            eq. 11.5-22



because then the substituted integral will become






   



   dx
         =
 1 + x2



  



 (1 + f2(u)) du
                 =
    1 + f2(u)



  



 du                        eq. 11.5-23





Once again, I just happen to know of such a function. Go to the
table of common derivatives
and look up the derivative of the tangent function. Can you see that if
we make the substitution






   tan(u)  =  x




and



(1 + tan2(u))




 du
     =  1
 dx



or



(1 + tan2(u)) du  =  dx






                                                                  eq. 11.5-24





then all that's left in the substituted integral is






   



   dx
         =
 1 + x2



  




 du  =  u + C                                    eq. 11.5-25a






Then substituting back, we have to apply the inverse function of our original
substitution in order to back-substitute the bare u.






   



   dx
         =
 1 + x2




  arctan(x) + C                                     eq. 11.5-25b






Once again we have gone to some trouble to integrate something we already
knew the integral of. But what happens when you have a more complicated
quadratic in the denominator? Suppose we had






   



       dx
                                                              eq. 11.5-26
 40 + 12x + x2





The first thing to do is to complete the square.
Adding half the middle coefficient to x gives




   v  =  x + 6



and


   v2  =  x2 + 12x + 36                    eq. 11.5-27a





Now what do we add to or subtract from v2 to make it
be equal to the original  q(x)? 


   v2        =  x2  +  12x  +  36
       +  4  =              +   4
                                 
   v2  +  4  =  x2  +  12x  +  40



Taking the derivative of  v = x + 6,  we
find that  dv = dx. 




So the completed square integral is






   



       dx
                =
 40 + 12x + x2



  



   dv
                                          eq. 11.5-28a
 4 + v2





Note that trig substitution using the tangent function will work
only if the denominator has no real roots.
That is, at this point you must have ended up with a denominator in the form
of a positive quantity added to v2.
If you end up with the difference between v2
and a positive quantity, this substitution won't work. We will deal
with that case in a later section.




Now factor out the 4 from the denominator:






   



    dv
          =
  4 + v2



  1
   
  4



 



     dv
                                              eq. 11.5-28b
 1 + (v/2)2





Now we make our trig substitution:




              v
   tan(u)  =   
              2



and




   (1 + tan2(u))




 du     1
     =   
 dx     2



or



2(1 + tan2(u)) du  =  dx






Substituting and integrating:






   1
    
   4



 



     dv
             =
 1 + (v/2)2



  1
   
  2



 



        1
 du  =    u + C                          eq 11.5-29
        2





Now substitute back to v using the inverse function






   1
    
   4



 



     dv
             =
 1 + (v/2)2



  1
    arctan
  2







v
 
2








 + C                             eq. 11.5-30a






And finally substitute back to x using
 v = x + 6. 






   



       dx
                =
 40 + 12x + x2



  1
    arctan
  2







x + 6
     
  2








 + C                        eq. 11.5-30b









A Broader Class of Integrands




The trig substitutions we have explored so far combined with
completing the square
can integrate






   



      dx
            
 b + ax + x2





whenever the quadratic has no real roots (which is the case if
and only if  a2 < 4b)  and






   



      dx
  ___________
 √b - ax - x2



and







  ___________
 √b - ax - x2 dx






whenever the quadratic under the radical does
have real roots (which is the case if and only if
a2 + 4b ≥ 0). 
The truth is that the method still addresses only a narrow class of
functions. In this paragraph, we show how we can still use this method
whenever the integrand is any of the functions listed above times
some polynomial function.




Suppose we needed to integrate






   



 15 - 7x + 3x2
               dx                                             eq. 11.5-31
    1 + x2





The strategy is to do some algebraic munging to put this into a form
that we do know how to integrate. And the algebraic tool that we use
to do this is polynomial long
division.


          3            
   x2 + 1)3x2 - 7x + 15
          3x2 + 0x +  3
                       
               -7x + 12



That is we get a quotient of 3 and a remainder
of  -7x + 12. 

You need to remember
the following very important rule
concerning polynomial long division:
The Quotient added to the Remainder over the Denominator is always
equal to the Numerator over the Denominator.





            R(x)     N(x)
   Q(x)  +        =                                               eq. 11.5-32
            D(x)     D(x)






In the case here, you have
 Q(x) = 3,   
 R(x) = -7x + 12,   
 D(x) = x2 + 1,  and
 N(x) = 3x2 - 7x + 15. 
Hence we can rewrite the integral as






   



 15 - 7x + 3x2
               dx  =
    1 + x2



  







      -7x + 12
3  +          
       1 + x2








 dx  =










                 3




 




 dx  -  7




 



  x dx
         +  12
 1 + x2



 



   dx
                  eq. 11.5-33
 1 + x2





All three of these integrals you know how to do. The first is trivial.
On the second you apply simple substitution.
And on the third you apply trig substitution (or you find it on the
table). We find that






   



 15 - 7x + 3x2
               dx  =
    1 + x2






                7
         3x  -    ln|1 + x2|  +  12 arctan(x)  +  C               eq. 11.5-34
                2







Let's try a more difficult one. Integrate






   



 3 - 2x - 14x2 + x3
                                                              eq. 11.5-35
    85 + 18x + x2





There are several ways you can attack this that will work. I prefer
doing the polynomial long division first (you could complete the
square first, but then you'd have to apply the substitution that
arises from that to the cubic in the numerator).



Step 1: Do the long division. The long division is


                         x  - 32                
         x2 + 18x + 85 ) x3 - 14x2 -   2x +    3
                         x3 + 18x2 +  85x
                                         
                            - 32x2 -  87x +    3
                            - 32x2 - 576x - 2720
                                                
                                     489x + 2723



If the polynomial long division here
confuses you, you should review that algebra topic until you are comfortable
with it. With the quotient of  x - 32 
and the remainder of  489x + 2723 
shown here, and applying
equation 11.5-32, you can break
the original integral up into:






   



 3 - 2x - 14x2 + x3
                     =                                        eq. 11.5-36
    85 + 18x + x2








                          




(x - 32) dx  +




  



  489x + 2723
                dx
 85 + 18x + x2





The left-hand piece is trivial to  integrate:






   




(x - 32) dx  =




  1
   
  2




 x2  -  32x  +  C                            eq. 11.5-37






But to do the other two pieces, you have to go to




Step 2: Complete the square.
When you add
half the middle coefficient to x you get the substitution of




   v  =  x + 9



and


v2  =  x2 + 18x + 81





Clearly to get this back to the original denominator, you have to add 4 to
v2:


   v2 + 4  =  x2 + 18x + 85



and since  v = x + 9,  then  dv = dx. 
But remember that you also have to put the  v = x + 9 
substitution into the numerator of the remaining integral.
To do this you have to convert the
substitution to  v - 9 = x. 
With these v substitutions the remaining integral becomes






   



  489x + 2723
               dx  =
 85 + 18x + x2



  



 489(v - 9) + 2723
                   dv                 eq. 11.5-38a
      4 + v2





Step 3: Simplify the numerator. That is, multiply out the
 489(v - 9)  term and then gather like
terms. You have  489 × -9 = -4401  and
 2723 - 4401 = -1678.  Using these results for the
numerator and separating the sum into two integrals according to equation
11.2-13b, we get that
equation 11.5-38a is the same  as






   



 489v - 1678
             dv  =
    4 + v2




  489




 



  v dv
         -  1678
 4 + v2



 



   dv
                  eq. 11.5-38b
 4 + v2





Step 4: Substitute.
These are integrals you know how to do. The left-hand summand succumbs to
simple substitution. For the one
on the right, divide the 4 out of the denominator to  get







   -1678




 



   dv
         =
 4 + v2



    1678
  -     
      4



 



     dv
                                   eq. 11.5-39
 1  + (v/2)2





Then substitute




              v
   tan(u)  =   
              2



and




   (1 + tan2(u)) du  =




  dv
    
   2



or



2(1 + tan2(u)) du  =  dv






Step 5: Integrate.
When you do all of the above, and integrate, and substitute back (which is
really Step 6), and put together the pieces
from equations
11.5-37,
11.5-38b, and
11.5-39, you get:






   



 3 - 2x - 14x2 + x3
                     =                                        eq. 11.5-40
    85 + 18x + x2








      1
       
      2




 x2  -  32x  +




  489
     
   2




 ln|85 + 18x + x2|  -




  1678
       arctan
    2







x + 9
     
  2








  +  C






If you had to do the above example yourself from beginning to end on
a homework problem or an exam, it would be naive to believe you wouldn't
make any mistakes in arithmetic or carrying the correct signs through
all the algebra. I, myself, will admit to having made a couple of
mistakes the first time I ran through this one. But I caught them because
at the end, I took the derivative of my answer, and it held manifest
clues where I went wrong. I'll agree that taking the derivative of this
answer and manipulating it back to the original integrand
involves lots of fiddly work of putting things over common denominators
and such. But it pays off in the end with a correct answer.


I have just one more trick to show you before we move on to hyperbolic
substitution (which will involve again all the tricks you learned in this section).
Once more we will find a way to algebraically munge something that we don't know how
to integrate into something that we do.
Suppose you wanted to integrate






   



 17 - 2x + 3x2
     ______    dx                                              eq. 11.5-41a
    √1 - x2





You can't very well do polynomial long division on this quotient because
the denominator is not a polynomial. The trick here is to make it a polynomial.
And you do that by multiplying by


    ______
   √1 - x2
    ______  =  1
   √1 - x2



As you know, multiplying by 1 doesn't change the function under the
integral at all, but it does change its form:






   



 17 - 2x + 3x2
     ______    dx  =
    √1 - x2



  



 17 - 2x + 3x2
              
     1 - x2



  ______
 √1 - x2 dx               eq. 11.5-41b






Now we have one factor that is the quotient of polynomials, hence
we can do polynomial long division
on that.


            -3                
   -x2 + 1 ) 3x2  -  2x  +  17
             3x2  +  0x  -   3
                              
                  -  2x  +  20



This gives a quotient of -3 and a remainder of
 -2x + 20. 
According to equation 11.5-32, our integral is the same as






   



 17 - 2x + 3x2
              
     1 - x2



  ______
 √1 - x2 dx  =




  








-3 +




 -2x + 20
         
  1 - x2







  ______
 √1 - x2 dx      eq. 11.5-41c






You can now multiply the radical through using the distributive law and
you have an integral of a sum. When you integrate a sum, you can apply
equation 11.2-13b to make it into a
sum of integrals:






   








-3 +




 -2x + 20
         
  1 - x2







  ______
 √1 - x2 dx  =










               -3





 




  ______
 √1 - x2 dx  -  2





 




   x dx
  ______  +  20
 √1 - x2




 




    dx
  ______    eq. 11.5-41d
 √1 - x2





We have done integrals of each of these three forms before. The middle one
is easily done by simple substitution.
The other two fall to a substitution of  sin(u) = x. 
I will leave the details of the attack to you. At the end you should end up with






   



 17 - 2x + 3x2
     ______    dx  =
    √1 - x2







                 3
               -  
                 2



   ______
 x√1 - x2  +




    ______
  2√1 - x2  +




  37
     arcsin(x)  +  C    eq. 11.5-41e
   2





(Pssst! Click here if you need a reminder on how to integrate
the square of the cosine).



Note: We will be expanding the notion of trigonometric substitution
in a later section. But before you get to that, we need to cover
Partial Fractions first. In the mean time,
all the integrands that you can integrate using the expanded trig substitutions will also yield to
Hyperbolic Substitution, which is the
next section. Some calculus courses do not cover hyperbolic functions. If you
are in that boat, you can consider the treatment of them here to be optional and
skip from here (after you do the exercise below) right to
Partial Fractions. Or you might want to
learn about hyperbolics anyway.





Another Exercise




4) You're going to try a hard one now that will drill you
on most of the stuff you have been learning in this unit as well as
making you stretch to apply your knowledge to a form that is not quite
the same as what you've seen before. Find the indefinite integral of






   




 x2




  ____________
 √15 + 2x - x2 dx






Your recommended strategy is:

    

  1. Complete the square and make a substitution using the
relationship that arises. I used the substitution variable, v, for this, so
if you do the same, your development will look like mine.
Note that you will have to substitute for that leading
x2 term as well, which means turning your substitution around
so that it expresses x in terms of your substitution variable.
    2. 


  2. It behooves you at this point to factor the constant out of the radical
so that under the radical you have something in the form of
 1 - (v/a)2,  where a is
a constant. Pull that factor to outside the integral (remembering that when
it passes from inside the radical to outside the radical, it gets square-rooted).
    3. 


  3. Look at the substituted integral that results and see how you can break it up
into the sum of three separate integrals.
    4. 


  4. Examine each of the three separately and determine which method holds promise
for integrating each piece. Note that for one of them it will not be trig
substitution.
    5. 


  5. Determine the substitutions for each of the pieces. Carefully take the derivatives
of each of the substitution equations to determine how you are going to substitute
the dv in each piece. There is plenty of opportunity for sign errors or errors
in multiplier-constants in this step. Take it slowly and keep your eyes open.
    6. 

  6. Make your substitutions and rewrite the three integrals in your new variables.
    7. 

  7. You will have to apply trig identities to
whip the integrals you have at this point into a shape that you can integrate. One
of them will require more than one trig identity application.
    8. 

  8. Integrate.
    9. 

  9. Before substituting back to v it will be necessary to apply trig
identities again. And again one of the terms will require multiple applications.
    10. 

  10. Substitute the results back to an expression in v.
    11. 

  11. Last step -- substitute back to an expression in x.



Work through as much of this on your own as possible before looking at the
solution. You might get all the way through.
If you can't, go to the solution and follow along as far as you did get.
From there do each step as recommended in the solution before clicking ahead
to see the intermediate result. Then compare your intermediate result with mine.




If you work carefully and diligently, this problem will probably take you over
an hour. If you then check it by taking the derivative of your answer,
which is not trivial either, you will probably find
that you made a mistake somewhere. Tracking down your mistake or mistakes will
take more time. Be patient with yourself. Just for reference, this integral filled
two full pages of my notebook to do and then check, not counting the pages I tore out
and threw away on it.






Return to Table of Contents


Move on to Hyperbolic Substitution



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