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MA101 Calculus - Outline Notes: Integration - Further Substitutions
In a previous section we revised a few basic substitutions. In this section we shall explore
some of the substitutions discussed in Section 6.2 of Adams. We shall concentrate on classes
of integrals rather than "tricks" for individual integrals.
TYPE I Square roots of quadratics
These can be simplified by means of trigonometric substitutions, and there are several
examples in Adams, on pages 353-355. A prerequisite is that you should be fluent in the
algebraic technique of completing the square
Example 1
Evaluate I 

4 x 2  16 x  52 dx.
Step1 Make the coefficient of x 2 equal to 1.

x 2  4 x  13 dx.

( x  2) 2  9 dx.

( x  2) 2  32 dx.
I 2
Step 2 Complete the square.
I 2
Step 3 Write the constant as a square
I 2
Step 4 Substitute u  x  2 i.e. replace the bracket involving x with a new variable u.
I 2

u 2  32 du.
Step 5 Make an appropriate trigonometric substitution which converts the sum of two
squares into a single square term, thereby getting rid of the square root.
In this case (page 354) let u  3 tan , so du  3 sec 2  d. We then obtain
I 2


32 tan 2   32  3 sec 2  d  2  3  3 sec 3  d.
We have used the trigonometric identity tan   1  sec .
2
2
Step 6 Evaluate the trigonometric integral (this is example 3 on page 348), giving
I  18 12 sec  tan   12 ln sec   tan   C.


Step 7 Express the answer in terms of u using the right-angled triangle technique
u 2  32

3
u
The diagram is constructed from the substitution u  3 tan , and then Pythagoras' Theorem.
From this we can work out any other trigonometric function in terms of u.
u 2  32
. So we can express the answer in terms of u as
3
In this case we need sec  
I 9
u 2  32 u
  9 ln
3
3
u 2  32 u
  C.
3
3
Step 8 Replace u in terms of x and simplify where possible.
I  ( x  2) 2  32 .( x  2)  9 ln ( x  2) 2  32  ( x  2)  9 ln 3  C
 ( x  2) x 2  4 x  13  9 ln x 2  4 x  13  ( x  2)  K .
An alternative to some trigonometric substitutions is to use hyperbolic substitutions. These
are sometimes more straightforward. We shall look at the types dealt with on pages 354-355
by this method as an alternative to the trigonometric method illustrated there.
TYPE II
Integrals involving
x2  a2 .
In fact this is not really distinct from TYPE I, and they can be done using x  a tan .
However we shall use a hyperbolic substitution. Again we look for an appropriate substitution
which converts the sum of two squares into a single square term, thereby getting rid of the
square root, and we use cosh 2 t  1  sinh 2 t. So to evaluate I 
substitution x  a sinh t ; dx  a cosh t dt. This then gives
I  a2


x 2  a 2 dx we use the

sinh 2 t  1  cosh t dt  a 2 cosh 2 t dt.
We can evaluate the latter integral by using a hyperbolic identity, as it is similar to the method
for integrating sin 2 t.
We use the identity cosh 2t  2 cosh 2 t  1, giving cosh 2 t  12 (1  cosh 2t ). We then have


a2
a2
t  12 sinh 2t 
(t  sinh t cosh t )
2
2
a2 
a 2 
x 2 
2 
1  x  x

sinh   
1 2 .
 t  sinh t 1  sinh t  
 2 
2 
a a

a 

I  a2

1 (1  cosh 2t ) dt
2

This looks different from the answer we obtained in example 1 above, until we remember that
there are logarithmic equivalents of the inverse hyperbolic functions. We can also evaluate the
final integral as follows, obtaining an answer in agreement with the calculations above.

 e t  e t
cosh t dt  

2


1  e 2t
e  2t
 2t 
4  2
2
2

2

 dt 


1
4
 e
2t

 2  e 2t dt
 t 1  e 2t  e  2t
  
 2 4
2


 t 1
   sinh 2t
 2 4

x2  a2
TYPE III Integrals involving
(x2  a 2 )
These are dealt with on page 355 using the substitution x  a sec , but we can also make use
of a hyperbolic substitution, derived from the identity cosh 2 t  sinh 2 t  1.
So to find


x 2  a 2 dx we let x  a cosh t ; dx  a sinh t dt. Substituting gives
x 2  a 2 dx 


a 2 (cosh 2 t  1)  a sinh t dt  a 2 sinh 2 t dt , which we can evaluate using

similar methods to those for a 2 cosh 2 t dt which we did in TYPE II above.
TYPE IV Integrals involving
a2  x2
(a 2  x 2 )
These are dealt with on page 352-353 using the substitution x  a sin , and in this case there
is not a corresponding hyperbolic substitution.
The integral
x
2

a 2  x 2 dx is done as Example 2 on page353, so instead we shall evaluate
a 2  x 2 dx. Using the substitution x  a sin ; dx  a cos  d gives
x

a 2  x 2 dx  a 2 sin 2  a 2 (1  sin 2 )  a cos  d
2

 a 4 sin 2  cos 2  d 

a4
8
a4

8

a4
a4
sin 2 2 d 
(1  cos 4) d
4
8




sin 4 
a4




C

  sin  cos (1  2 sin 2 )  C


4 
8

x
x
x 2 
x 2 
sin 1   
1 2 1 2 2   C

a a
a 
a 


a4
 x 1
sin 1    x a 2  x 2 a 2  2 x 2  C.
8
a 8
TYPE V Integrals involving
ax  b .
These are dealt with using the substitution u 2  ax  b, as in the examples on page 357.
TYPE VI Rational functions of cos  and sin .

These are discussed on page 358. The substitution x  tan 2 reduces such integrals to
integrals involving rational functions of x, which we shall deal with in the next section of
these notes. We need various identities in connection with this substitution, and these are
derived on page 358 using trigonometric identities. We shall derive them here using a mixture
of trigonometric identities and a right-angled triangle.
1 x2
x

2
1
x  tan
In the right angled triangle we have
2 .
dx  12 sec 2
So
2 d,
giving
2 dx  2dx2 using the right-angled triangle. From the triangle we can also see
1 x

the value of sin 2  in terms of x, and so we have
2
1 x2
cos   2 cos 2 2   1 

1

,
2
2
d  2 cos 2
1 x
sin   2 sin
2 cos2   2
1
1 x2

1 x
x
1 x2

2x
1 x2
Example 2
Find

d
.
2  sin 
We use the substitution x  tan
2  and so the identities above give
d
1
2dx

2x
1 x2
2
1 x2
1
1

dx  
dx
3
1 2
algebra  x 2  x  1
complete


x


the square
2
4
 2  sin   
 x  1  
 2x  1
2
2 

tan 
C 
tan 1 
  C
3
3


3
3


4
4


 2 tan  2   1 
2
  C.

tan 1 
3
3


1
1
.