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TECHNIQUES OF INTEGRATION
7.3
Trigonometric Substitution
In this section, we will learn about:
The various types of trigonometric substitutions.
TRIGONOMETRIC SUBSTITUTION
In finding the area of a circle or an ellipse, an
integral of the form a 2  x 2 dx arises,
where a > 0.
 If it were  x a 2  x 2 dx , the substitution u  a 2  x 2
would be effective.
 However, as it stands,

a 2  x 2 dx is more difficult.
TRIGONOMETRIC SUBSTITUTION
If we change the variable from x to θ by the
substitution x = a sin θ, the identity 1 – sin2θ =
cos2θ lets us lose the root sign.
 This is because:
a 2  x 2  a 2  a 2 sin 2 
 a (1  sin  )
2
 a 2 cos 2 
 a cos 
2
TRIGONOMETRIC SUBSTITUTION
Notice the difference between the substitution
u = a2 – x2 and the substitution x = a sin θ.
 In the first, the new variable is a function of
the old one.
 In the second, the old variable is a function of
the new one.
TRIGONOMETRIC SUBSTITUTION
In general, we can make a substitution of the form
x = g(t) by using the Substitution Rule in reverse.
 To make our calculations simpler, we assume g
has an inverse function, that is, g is one-to-one.
INVERSE SUBSTITUTION
Here, if we replace u by x and x by t in the
Substitution Rule, we obtain:
 f ( x)dx   f ( g (t )) g '(t )dt
 This kind of substitution is called inverse substitution.
INVERSE SUBSTITUTION
We can make the inverse substitution x = asinθ,
provided that it defines a one-to-one function.
 This can be accomplished by restricting θ
to lie in the interval [-π/2, π/2].
TABLE OF TRIGONOMETRIC SUBSTITUTIONS
Here, we list trigonometric substitutions that are effective
for the given radical expressions because of the specified
trigonometric identities.
TABLE OF TRIGONOMETRIC SUBSTITUTIONS
The restriction on θ is imposed to ensure that the function
that defines the substitution is 1-1.
 These are the same intervals used in defining the inverse
functions.
TRIGONOMETRIC SUBSTITUTION
Evaluate

9x
x
2
Example 1
2
dx
 Let x = 3 sin θ, where –π/2 ≤ θ ≤ π/2.
 Then, dx = 3 cos θ dθ and
9  x2  9  9sin 2   9cos2   3 cos   3cos 
 Note that cos θ ≥ 0 because –π/2 ≤ θ ≤ π/2.
TRIGONOMETRIC SUBSTITUTION
Example 1
Thus, the Inverse Substitution Rule gives:

9  x2
3cos 
dx  
3cos  d
2
2
x
9sin 
2
cos 

d
2
sin 
  cot  d
2
  (csc2   1) d
  cot     C
TRIGONOMETRIC SUBSTITUTION
Example 1
As this is an indefinite integral, we must return to
the original variable x.
This can be done in either of two ways.
One, we can use trigonometric identities to express
cot θ in terms of sin θ = x/3.
TRIGONOMETRIC SUBSTITUTION
Example 1
Two, we can draw a diagram, where θ is
interpreted as an angle of a right triangle.
TRIGONOMETRIC SUBSTITUTION
Example 1
Since sin θ = x/3, we label the opposite side and the
hypotenuse as having lengths x and 3.
TRIGONOMETRIC SUBSTITUTION
Example 1
Then, the Pythagorean Theorem gives the length of
the adjacent side as:
9 x
2
TRIGONOMETRIC SUBSTITUTION
Example 1
So, we can simply read the value of cot θ from the
figure:
cot   9  x
 Although θ > 0 here,
this expression for
cot θ is valid even
when θ < 0.
2
x
TRIGONOMETRIC SUBSTITUTION
Example 1
As sin θ = x/3, we have θ = sin-1(x/3).
Hence,

9 x
9 x
1  x 
dx  
 sin    C
2
2
x
x
3
2
2
TRIGONOMETRIC SUBSTITUTION
Example 2
Find the area enclosed by the ellipse
2
2
x
y


1
2
2
a
b
TRIGONOMETRIC SUBSTITUTION
Example 2
Solving the equation of the ellipse for y, we get
y
x
a x
 1 2 
2
2
b
a
a
2
2
2
or
b 2
2
y
a x
a
2
TRIGONOMETRIC SUBSTITUTION
Example 2
As the ellipse is symmetric with respect to both
axes, the total area A is four times the area in the
first quadrant.
TRIGONOMETRIC SUBSTITUTION
Example 2
The part of the ellipse in the first quadrant is given
by the function
b 2
2
y
a x
a
 Hence,
1
4
A
a
0
b 2
a  x 2 dx
a
0 xa
TRIGONOMETRIC SUBSTITUTION
Example 2
To evaluate this integral, we substitute x = a sin θ.
Then, dx = a cos θ dθ.
To change the limits of integration, we note that:
 When x = 0, sin θ = 0; so θ = 0
 When x = a, sin θ = 1; so θ = π/2
TRIGONOMETRIC SUBSTITUTION
Example 2
Also, since 0 ≤ θ ≤ π/2,
a  x  a  a sin 
2
2
2
2
 a cos 
2
 a cos 
 a cos 
2
2
TRIGONOMETRIC SUBSTITUTION
Example 2
Therefore,
b a 2
b  /2
2
A  4  a  x dx  4  a cos   a cos  d
a 0
a 0
 4ab 
 /2
0
cos  d  4ab 
2
 /2
0
 /2
 2ab[  sin 2 ]0
1
2


 2ab   0  0    ab
2

1
2
(1  cos 2 ) d
TRIGONOMETRIC SUBSTITUTION
Example 2
We have shown that the area of an ellipse with
semiaxes a and b is πab.
 In particular, taking a = b = r, we have proved
the famous formula that the area of a circle with
radius r is πr2.
TRIGONOMETRIC SUBSTITUTION
Note
The integral in Example 2 was a definite integral.
So, we changed the limits of integration, and did
not have to convert back to the original variable x.
TRIGONOMETRIC SUBSTITUTION
Find
x
1
2
x 4
2
Example 3
dx
 Let x = 2 tan θ, –π/2 < θ < π/2.
 Then, dx = 2 sec2 θ dθ and
x 2  4  4(tan 2   1)
 4sec 2 
 2 sec 
 2sec 
TRIGONOMETRIC SUBSTITUTION
Example 3
Thus, we have:
x
2sec  d

2
2
4 tan   2sec 
x 4
1 sec 
 
d
2
4 tan 
dx
2
2
TRIGONOMETRIC SUBSTITUTION
Example 3
To evaluate this trigonometric integral, we put
everything in terms of sin θ and cos θ:
sec 
1 cos  cos 

 2 
2
2
tan  cos  sin  sin 
2
TRIGONOMETRIC SUBSTITUTION
Example 3
Therefore, making the substitution u = sin θ, we
have:
x
1 cos 
  2 d
x 2  4 4 sin 
1 du 1  1 
  2   C
4 u
4 u 
1
csc 

C  
C
4sin 
4
dx
2
TRIGONOMETRIC SUBSTITUTION
Example 3
We use the figure to
determine that:
csc  x  4 / x
2
 Hence,
x
x2  4

C
2
4x
x 4
dx
2
TRIGONOMETRIC SUBSTITUTION
Find

x
x 4
2
Example 4
dx
 It would be possible to use the trigonometric substitution
x = 2 tan θ (as in Example 3).
TRIGONOMETRIC SUBSTITUTION
Example 4
However, the direct substitution u = x2 + 4
is simpler. This is because, then, du = 2x dx and

x
1 du
dx  
2
2
u
x 4
 u C
 x2  4  C
TRIGONOMETRIC SUBSTITUTION
Note
Example 4 illustrates the fact that, even when
trigonometric substitutions are possible, they may
not give the easiest solution.
 You should look for a simpler method first.
TRIGONOMETRIC SUBSTITUTION
Evaluate

where a > 0.
dx
x a
2
2
Example 5
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
We let x = a sec θ, where 0 < θ < π/2 or
π < θ < π/2.
Then, dx = a sec θ tan θ dθ and
x  a  a (sec   1)
2
2
2
2
 a tan 
2
2
 a tan   a tan 
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
Therefore,

a sec  tan 

d
2
2
a tan 
x a
dx
  sec  d
 ln sec   tan   C
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
The triangle in the figure gives:
tan   x  a / a
2
2
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 1
So, we have:

x
x a
 ln 
C
2
2
a
a
x a
dx
2
2
 ln x  x  a  ln a  C
2
2
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Sol. 1 (For. 1)
Writing C1 = C – ln a, we have:

dx
x a
2
 ln x  x  a  C1
2
2
2
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 2
For x > 0, the hyperbolic substitution x = a cosh t
can also be used.
 Using the identity cosh2y – sinh2y = 1, we have:
x 2  a 2  a 2 (cosh 2 t  1)
 a sinh t
2
 a sinh t
2
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Solution 2
Since dx = a sinh t dt, we obtain:

dx
a sinh t dt

2
2
a sinh t
x a
  dt
t C
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Sol. 2 (For. 2)
Since cosh t = x/a, we have t = cosh-1(x/a) and

x
 cosh    C
2
2
a
x a
dx
1
TRIGONOMETRIC SUBSTITUTION
E. g. 5—Sol. 2 (For. 2)
Although Formulas 1 and 2 look quite different,
they are actually equivalent by Formula 4 in
Section 3.11
TRIGONOMETRIC SUBSTITUTION
Note
As Example 5 illustrates, hyperbolic substitutions
can be used instead of trigonometric substitutions,
and sometimes they lead to simpler answers.
 However, we usually use trigonometric substitutions,
because trigonometric identities are more familiar
than hyperbolic identities.
TRIGONOMETRIC SUBSTITUTION
Find

3 3/2
0
Example 6
3
x
dx
2
3/ 2
(4 x  9)
 First, we note that (4 x 2  9)3/ 2  ( 4 x 2  9)3
 So, trigonometric substitution is appropriate.
TRIGONOMETRIC SUBSTITUTION
Example 6
4 x  9 is not quite one of the expressions in the
2
table of trigonometric substitutions.
 However, it becomes one if we make
the preliminary substitution u = 2x.
TRIGONOMETRIC SUBSTITUTION
Example 6
When we combine this with the tangent substitution,
we have x
 32 tan  .
This gives dx  32 sec 2  d and
4 x  9  9 tan   9
2
2
 3sec 
TRIGONOMETRIC SUBSTITUTION
Example 6
When x = 0, tan θ = 0; so θ = 0.
When x = 3 3 / 2 , tan θ = 3; so θ = π/3.
TRIGONOMETRIC SUBSTITUTION

3 3/2
0
3
tan  3 2
2 sec  d
3
27 sec 
 / 3 27
x
dx  
2
3/ 2
0
(4 x  9)
Example 6
3
8
tan 
 
d
0
sec 
3
 / 3 sin 
 163 
d
2
0
cos 
2
 / 3 1  cos 
 163 
sin  d
2
0
cos 
3
16
 /3
3
TRIGONOMETRIC SUBSTITUTION
Example 6
Now, we substitute u = cos θ so that du = - sin θ dθ.
 When θ = 0, u = 1.
 When θ = π/3, u = ½.
TRIGONOMETRIC SUBSTITUTION
Example 6
Therefore,

3 3/2
0

3
x
dx
2
3/ 2
(4 x  9)

1/ 2
3
16 1
1/ 2
1 u
2
3
du

(1

u
) du
16 1
2
u
2
1/ 2

3
16
1

3
3
1


u



2

(1

1)


16  2
 32
 u 
1
TRIGONOMETRIC SUBSTITUTION
Evaluate

x
3  2x  x
2
Example 7
dx
 We can transform the integrand into a function
for which trigonometric substitution is appropriate,
by first completing the square under the root sign:
3  2 x  x 2  3  ( x 2  2 x)
 3  1  ( x 2  2 x  1)
 4  ( x  1) 2
TRIGONOMETRIC SUBSTITUTION
Example 7
This suggests we make the substitution
u = x + 1.
 Then, du = dx and x = u – 1.
 So,

x
3  2 x  x2
dx  
u 1
4  u2
du
TRIGONOMETRIC SUBSTITUTION
Example 7
We now substitute u  2sin  .
This gives
du  2 cos  d and
4  u  2cos
2
TRIGONOMETRIC SUBSTITUTION
So,

Example 7
2sin   1
dx  
2 cos  d
2
2 cos 
3  2x  x
x
  (2sin   1) d
 2 cos     C
u
  4  u  sin    C
2
2
1  x  1 
  3  2 x  x  sin 
C
 2 
2
1
TRIGONOMETRIC SUBSTITUTION
The figure shows the graphs of the integrand in
Example 7 and its indefinite integral (with C = 0).
 Which is which?