Section 8.4b
Trigonometric Substitutions
How do we evaluate this integral?

3
x dx
9 x
2
These trigonometric substitutions allow us to replace
binomials of the form
a x
2
2
a x
2
2
x a
2
2
by single squared terms, and thereby transform a number
of integrals (like the one above) into ones we can evaluate
directly or find in a table of integrals.
Trigonometric Substitutions
The most common substitutions are based on the following
reference triangles:
a x
2
x  a tan 
2
x
a  x  a sec 
2
2

a
With x  a tan  ,
2
2
2
2
2
2
2
a  x  a  a tan   a 1  tan  
2
2
 a sec 
Trigonometric Substitutions
The most common substitutions are based on the following
reference triangles:
x  a sin 
a
x
a  x  a cos 
2
2

a x
With x  a sin  ,
2
2
2
2
2
2
2
a  x  a  a sin   a 1  sin  
2
2
 a cos 
2
2
Trigonometric Substitutions
The most common substitutions are based on the following
reference triangles:
x
x a
2
2
x  a sec 

x  a  a tan 
a
With x  a sec  ,
2
2
2
2
2
2
2
x  a  a sec   a  a  sec   1
2
2
 a tan 
2
2
Trigonometric Substitutions
1. x  a tan  replaces a  x with a sec

2
2
2
2
a  x with a cos 
2
2
2
2
x  a with a tan 
2
2. x  a sin  replaces
3. x  a sec  replaces
2
2
2
Also, we want any substitution to be reversible so we can
change back to the original variable afterward. For example:
x  a tan 
requires
x
  tan  
a
Essentially, keep positives
with any absolute values…
1
with


2
 

2
Trigonometric Substitutions
Evaluate

x  3sin 
dx  3cos  d
2
2
9  x  9 cos 
Set
3
x dx

9 x
 3sin  
3
2
3cos  d
9 cos 
2
27 sin  3cos  d

3cos 
3
 27  sin  d  27  1  cos   sin  d
3
2
 27   sin   sin  cos   d
2
 27 cos   9 cos   C
3
3

32  x 2
x
Trigonometric Substitutions
Evaluate
x  3sin 
dx  3cos  d
2
2
9  x  9 cos 
Set
3
x dx

9 x
2
 27 cos   9 cos   C
3
3
 27
2

9 x
9 x
 9
 3
3

2
9 x 



 C


2 32
 9 9  x
2
3
3
C

32  x 2
x
Trigonometric Substitutions
Evaluate


x  3sec 
dx  3sec  tan  d
2
2
x  9  9 tan 
dx
x 9
2
3sec  tan  d
9 tan 
2
3sec  tan  d
  sec  d

3 tan 
 ln sec   tan   C
Appendix A7, Formula 88 (p.631)
x
x 9
 ln 
C
3
3
2
x
x 3
2

3
2
Trigonometric Substitutions
Evaluate
x
dx
2
x 1
2
sec  d
x  tan 
2
dx  sec  d
2
2
x  1  sec 
sec  d
sec  d



2
2
2
2
tan  sec 
tan 
tan  sec 
2
2
cos  d
cos  d
2


   sin   cos  d
2
2
cos  sin 
sin 
2
   sin    C
1
Trigonometric Substitutions
Evaluate
x
dx
x 1
2
2
   sin    C
x  tan 
2
dx  sec  d
2
2
x  1  sec 
1
  csc   C
1 x

C
x
2
12  x 2
x

1