Download 8 Extra Trigonometric Substitutions

Ch.8 Extra Topic:
Trigonometric Substitutions
Monticello (Thomas Jefferson’s home), Alexandria, VA
Photo by Vickie Kelly, 2004
Greg Kelly, Hanford High School, Richland, Washington
We can use right triangles and the pythagorean
theorem to simplify some problems.
1

dx
4 x
2
These are
in the same
form.
a4 xx
ln sec  tan   C
4 x
x
 C
2
2
2
ln
x

2sec2  d
 2sec
 sec d
22
2
a2
4  x2
sec 
2
2sec  4  x
2
x
tan  
2
2 tan   x
2sec 2  d  dx

We can use right triangles and the pythagorean
theorem to simplify some problems.
1

dx
4  x2
2sec2  d
 2sec
4  x2  x
C
2
ln
ln
4  x 2  x  ln 2  C
 sec d
ln sec  tan   C
ln
4  x2 x
 C
2
2
This is a constant.
ln
4  x2  x  C

This method is called Trigonometric Substitution.
a x
2
a x ,
2
If the integral contains
2
2
x

we use the triangle at right.
a
If we need
a2  x2 , we
move a to the hypotenuse.
If we need
move x to the hypotenuse.
x
a
x

a x
2
x2  a2 , we
2
x2  a2

a

2

x 2 dx
9  x2
3

9  x2
9sin 2   3cos d

3cos
1  cos 2
9
d
2
9
1  cos 2 d

2
9
9 1
   sin 2  C
2
2 2
x
x sin   3
3sin  x
9  x2
cos 
3
3cos  9  x2
3cos d  dx
x
sin  
3
double angle
x
  sin 1
formula
3
9 1 x 9
sin
  2sin  cos   C
2
3 4
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3

2

x 2 dx
9  x2
3

9  x2
x
x sin   3
3sin  x
9  x2
cos 
3
3cos  9  x2
3cos d  dx
x
sin  
3
double angle
x
  sin 1
formula
3
9 1  x  x
sin   
9  x2  C
2
3 2
9 1 x 9
sin
  2sin  cos   C
2
3 4
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3

5


dx
2 x  x2
2x  x 2
dx
1   x  1

2
Let u  x 1


 x2  2 x


du  dx
 x2  2x  1  1
du
1 u2
cos  d
 cos
 d
We can get 2x  x 2 into the necessary
form by completing the square.
  x  1  1
2
1   x  1
2
1

sin
u C
  C
 sin 1  x 1  C
1
u

1 u2
1 u2
 1 u2
cos 
1
sin   u cos d  du

6
dx
 4x2  4x  2
4 x2  4 x  2
4 x2  4 x  1  1
dx
  2 x  1
Complete the square:
2
1
1 du
2  u2 1
1 sec2  d
2  sec2 
 2 x  1
2
1
Let u  2x 1
du  2 dx
1
du  dx
2
1
1
1
1

tan
u C



C
d

2
2
2
1

tan 1  2 x  1  C
2
u2 1
u

1
tan  u
sec2  d  du
sec  u 2  1
sec2   u 2  1

Here are a couple of shortcuts that are result from
Trigonometric Substitution:
du
1
1 u
 u 2  a 2  a tan a  C

du
u
 sin
C
a
a2  u 2
1
These are on your list of
formulas. They are not
really new.
p