Inverse Trigonometry Integrals
Derivative and Antiderivatives that Deal
with the Inverse Trigonometry
We know the following to be true:
d
u'
arcsin u 
2
dx
1 u
This shows the following indefinite integral:

u'
1 u
2
dx  arcsin u  C
But, what if the value in the square root is not 1? Can we still use
this antiderivative?
Derivative and Antiderivatives that Deal
with the Inverse Trigonometry
Investigate the following:
1
1
d
a
u
'
u'
a u'
u
a
arcsin a 
 


u
2
dx
a
u2
u 2
1

a 1  2 
1  
2
a2
a
a
u'
a2  u2
This shows the following indefinite integral:

du
a u
2
2
 arcsin  C
u
a
Now investigate arccos(x).
Derivative and Antiderivatives that Deal
with the Inverse Trigonometry
Investigate the following:
1
1

a

u
'
d
u '
a u'
a
u
 
arccos a 


u
2
a
u2
dx
u 2
1

a 1  2 
1  
2
a
a2
a
u '
a2  u2
This shows the following indefinite integral:

du
a 2 u 2
  arccos  C
u
a
This only differs by a minus sign from arcsin(x). It will be
omitted from our list.
Derivative and Antiderivatives that Deal
with the Inverse Trigonometry
Investigate the following:
1
a
u'
d
u
arctan a 

2
u
dx
 a  1
1
a
2
u' a
u'


a

2
2
2
2
u
u

a
2 1 a
a
This shows the following indefinite integral:

du
a2 u 2
 tan
1
a
1 u
a
C
Arccot(x) will only differs by a minus sign from this.
It will be omitted from our list.
Integrals Involving Inverse
Trigonometric Functions
If u(x) is a differentiable function and a > 0, then



du
a 2 u 2
du
2
2
a u
du
u u 2 a2
 sin
1 u
a
C
1 u
a
C
 tan
1
a
 sec
1
a
1 u
a
C
Arcsec(x) is challenging to prove due to sign changes.
Example 1
6dx
Evaluate:
 x 2  16
Rewrite the
integral to
resemble the
Rule
a4
dx
6dx
6

 x 2  16  x 2  42
Use the Rule
1
x
 6  arctan    C
4
4
3
x
 arctan    C
2
4
Example 2
Evaluate:
Rewrite the
integral to
resemble the
Rule
x
x
dx
u  2x du  2  dx
4 x  25
2
dx
4 x 2  25


1
2
1
2

dx
x
 2x
2
 52 a  5
Still missing
things…
du
u u 2  52
du
u u 2  52
1 1  u 
1 1  2 x 
 sec    C  sec 
C
5
5
5
 5 
Example 2
2x  5
Evaluate:
 x 2  2 x  2 dx
2x  5
 x 2  2 x  2 dx 
Manipulate the
Numerator so
it contains the
derivative of
the base.
2x  2  7
x
2
 2x  2
dx
2x  2
7
 2
dx   2
dx
x  2x  2
x  2x  2
Complete the
square.
1
2
 ln x  2 x  2  C  7 
dx
2
 x  1  1
 ln x 2  2 x  2  7 arctan  x  1  C
1980 AB Free Response 4