Calculus II
MAT 146
Derivatives and Integrals Involving Inverse Trig Functions
As part of a first course in Calculus, you may or may not have learned about
derivatives and integrals of inverse trigonometric functions. These notes are
intended to review these concepts as we come to rely on this information in
second-semester calculus.
Derivatives of Inverse Trig Functions
One way to translate into words the meaning of the function y = sin(x) is as
follows, based on right-triangle trigonometry:
“y represents the sine ratio for an angle of measure x.”
We write the inverse of this function as y = sin !1 (x) and translate this
equation into words as
“y is the angle measure whose sine ratio is x.”
The right triangle shown here illustrates this, for the sine of angle y is
opposite/hypotenuse = x/1 = x.
For use in the derivation that
follows, we note that in the
triangle here, ?2 + x2 = 12.
Solving for the unknown, we
have ? = 12 ! x 2 . From that we
deduce that
1 ! x2
cos(y) =
= 1! x 2 .
1
1
y
x
?
A
Our goal is to determine the
derivative of y = sin !1 (x) . We start that process by noting that an equivalent
form for y = sin !1 (x) is sin( y) = x . We now apply implicit differentiation to
this equivalent form:
sin( y) = x
d
d
(sin( y)) = ( x )
dx
dx
dy
! cos( y ) "
=1
dx
dy
1
!
=
dx cos( y)
!
dy
1
=
dx
1 # x2
d
1
! sin #1 (x) =
2
dx
1# x
We now have the derivative of the inverse sine function and, because
differentiation and integration are inverse operations, we have an integral as
well:
1
!1
dx
=
sin
(x) + C
"
1! x2
Through a similar derivation as just presented, we can show that
!
as well as
d
1
!1
and "
cos!1 (x) = !
dx = cos!1(x) + C
2
2
dx
1! x
1! x
d
1
1
!1
!1
tan (x) =
2 and "
2 dx = tan (x) + C .
dx
1+ x
1+ x
Likewise, we can generate derivatives for three other inverse trig functions:
d
1
d
1
d
1
!1
cot !1 (x) = !
sec !1 (x) =
,
csc
(x)
=
!
2 ,
dx
dx
1+ x
x x 2 !1 dx
x x 2 !1
We use these derivatives and integrals to solve additional problems
involving inverse trig functions. Here are some examples, first involving
derivatives and then involving integrals.
(
!1
)
2
Example #1: Determine the derivative of y = cos (x) .
We use derivative rules we already know—in this case, the chain rule—as
well as the new information about derivatives of inverse trig functions:
dy d
=
cos!1 (x)
dx dx
(
(
)
(
)
)
= 2 cos!1 (x) "
= 2 cos!1 (x) "
=
!2
1! x
2
2
d
cos!1 (x)
dx
!1
(
)
1! x2
(cos!1(x))
(
))
(
Example 2: Determine the derivative of y = tan !1 5x 2 ! 3 .
(
))
(
y = tan !1 5x 2 ! 3
dy d
=
tan !1 5x 2 ! 3
dx dx
1
d
=
"
5x 2 ! 3
2
2
1 + 5x ! 3 dx
(
(
=
))
(
)
1
(
2
)
2
)
2
1 + 5x ! 3
=
(
)
" 10x
10 x
(
1 + 5x 2 ! 3
Example 3: Determine the derivative of y = (arcsin (cos( x ))) .
Here we recognize that arcsin(u) = sin-1(u). We also use the trig identity
sin2(x) + cos2(x) = 1.
y = (arcsin (cos( x )))
dy d
=
(arcsin(cos(x )))
dx dx
d
=
sin !1 (cos(x ))
dx
1
=
" (!sin (x ))
2
1 ! (cos( x ))
(
)
!1
" (sin( x ))
sin( x )
Depending on the value of x, this last expression will simplify to either 1 or
–1. On the interval -π < x < π, the ratio is –1 when x < 0 and is 1 when x > 0.
=
3
Example 4: Evaluate "
2
dx .
1! x
We can factor 3 from the integral expression to create an integral that can be
evaluated by inspection:
"
3
1! x2
Example 5: Evaluate "
dx = 3 "
8
1 ! 4x 2
The integral looks similar to
1
1! x2
dx = 3sin !1 (x) + C
dx .
1
dx , so we attempt a substitution that
1! x2
will mirror that format. If we let u = 2x, so that u2 = 4x2, we get du = 2 dx, or
4 du = 8 dx. We now have
8
1
dx
=
4
du
"
"
2
2
1 ! 4x
1!u
This is similar to Example 4 and we proceed with a solution by inspection:
1
4"
du = 4 sin !1 (u) + C = 4 sin !1 (2x ) + C .
2
1!u
1
Example 6: Evaluate
"
0
"
dx
1! x2
.
This is similar to previous problems with the added task of evaluating a
definite integral.
1
"
dx
1
= sin !1( x ) 0 = sin !1(1) ! sin !1 (0 ) =
#
#
!0=
2
2
1! x2
Remember that the function y = sin !1 (x) represents angle measures (or,
# " "&
more generally, real numbers) over the interval %! , ( .
$ 2 2'
0