Math documents

Integrals of inverse functions


On this page: Integral of: sin-1 x cos-1 x tan-1 x cot-1 x sec-1 x csc-1 x
  Integral of: sinh-1 x cosh-1 x tanh-1 x coth-1 x sech-1 x csch-1 x

The Inverse Trigonometric and Hyperbolic functions can be integrated using Integration by Parts as shown for each function.
 

Integral of sin-1 x

sin-1 x dx  =  x sin-1 x + 
 
1-x2
 + C     [ Verify with the derivative calculator ]

Derivation:

 U = sin-1
 
dV = dx
dU =
 
1
 
1-x2
 dx 
 
 V = x

sin-1 x dx  =  x sin-1 - 
x
 
1-x2
 dx

Integrate the  VdU  integral using a U-substitution:

U2 = 1-x2
2UdU = -2xdx
From 2UdU,  xdx = -UdU
 
x
 
1-x2
 dx = 
-UdU
U
 =  -dU = -U = -
 
1-x2
 

The final result is

sin-1 x dx  =  x sin-1 x + 
 
1-x2
 + C

Integral of cos-1 x

cos-1 x dx  =  x cos-1 - 
 
1-x2
 + C     [ Verify with the derivative calculator ]

Derivation:

 U = cos-1
 
dV = dx
dU =
 
-1
 
1-x2
 dx 
 
 V = x

cos-1 x dx  =  x cos-1 x + 
x
 
1-x2
 dx

Integrate the  VdU  integral using a U-substitution (done differently from sin-1 x just by choice):

 U = 1-x2
dU = -2xdx
From dU,  xdx = 
-dU
2
 
 
x
 
1-x2
 dx = 
-1
2
 
U
 dU = - 
1
2
U-1/2dU = - 
1
2
·
U1/2
(
1
2
)
 = -
 
U
 = -
 
1-x2
 

The final result is

cos-1 x dx  =  x cos-1 - 
 
1-x2
 + C

Integral of tan-1 x

tan-1 x dx  =  x tan-1 - 
1
2
 ln 
x2+1
 + C     [ Verify with the derivative calculator ]

Derivation:

 U = tan-1
 
dV = dx
dU = 
1
x2+1
 dx 
 
 V = x

tan-1 x dx  =  x tan-1 - 
x
x2+1
 dx

Integrate the  VdU  integral using a U-substitution:

 U = x2+1
dU = 2xdx
From dU,  xdx = 
dU
2
 
 
x
x2+1
 dx = 
1
2U
 dU = 
1
2
ln 
U
 = 
1
2
ln 
x2+1
 

The final result is

tan-1 x dx  =  x tan-1 - 
1
2
 ln 
x2+1
 + C

Integral of cot-1 x

cot-1 x dx  =  x cot-1 x + 
1
2
 ln 
x2+1
 + C     [ Verify with the derivative calculator ]

Derivation:

 U = cot-1
 
dV = dx
dU = 
-1
x2+1
 dx 
 
 V = x

cot-1 x dx  =  x cot-1 x + 
x
x2+1
 dx

Integrate the  VdU  integral using a U-substitution:

 U = x2+1
dU = 2xdx
From dU,  xdx = 
dU
2
 
 
x
x2+1
 dx = 
1
2U
 dU = 
1
2
ln 
U
 = 
1
2
ln 
x2+1
 

The final result is

cot-1 x dx  =  x cot-1 x + 
1
2
 ln 
x2+1
 + C

Integral of sec-1 x

sec-1 x dx  =   {
x sec-1 - ln 
x+
 
x2-1
 + C     [ Verify with the derivative calculator ]
 
x sec-1 - cosh-1 x
 

Derivation:

 U = sec-1
 
dV = dx
dU =
 
1
x
 
x2-1
 dx 
 
 V = x

sec-1 x dx  =  x sec-1 - 
1
 
x2-1
 dx

Integrate the  VdU  integral using a trigonometric substitution:

 x = secθ
dx = secθ tanθ dθ
 
1
 
x2-1
 dx =  secθ dθ = ln 
secθ + tanθ
 = ln 
secθ+
 
sec2θ-1
 = ln 
x+
 
x2-1
 

This logarithmic expression is the definition of cosh-1 x, so the shorthand cosh-1 x notation may also be used.

The final result is

sec-1 x dx  =   {
x sec-1 - ln 
x+
 
x2-1
 + C
 
x sec-1 - cosh-1 x
 

Integral of csc-1 x

csc-1 x dx  =   {
x csc-1 x + ln 
x+
 
x2-1
 + C     [ Verify with the derivative calculator ]
 
x csc-1 x + cosh-1 x
 

Derivation:

 U = csc-1
 
dV = dx
dU =
 
-1
x
 
x2-1
 dx 
 
 V = x

csc-1 x dx  =  x csc-1 x + 
1
 
x2-1
 dx

Integrate the  VdU  integral using a trigonometric substitution:

 x = secθ
dx = secθ tanθ dθ
 
1
 
x2-1
 dx =  secθ dθ = ln 
secθ + tanθ
 = ln 
secθ+
 
sec2θ-1
 = ln 
x+
 
x2-1
 

This logarithmic expression is the definition of cosh-1 x, so the shorthand cosh-1 x notation may also be used.

The final result is

csc-1 x dx  =   {
x csc-1 x + ln 
x+
 
x2-1
 + C
 
x csc-1 x + cosh-1 x
 

Integral of sinh-1 x

sinh-1 x dx  =  x sinh-1 - 
 
1+x2
 + C     [ Verify with the derivative calculator ]

Derivation:

 U = sinh-1
 
dV = dx
dU =
 
1
 
1+x2
 dx 
 
 V = x

sinh-1 x dx  =  x sinh-1 - 
x
 
1+x2
 dx

Integrate the  VdU  integral using a U-substitution:

U2 = 1+x2
2UdU = 2xdx
From 2UdU,  xdx = UdU
 
x
 
1+x2
 dx = 
UdU
U
 =  dU = U = 
 
1+x2
 

The final result is

sinh-1 x dx  =  x sinh-1 - 
 
1+x2
 + C

Integral of cosh-1 x

cosh-1 x dx  =  x cosh-1 - 
 
x2-1
 + C     [ Verify with the derivative calculator ]

Derivation:

 U = cosh-1
 
dV = dx
dU =
 
1
 
x2-1
 dx 
 
 V = x

cosh-1 x dx  =  x cosh-1 - 
x
 
x2-1
 dx

Integrate the  VdU  integral using a U-substitution:

U2 = x2-1
2UdU = 2xdx
From 2UdU,  xdx = UdU
 
x
 
x2-1
 dx = 
UdU
U
 =  dU = U = 
 
x2-1
 

The final result is

cosh-1 x dx  =  x cosh-1 - 
 
x2-1
 + C

Integral of tanh-1 x

tanh-1 x dx  =  x tanh-1 x + 
1
2
 ln 
1-x2
 + C     [ Verify with the derivative calculator ]

Derivation:

 U 
= tanh-1 x  
dV = dx
dU* 
1
1-x2
 dx 
 V = x

* Using this form of denominator because |x| < 1

tanh-1 x dx  =  x tanh-1 - 
x
1-x2
 dx

Integrate the  VdU  integral using a U-substitution:

U2 = 1-x2
2UdU = -2xdx
From 2UdU,  xdx = -UdU
 
x
1-x2
 dx = 
-UdU
U2
 = -
dU
U
 = -ln U = -ln
 
1-x2
 

The final result is

tanh-1 x dx  =  x tanh-1 x + 
1
2
 ln 
1-x2
 + C

Integral of coth-1 x

coth-1 x dx  =  x coth-1 x + 
1
2
 ln 
x2-1
 + C     [ Verify with the derivative calculator ]

Derivation:

 U 
= coth-1 x  
dV = dx
dU* 
-1
x2-1
 dx 
 V = x

* Using this form of denominator because |x| > 1

coth-1 x dx  =  x coth-1 x + 
x
x2-1
 dx

Integrate the  VdU  integral using a U-substitution:

U2 = x2-1
2UdU = 2xdx
From 2UdU,  xdx = UdU
 
x
x2-1
 dx = 
UdU
U2
 = 
dU
U
 = ln U = ln
 
x2-1
 

The final result is

coth-1 x dx  =  x coth-1 x + 
1
2
 ln 
x2-1
 + C

Integral of sech-1 x

sech-1 x dx  =  x sech-1 x + sin-1 x + C     [ Verify with the derivative calculator ]

Derivation:

 U = sech-1
 
dV = dx
dU =
 
-1
x
 
1-x2
 dx 
 
 V = x

sech-1 x dx  =  x sech-1 x + 
1
 
1-x2
 dx

Integrate the  VdU  integral using a trigonometric substitution:

 x = sinθ
dx = cosθ dθ

1
 
1-x2
 dx = 
cosθ dθ
cosθ
 = θ = sin-1 x

The  VdU  integral can also be integrated using a hyperbolic substitution, but the result is more complicated and requires extra work to obtain a simplified result:

 x  = sech U
dx = -sech U tanh U dU

1
 
1-x2
 dx 
-sech U dU = - tan-1(sinh U) = - tan-1
 
1
sech2 U
 -1
 
 
 
= - tan-1
 
1
x2
 - 1
 = - tan-1
 
1 - x2
x2
 = - tan-1 (
 
1-x2
x
)  
 
 
= - 
π
2
 + cot-1 (
 
1-x2
x
)  = - 
π
2
 + tan-1 (
x
 
1-x2
)  = - 
π
2
 + sin-1 x

The  -π/2  will be combined with the constant and will disappear.

The final result is

sech-1 x dx  =  x sech-1 x + sin-1 x + C

Integral of csch-1 x

csch-1 x dx  =   {
x csch-1 x + ln 
x+
 
x2+1
 + C     [ Verify with the derivative calculator ]
 
x csch-1 x + sinh-1 x
 

Derivation:

 U = csch-1
 
dV = dx
dU =
 
-1
x
 
x2+1
 dx 
 
 V = x

csch-1 x dx  =  x csch-1 x + 
1
 
x2+1
 dx

Integrate the  VdU  integral using a trigonometic substitution:

 x = tanθ
dx = sec2θ dθ
 
1
 
x2+1
 dx =  secθ dθ = ln 
secθ + tanθ
 = ln 
secθ+
 
sec2θ-1
 = ln 
x+
 
x2+1
 

Alternatively, the  VdU  integral can be integrated using a hyperbolic substitution:

 x = sinh U
dx = cosh U dU

1
 
x2+1
 dx = 
cosh U dU
cosh U
 dx = U = sinh-1 x = ln 
x+
 
x2+1
 

This logarithmic expression is the definition of sinh-1 x, so the shorthand sinh-1 x notation may also be used.

The final result is

csch-1 x dx  =   {
x csch-1 x + ln 
x+
 
x2+1
 + C
 
x csch-1 x + sinh-1 x
 


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