Equivalent infinitesimal
edit
When two variables
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
lim
β
α
=
1
{\displaystyle \textstyle \lim {\frac {\beta }{\alpha }}=1}
equivalent infinitesimal (equiv.
α
∼
β
{\displaystyle \alpha \sim \beta }
Moreover, if variables
α
′
{\displaystyle \alpha '}
β
′
{\displaystyle \beta '}
α
∼
α
′
{\displaystyle \alpha \sim \alpha '}
β
∼
β
′
{\displaystyle \beta \sim \beta '}
lim
β
α
=
lim
β
′
α
′
{\displaystyle \lim {\frac {\beta }{\alpha }}=\lim {\frac {\beta '}{\alpha '}}}
Here is a brief proof:
Suppose there are two equivalent infinitesimals
α
∼
α
′
{\displaystyle \alpha \sim \alpha '}
β
∼
β
′
{\displaystyle \beta \sim \beta '}
lim
β
α
=
lim
β
β
′
α
′
β
′
α
′
α
=
lim
β
β
′
lim
α
′
α
lim
β
′
α
′
=
lim
β
′
α
′
{\displaystyle \lim {\frac {\beta }{\alpha }}=\lim {\frac {\beta \beta '\alpha '}{\beta '\alpha '\alpha }}=\lim {\frac {\beta }{\beta '}}\lim {\frac {\alpha '}{\alpha }}\lim {\frac {\beta '}{\alpha '}}=\lim {\frac {\beta '}{\alpha '}}}
For the evaluation of the indeterminate form
0
/
0
{\displaystyle 0/0}
infinitesimals (e.g.,
x
∼
sin
x
{\displaystyle x\sim \sin x}
x becomes closer to zero):[1]
x
∼
sin
x
,
{\displaystyle x\sim \sin x,}
x
∼
arcsin
x
,
{\displaystyle x\sim \arcsin x,}
x
∼
sinh
x
,
{\displaystyle x\sim \sinh x,}
x
∼
tan
x
,
{\displaystyle x\sim \tan x,}
x
∼
arctan
x
,
{\displaystyle x\sim \arctan x,}
x
∼
ln
(
1
+
x
)
,
{\displaystyle x\sim \ln(1+x),}
1
−
cos
x
∼
x
2
2
,
{\displaystyle 1-\cos x\sim {\frac {x^{2}}{2}},}
cosh
x
−
1
∼
x
2
2
,
{\displaystyle \cosh x-1\sim {\frac {x^{2}}{2}},}
a
x
−
1
∼
x
ln
a
,
{\displaystyle a^{x}-1\sim x\ln a,}
e
x
−
1
∼
x
,
{\displaystyle e^{x}-1\sim x,}
(
1
+
x
)
a
−
1
∼
a
x
.
{\displaystyle (1+x)^{a}-1\sim ax.}
For example:
lim
x
→
0
1
x
3
[
(
2
+
cos
x
3
)
x
−
1
]
=
lim
x
→
0
e
x
ln
2
+
cos
x
3
−
1
x
3
=
lim
x
→
0
1
x
2
ln
2
+
cos
x
3
=
lim
x
→
0
1
x
2
ln
(
cos
x
−
1
3
+
1
)
=
lim
x
→
0
cos
x
−
1
3
x
2
=
lim
x
→
0
−
x
2
6
x
2
=
−
1
6
{\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {1}{x^{3}}}\left[\left({\frac {2+\cos x}{3}}\right)^{x}-1\right]&=\lim _{x\to 0}{\frac {e^{x\ln {\frac {2+\cos x}{3}}}-1}{x^{3}}}\\&=\lim _{x\to 0}{\frac {1}{x^{2}}}\ln {\frac {2+\cos x}{3}}\\&=\lim _{x\to 0}{\frac {1}{x^{2}}}\ln \left({\frac {\cos x-1}{3}}+1\right)\\&=\lim _{x\to 0}{\frac {\cos x-1}{3x^{2}}}\\&=\lim _{x\to 0}-{\frac {x^{2}}{6x^{2}}}\\&=-{\frac {1}{6}}\end{aligned}}}
In the 2nd equality,
e
y
−
1
∼
y
{\displaystyle e^{y}-1\sim y}
y
=
x
ln
2
+
cos
x
3
{\displaystyle y=x\ln {2+\cos x \over 3}}
y become closer to 0 is used, and
y
∼
ln
(
1
+
y
)
{\displaystyle y\sim \ln {(1+y)}}
y
=
cos
x
−
1
3
{\displaystyle y={{\cos x-1} \over 3}}
1
−
cos
x
∼
x
2
2
{\displaystyle 1-\cos x\sim {x^{2} \over 2}}
Differentiation rules
edit
Unless otherwise stated, all functions are functions of real numbers (R ) that return real values; although more generally, the formulae below apply wherever they are well defined [2] [3] complex numbers (C ) .[4]
Constant term rule
edit
For any value of
c
{\displaystyle c}
c
∈
R
{\displaystyle c\in \mathbb {R} }
f
(
x
)
{\displaystyle f(x)}
f
(
x
)
=
c
{\displaystyle f(x)=c}
d
f
d
x
=
0
{\displaystyle {\frac {df}{dx}}=0}
[5]
Let
c
∈
R
{\displaystyle c\in \mathbb {R} }
f
(
x
)
=
c
{\displaystyle f(x)=c}
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
(
c
)
−
(
c
)
h
=
lim
h
→
0
0
h
=
lim
h
→
0
0
=
0
{\displaystyle {\begin{aligned}f'(x)&=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\\&=\lim _{h\to 0}{\frac {(c)-(c)}{h}}\\&=\lim _{h\to 0}{\frac {0}{h}}\\&=\lim _{h\to 0}0\\&=0\end{aligned}}}
This shows that the derivative of any constant function is 0.
Intuitive (geometric) explanation
edit
The derivative of the function at a point is the slope of the line tangent to the curve at the point. Slope of the constant function is zero, because the tangent line to the constant function is horizontal and it's angle is zero.
In other words, the value of the constant function, y, will not change as the value of x increases or decreases.
At each point, the derivative is the slope of a line that is tangent to the curve at that point. Note: the derivative at point A is positive where green and dash–dot, negative where red and dashed, and zero where black and solid. Differentiation is linear
edit
For any functions
f
{\displaystyle f}
g
{\displaystyle g}
a
{\displaystyle a}
b
{\displaystyle b}
h
(
x
)
=
a
f
(
x
)
+
b
g
(
x
)
{\displaystyle h(x)=af(x)+bg(x)}
x
{\displaystyle x}
h
′
(
x
)
=
a
f
′
(
x
)
+
b
g
′
(
x
)
.
{\displaystyle h'(x)=af'(x)+bg'(x).}
In Leibniz's notation this is written as:
d
(
a
f
+
b
g
)
d
x
=
a
d
f
d
x
+
b
d
g
d
x
.
{\displaystyle {\frac {d(af+bg)}{dx}}=a{\frac {df}{dx}}+b{\frac {dg}{dx}}.}
Special cases include:
The constant factor rule
(
a
f
)
′
=
a
f
′
{\displaystyle (af)'=af'}
The sum rule
(
f
+
g
)
′
=
f
′
+
g
′
{\displaystyle (f+g)'=f'+g'}
The difference rule
(
f
−
g
)
′
=
f
′
−
g
′
.
{\displaystyle (f-g)'=f'-g'.}
The product rule
edit
For the functions
f
{\displaystyle f}
g
{\displaystyle g}
h
(
x
)
=
f
(
x
)
g
(
x
)
{\displaystyle h(x)=f(x)g(x)}
x
{\displaystyle x}
h
′
(
x
)
=
(
f
g
)
′
(
x
)
=
f
′
(
x
)
g
(
x
)
+
f
(
x
)
g
′
(
x
)
.
{\displaystyle h'(x)=(fg)'(x)=f'(x)g(x)+f(x)g'(x).}
d
(
f
g
)
d
x
=
g
d
f
d
x
+
f
d
g
d
x
.
{\displaystyle {\frac {d(fg)}{dx}}=g{\frac {df}{dx}}+f{\frac {dg}{dx}}.}
The chain rule
edit
The derivative of the function
h
(
x
)
=
f
(
g
(
x
)
)
{\displaystyle h(x)=f(g(x))}
h
′
(
x
)
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
.
{\displaystyle h'(x)=f'(g(x))\cdot g'(x).}
In Leibniz's notation, this is written as:
d
d
x
h
(
x
)
=
d
d
z
f
(
z
)
|
z
=
g
(
x
)
⋅
d
d
x
g
(
x
)
,
{\displaystyle {\frac {d}{dx}}h(x)=\left.{\frac {d}{dz}}f(z)\right|_{z=g(x)}\cdot {\frac {d}{dx}}g(x),}
d
h
(
x
)
d
x
=
d
f
(
g
(
x
)
)
d
g
(
x
)
⋅
d
g
(
x
)
d
x
.
{\displaystyle {\frac {dh(x)}{dx}}={\frac {df(g(x))}{dg(x)}}\cdot {\frac {dg(x)}{dx}}.}
Focusing on the notion of maps, and the differential being a map
D
{\displaystyle {\text{D}}}
[
D
(
f
∘
g
)
]
x
=
[
D
f
]
g
(
x
)
⋅
[
D
g
]
x
.
{\displaystyle [{\text{D}}(f\circ g)]_{x}=[{\text{D}}f]_{g(x)}\cdot [{\text{D}}g]_{x}\,.}
The inverse function rule
edit
If the function f has an inverse function g , meaning that
g
(
f
(
x
)
)
=
x
{\displaystyle g(f(x))=x}
f
(
g
(
y
)
)
=
y
,
{\displaystyle f(g(y))=y,}
g
′
=
1
f
′
∘
g
.
{\displaystyle g'={\frac {1}{f'\circ g}}.}
In Leibniz notation, this is written as
d
x
d
y
=
1
d
y
d
x
.
{\displaystyle {\frac {dx}{dy}}={\frac {1}{\frac {dy}{dx}}}.}
The polynomial or elementary power rule
edit
If
f
(
x
)
=
x
r
{\displaystyle f(x)=x^{r}}
r
≠
0
,
{\displaystyle r\neq 0,}
f
′
(
x
)
=
r
x
r
−
1
.
{\displaystyle f'(x)=rx^{r-1}.}
When
r
=
1
,
{\displaystyle r=1,}
f
(
x
)
=
x
,
{\displaystyle f(x)=x,}
f
′
(
x
)
=
1.
{\displaystyle f'(x)=1.}
Combining the power rule with the sum and constant multiple rules permits the computation of the derivative of any polynomial.
The reciprocal rule
edit
The derivative of
h
(
x
)
=
1
f
(
x
)
{\displaystyle h(x)={\frac {1}{f(x)}}}
f
h
′
(
x
)
=
−
f
′
(
x
)
(
f
(
x
)
)
2
{\displaystyle h'(x)=-{\frac {f'(x)}{(f(x))^{2}}}}
f In Leibniz's notation, this is written
d
(
1
/
f
)
d
x
=
−
1
f
2
d
f
d
x
.
{\displaystyle {\frac {d(1/f)}{dx}}=-{\frac {1}{f^{2}}}{\frac {df}{dx}}.}
The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule.
The quotient rule
edit
If f g
(
f
g
)
′
=
f
′
g
−
g
′
f
g
2
{\displaystyle \left({\frac {f}{g}}\right)'={\frac {f'g-g'f}{g^{2}}}\quad }
g This can be derived from the product rule and the reciprocal rule.
Generalized power rule
edit
The elementary power rule generalizes considerably. The most general power rule is the functional power rule : for any functions f g
(
f
g
)
′
=
(
e
g
ln
f
)
′
=
f
g
(
f
′
g
f
+
g
′
ln
f
)
,
{\displaystyle (f^{g})'=\left(e^{g\ln f}\right)'=f^{g}\left(f'{g \over f}+g'\ln f\right),\quad }
wherever both sides are well defined.
Special cases
If
f
(
x
)
=
x
a
{\textstyle f(x)=x^{a}\!}
f
′
(
x
)
=
a
x
a
−
1
{\textstyle f'(x)=ax^{a-1}}
a x
The reciprocal rule may be derived as the special case where
g
(
x
)
=
−
1
{\textstyle g(x)=-1\!}
d
d
x
(
c
a
x
)
=
a
c
a
x
ln
c
,
c
>
0
{\displaystyle {\frac {d}{dx}}\left(c^{ax}\right)={ac^{ax}\ln c},\qquad c>0}
the equation above is true for all c , but the derivative for
c
<
0
{\textstyle c<0}
d
d
x
(
e
a
x
)
=
a
e
a
x
{\displaystyle {\frac {d}{dx}}\left(e^{ax}\right)=ae^{ax}}
d
d
x
(
log
c
x
)
=
1
x
ln
c
,
c
>
1
{\displaystyle {\frac {d}{dx}}\left(\log _{c}x\right)={1 \over x\ln c},\qquad c>1}
the equation above is also true for all c
c
<
0
{\textstyle c<0\!}
d
d
x
(
ln
x
)
=
1
x
,
x
>
0.
{\displaystyle {\frac {d}{dx}}\left(\ln x\right)={1 \over x},\qquad x>0.}
d
d
x
(
ln
|
x
|
)
=
1
x
,
x
≠
0.
{\displaystyle {\frac {d}{dx}}\left(\ln |x|\right)={1 \over x},\qquad x\neq 0.}
d
d
x
(
W
(
x
)
)
=
1
x
+
e
W
(
x
)
,
x
>
−
1
e
.
{\displaystyle {\frac {d}{dx}}\left(W(x)\right)={1 \over {x+e^{W(x)}}},\qquad x>-{1 \over e}.\qquad }
W
(
x
)
{\displaystyle W(x)}
Lambert W function
d
d
x
(
x
x
)
=
x
x
(
1
+
ln
x
)
.
{\displaystyle {\frac {d}{dx}}\left(x^{x}\right)=x^{x}(1+\ln x).}
d
d
x
(
f
(
x
)
g
(
x
)
)
=
g
(
x
)
f
(
x
)
g
(
x
)
−
1
d
f
d
x
+
f
(
x
)
g
(
x
)
ln
(
f
(
x
)
)
d
g
d
x
,
if
f
(
x
)
>
0
,
and if
d
f
d
x
and
d
g
d
x
exist.
{\displaystyle {\frac {d}{dx}}\left(f(x)^{g(x)}\right)=g(x)f(x)^{g(x)-1}{\frac {df}{dx}}+f(x)^{g(x)}\ln {(f(x))}{\frac {dg}{dx}},\qquad {\text{if }}f(x)>0,{\text{ and if }}{\frac {df}{dx}}{\text{ and }}{\frac {dg}{dx}}{\text{ exist.}}}
d
d
x
(
f
1
(
x
)
f
2
(
x
)
(
.
.
.
)
f
n
(
x
)
)
=
[
∑
k
=
1
n
∂
∂
x
k
(
f
1
(
x
1
)
f
2
(
x
2
)
(
.
.
.
)
f
n
(
x
n
)
)
]
|
x
1
=
x
2
=
.
.
.
=
x
n
=
x
,
if
f
i
<
n
(
x
)
>
0
and
{\displaystyle {\frac {d}{dx}}\left(f_{1}(x)^{f_{2}(x)^{\left(...\right)^{f_{n}(x)}}}\right)=\left[\sum \limits _{k=1}^{n}{\frac {\partial }{\partial x_{k}}}\left(f_{1}(x_{1})^{f_{2}(x_{2})^{\left(...\right)^{f_{n}(x_{n})}}}\right)\right]{\biggr \vert }_{x_{1}=x_{2}=...=x_{n}=x},{\text{ if }}f_{i<n}(x)>0{\text{ and }}}
d
f
i
d
x
exists.
{\displaystyle {\frac {df_{i}}{dx}}{\text{ exists. }}}
Logarithmic derivatives
edit
The logarithmic derivative is another way of stating the rule for differentiating the logarithm of a function (using the chain rule):
(
ln
f
)
′
=
f
′
f
{\displaystyle (\ln f)'={\frac {f'}{f}}\quad }
f Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative.[citation needed
Logarithms can be used to remove exponents, convert products into sums, and convert division into subtraction — each of which may lead to a simplified expression for taking derivatives.
d
d
x
sin
x
=
cos
x
{\displaystyle {\frac {d}{dx}}\sin x=\cos x}
d
d
x
arcsin
x
=
1
1
−
x
2
{\displaystyle {\frac {d}{dx}}\arcsin x={\frac {1}{\sqrt {1-x^{2}}}}}
d
d
x
cos
x
=
−
sin
x
{\displaystyle {\frac {d}{dx}}\cos x=-\sin x}
d
d
x
arccos
x
=
−
1
1
−
x
2
{\displaystyle {\frac {d}{dx}}\arccos x=-{\frac {1}{\sqrt {1-x^{2}}}}}
d
d
x
tan
x
=
sec
2
x
=
1
cos
2
x
=
1
+
tan
2
x
{\displaystyle {\frac {d}{dx}}\tan x=\sec ^{2}x={\frac {1}{\cos ^{2}x}}=1+\tan ^{2}x}
d
d
x
arctan
x
=
1
1
+
x
2
{\displaystyle {\frac {d}{dx}}\arctan x={\frac {1}{1+x^{2}}}}
d
d
x
csc
x
=
−
csc
x
cot
x
{\displaystyle {\frac {d}{dx}}\csc x=-\csc {x}\cot {x}}
d
d
x
arccsc
x
=
−
1
|
x
|
x
2
−
1
{\displaystyle {\frac {d}{dx}}\operatorname {arccsc} x=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}}
d
d
x
sec
x
=
sec
x
tan
x
{\displaystyle {\frac {d}{dx}}\sec x=\sec {x}\tan {x}}
d
d
x
arcsec
x
=
1
|
x
|
x
2
−
1
{\displaystyle {\frac {d}{dx}}\operatorname {arcsec} x={\frac {1}{|x|{\sqrt {x^{2}-1}}}}}
d
d
x
cot
x
=
−
csc
2
x
=
−
1
sin
2
x
=
−
1
−
cot
2
x
{\displaystyle {\frac {d}{dx}}\cot x=-\csc ^{2}x=-{\frac {1}{\sin ^{2}x}}=-1-\cot ^{2}x}
d
d
x
arccot
x
=
−
1
1
+
x
2
{\displaystyle {\frac {d}{dx}}\operatorname {arccot} x=-{1 \over 1+x^{2}}}
The derivatives in the table above are for when the range of the inverse secant is
[
0
,
π
]
{\displaystyle [0,\pi ]\!}
[
−
π
2
,
π
2
]
.
{\displaystyle \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right].}
It is common to additionally define an inverse tangent function with two arguments ,
arctan
(
y
,
x
)
.
{\displaystyle \arctan(y,x).}
[
−
π
,
π
]
{\displaystyle [-\pi ,\pi ]}
(
x
,
y
)
.
{\displaystyle (x,y).}
x
>
0
{\displaystyle x>0}
arctan
(
y
,
x
>
0
)
=
arctan
(
y
/
x
)
.
{\displaystyle \arctan(y,x>0)=\arctan(y/x).}
∂
arctan
(
y
,
x
)
∂
y
=
x
x
2
+
y
2
and
∂
arctan
(
y
,
x
)
∂
x
=
−
y
x
2
+
y
2
.
{\displaystyle {\frac {\partial \arctan(y,x)}{\partial y}}={\frac {x}{x^{2}+y^{2}}}\qquad {\text{and}}\qquad {\frac {\partial \arctan(y,x)}{\partial x}}={\frac {-y}{x^{2}+y^{2}}}.}
Maclaurin Series
edit
Several important Maclaurin series expansions follow. All these expansions are valid for complex arguments x .
Exponential function
edit
The exponential function e x n + 1 The exponential function
e
x
{\displaystyle e^{x}}
e
e
x
=
∑
n
=
0
∞
x
n
n
!
=
1
+
x
+
x
2
2
!
+
x
3
3
!
+
⋯
.
{\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots .}
x .
The exponential generating function of the Bell numbers is the exponential function of the predecessor of the exponential function:
exp
(
exp
x
−
1
)
=
∑
n
=
0
∞
B
n
n
!
x
n
{\displaystyle \exp(\exp {x}-1)=\sum _{n=0}^{\infty }{\frac {B_{n}}{n!}}x^{n}}
Natural logarithm
edit
The natural logarithm (with base e [7]
ln
(
1
−
x
)
=
−
∑
n
=
1
∞
x
n
n
=
−
x
−
x
2
2
−
x
3
3
−
⋯
,
ln
(
1
+
x
)
=
∑
n
=
1
∞
(
−
1
)
n
+
1
x
n
n
=
x
−
x
2
2
+
x
3
3
−
⋯
.
{\displaystyle {\begin{aligned}\ln(1-x)&=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\cdots ,\\\ln(1+x)&=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots .\end{aligned}}}
The last series is known as Mercator series , named after Nicholas Mercator (since it was published in his 1668 treatise Logarithmotechnia ). Both of these series converge for
|
x
|
<
1
{\displaystyle |x|<1}
ln(1 − x ) converges for x = −1ln(1 + x ) converges for x = 1[7]
Geometric series
edit
The geometric series and its derivatives have Maclaurin series
1
1
−
x
=
∑
n
=
0
∞
x
n
1
(
1
−
x
)
2
=
∑
n
=
1
∞
n
x
n
−
1
1
(
1
−
x
)
3
=
∑
n
=
2
∞
(
n
−
1
)
n
2
x
n
−
2
.
{\displaystyle {\begin{aligned}{\frac {1}{1-x}}&=\sum _{n=0}^{\infty }x^{n}\\{\frac {1}{(1-x)^{2}}}&=\sum _{n=1}^{\infty }nx^{n-1}\\{\frac {1}{(1-x)^{3}}}&=\sum _{n=2}^{\infty }{\frac {(n-1)n}{2}}x^{n-2}.\end{aligned}}}
All are convergent for
|
x
|
<
1
{\displaystyle |x|<1}
binomial series given in the next section.
Binomial series
edit
The binomial series is the power series
(
1
+
x
)
α
=
∑
n
=
0
∞
(
α
n
)
x
n
{\displaystyle (1+x)^{\alpha }=\sum _{n=0}^{\infty }{\binom {\alpha }{n}}x^{n}}
whose coefficients are the generalized binomial coefficients
(
α
n
)
=
∏
k
=
1
n
α
−
k
+
1
k
=
α
(
α
−
1
)
⋯
(
α
−
n
+
1
)
n
!
.
{\displaystyle {\binom {\alpha }{n}}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}.}
(If n = 0empty product and has value 1.) It converges for
|
x
|
<
1
{\displaystyle |x|<1}
α .
When α = −1α = 1 / 2 α = −1 / 2 square root function and its inverse :
(
1
+
x
)
1
2
=
1
+
1
2
x
−
1
8
x
2
+
1
16
x
3
−
5
128
x
4
+
7
256
x
5
−
⋯
=
∑
n
=
0
∞
(
−
1
)
n
−
1
(
2
n
)
!
4
n
(
n
!
)
2
(
2
n
−
1
)
x
n
,
(
1
+
x
)
−
1
2
=
1
−
1
2
x
+
3
8
x
2
−
5
16
x
3
+
35
128
x
4
−
63
256
x
5
+
⋯
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
)
!
4
n
(
n
!
)
2
x
n
.
{\displaystyle {\begin{aligned}(1+x)^{\frac {1}{2}}&=1+{\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+{\frac {7}{256}}x^{5}-\cdots &=\sum _{n=0}^{\infty }{\frac {(-1)^{n-1}(2n)!}{4^{n}(n!)^{2}(2n-1)}}x^{n},\\(1+x)^{-{\frac {1}{2}}}&=1-{\frac {1}{2}}x+{\frac {3}{8}}x^{2}-{\frac {5}{16}}x^{3}+{\frac {35}{128}}x^{4}-{\frac {63}{256}}x^{5}+\cdots &=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{4^{n}(n!)^{2}}}x^{n}.\end{aligned}}}
When only the linear term is retained, this simplifies to the binomial approximation .
Trigonometric functions
edit
The usual trigonometric functions and their inverses have the following Maclaurin series:
sin
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
+
1
)
!
x
2
n
+
1
=
x
−
x
3
3
!
+
x
5
5
!
−
⋯
for all
x
cos
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
)
!
x
2
n
=
1
−
x
2
2
!
+
x
4
4
!
−
⋯
for all
x
tan
x
=
∑
n
=
1
∞
B
2
n
(
−
4
)
n
(
1
−
4
n
)
(
2
n
)
!
x
2
n
−
1
=
x
+
x
3
3
+
2
x
5
15
+
⋯
for
|
x
|
<
π
2
sec
x
=
∑
n
=
0
∞
(
−
1
)
n
E
2
n
(
2
n
)
!
x
2
n
=
1
+
x
2
2
+
5
x
4
24
+
⋯
for
|
x
|
<
π
2
arcsin
x
=
∑
n
=
0
∞
(
2
n
)
!
4
n
(
n
!
)
2
(
2
n
+
1
)
x
2
n
+
1
=
x
+
x
3
6
+
3
x
5
40
+
⋯
for
|
x
|
≤
1
arccos
x
=
π
2
−
arcsin
x
=
π
2
−
∑
n
=
0
∞
(
2
n
)
!
4
n
(
n
!
)
2
(
2
n
+
1
)
x
2
n
+
1
=
π
2
−
x
−
x
3
6
−
3
x
5
40
−
⋯
for
|
x
|
≤
1
arctan
x
=
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
x
2
n
+
1
=
x
−
x
3
3
+
x
5
5
−
⋯
for
|
x
|
≤
1
,
x
≠
±
i
{\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}&&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots &&{\text{for all }}x\\[6pt]\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}&&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots &&{\text{for all }}x\\[6pt]\tan x&=\sum _{n=1}^{\infty }{\frac {B_{2n}(-4)^{n}\left(1-4^{n}\right)}{(2n)!}}x^{2n-1}&&=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\sec x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}&&=1+{\frac {x^{2}}{2}}+{\frac {5x^{4}}{24}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\arcsin x&=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+\cdots &&{\text{for }}|x|\leq 1\\[6pt]\arccos x&={\frac {\pi }{2}}-\arcsin x\\&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&={\frac {\pi }{2}}-x-{\frac {x^{3}}{6}}-{\frac {3x^{5}}{40}}-\cdots &&{\text{for }}|x|\leq 1\\[6pt]\arctan x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}&&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots &&{\text{for }}|x|\leq 1,\ x\neq \pm i\end{aligned}}}
All angles are expressed in radians . The numbers Bk tan x are the Bernoulli numbers . The E k sec x are Euler numbers .
Hyperbolic functions
edit
The hyperbolic functions have Maclaurin series closely related to the series for the corresponding trigonometric functions:
sinh
x
=
∑
n
=
0
∞
x
2
n
+
1
(
2
n
+
1
)
!
=
x
+
x
3
3
!
+
x
5
5
!
+
⋯
for all
x
cosh
x
=
∑
n
=
0
∞
x
2
n
(
2
n
)
!
=
1
+
x
2
2
!
+
x
4
4
!
+
⋯
for all
x
tanh
x
=
∑
n
=
1
∞
B
2
n
4
n
(
4
n
−
1
)
(
2
n
)
!
x
2
n
−
1
=
x
−
x
3
3
+
2
x
5
15
−
17
x
7
315
+
⋯
for
|
x
|
<
π
2
arsinh
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
)
!
4
n
(
n
!
)
2
(
2
n
+
1
)
x
2
n
+
1
=
x
−
x
3
6
+
3
x
5
40
−
⋯
for
|
x
|
≤
1
artanh
x
=
∑
n
=
0
∞
x
2
n
+
1
2
n
+
1
=
x
+
x
3
3
+
x
5
5
+
⋯
for
|
x
|
≤
1
,
x
≠
±
1
{\displaystyle {\begin{aligned}\sinh x&=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{(2n+1)!}}&&=x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+\cdots &&{\text{for all }}x\\[6pt]\cosh x&=\sum _{n=0}^{\infty }{\frac {x^{2n}}{(2n)!}}&&=1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+\cdots &&{\text{for all }}x\\[6pt]\tanh x&=\sum _{n=1}^{\infty }{\frac {B_{2n}4^{n}\left(4^{n}-1\right)}{(2n)!}}x^{2n-1}&&=x-{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}-{\frac {17x^{7}}{315}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\operatorname {arsinh} x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&=x-{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}-\cdots &&{\text{for }}|x|\leq 1\\[6pt]\operatorname {artanh} x&=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{2n+1}}&&=x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+\cdots &&{\text{for }}|x|\leq 1,\ x\neq \pm 1\end{aligned}}}
The numbers Bk tanh x are the Bernoulli numbers .
Polylogarithmic functions
edit
The polylogarithms have these defining identities:
Li
2
(
x
)
=
∑
n
=
1
∞
1
n
2
x
n
Li
3
(
x
)
=
∑
n
=
1
∞
1
n
3
x
n
{\displaystyle {\begin{aligned}{\text{Li}}_{2}(x)&=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}x^{n}\\{\text{Li}}_{3}(x)&=\sum _{n=1}^{\infty }{\frac {1}{n^{3}}}x^{n}\end{aligned}}}
The Legendre chi functions are defined as follows:
χ
2
(
x
)
=
∑
n
=
0
∞
1
(
2
n
+
1
)
2
x
2
n
+
1
χ
3
(
x
)
=
∑
n
=
0
∞
1
(
2
n
+
1
)
3
x
2
n
+
1
{\displaystyle {\begin{aligned}\chi _{2}(x)&=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{2}}}x^{2n+1}\\\chi _{3}(x)&=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{3}}}x^{2n+1}\end{aligned}}}
And the formulas presented below are called inverse tangent integrals
Ti
2
(
x
)
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
+
1
)
2
x
2
n
+
1
Ti
3
(
x
)
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
+
1
)
3
x
2
n
+
1
{\displaystyle {\begin{aligned}{\text{Ti}}_{2}(x)&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{2}}}x^{2n+1}\\{\text{Ti}}_{3}(x)&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)^{3}}}x^{2n+1}\end{aligned}}}
In statistical thermodynamics these formulas are of great importance.
Elliptic functions
edit
The complete elliptic integrals of first kind K and of second kind E can be defined as follows:
2
π
K
(
x
)
=
∑
n
=
0
∞
[
(
2
n
)
!
]
2
16
n
(
n
!
)
4
x
2
n
2
π
E
(
x
)
=
∑
n
=
0
∞
[
(
2
n
)
!
]
2
(
1
−
2
n
)
16
n
(
n
!
)
4
x
2
n
{\displaystyle {\begin{aligned}{\frac {2}{\pi }}K(x)&=\sum _{n=0}^{\infty }{\frac {[(2n)!]^{2}}{16^{n}(n!)^{4}}}x^{2n}\\{\frac {2}{\pi }}E(x)&=\sum _{n=0}^{\infty }{\frac {[(2n)!]^{2}}{(1-2n)16^{n}(n!)^{4}}}x^{2n}\end{aligned}}}
The Jacobi theta functions describe the world of the elliptic modular functions and they have these Taylor series:
ϑ
00
(
x
)
=
1
+
2
∑
n
=
1
∞
x
n
2
ϑ
01
(
x
)
=
1
+
2
∑
n
=
1
∞
(
−
1
)
n
x
n
2
{\displaystyle {\begin{aligned}\vartheta _{00}(x)&=1+2\sum _{n=1}^{\infty }x^{n^{2}}\\\vartheta _{01}(x)&=1+2\sum _{n=1}^{\infty }(-1)^{n}x^{n^{2}}\end{aligned}}}
The regular partition number sequence P(n) has this generating function:
ϑ
00
(
x
)
−
1
/
6
ϑ
01
(
x
)
−
2
/
3
[
ϑ
00
(
x
)
4
−
ϑ
01
(
x
)
4
16
x
]
−
1
/
24
=
∑
n
=
0
∞
P
(
n
)
x
n
=
∏
k
=
1
∞
1
1
−
x
k
{\displaystyle \vartheta _{00}(x)^{-1/6}\vartheta _{01}(x)^{-2/3}{\biggl [}{\frac {\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}}{16\,x}}{\biggr ]}^{-1/24}=\sum _{n=0}^{\infty }P(n)x^{n}=\prod _{k=1}^{\infty }{\frac {1}{1-x^{k}}}}
The strict partition number sequence Q(n) has that generating function:
ϑ
00
(
x
)
1
/
6
ϑ
01
(
x
)
−
1
/
3
[
ϑ
00
(
x
)
4
−
ϑ
01
(
x
)
4
16
x
]
1
/
24
=
∑
n
=
0
∞
Q
(
n
)
x
n
=
∏
k
=
1
∞
1
1
−
x
2
k
−
1
{\displaystyle \vartheta _{00}(x)^{1/6}\vartheta _{01}(x)^{-1/3}{\biggl [}{\frac {\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}}{16\,x}}{\biggr ]}^{1/24}=\sum _{n=0}^{\infty }Q(n)x^{n}=\prod _{k=1}^{\infty }{\frac {1}{1-x^{2k-1}}}}
Integration rules
edit
C is used for an arbitrary constant of integration that can only be determined if something about the value of the integral at some point is known. Thus, each function has an infinite number of antiderivatives .
These formulas only state in another form the assertions in the table of derivatives .
Integrals with a singularity
edit
When there is a singularity in the function being integrated such that the antiderivative becomes undefined or at some point (the singularity), then C does not need to be the same on both sides of the singularity. The forms below normally assume the Cauchy principal value around a singularity in the value of C but this is in general, not necessary. For instance in
∫
1
x
d
x
=
ln
|
x
|
+
C
{\displaystyle \int {1 \over x}\,dx=\ln \left|x\right|+C}
antiderivative becomes infinite there. If the integral above were to be used to compute a definite integral between −1 and 1, one would get the wrong answer 0. This however is the Cauchy principal value of the integral around the singularity. If the integration is done in the complex plane the result depends on the path around the origin, in this case the singularity contributes −i π when using a path above the origin and i π for a path below the origin. A function on the real line could use a completely different value of C on either side of the origin as in:[14]
∫
1
x
d
x
=
ln
|
x
|
+
{
A
if
x
>
0
;
B
if
x
<
0.
{\displaystyle \int {1 \over x}\,dx=\ln |x|+{\begin{cases}A&{\text{if }}x>0;\\B&{\text{if }}x<0.\end{cases}}}
Rational functions
edit
∫
a
d
x
=
a
x
+
C
{\displaystyle \int a\,dx=ax+C}
The following function has a non-integrable singularity at 0 for n ≤ −1
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
C
(for
n
≠
−
1
)
{\displaystyle \int x^{n}\,dx={\frac {x^{n+1}}{n+1}}+C\qquad {\text{(for }}n\neq -1{\text{)}}}
Cavalieri's quadrature formula )
∫
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
+
C
(for
n
≠
−
1
)
{\displaystyle \int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\text{(for }}n\neq -1{\text{)}}}
∫
1
x
d
x
=
ln
|
x
|
+
C
{\displaystyle \int {1 \over x}\,dx=\ln \left|x\right|+C}
More generally,[15]
∫
1
x
d
x
=
{
ln
|
x
|
+
C
−
x
<
0
ln
|
x
|
+
C
+
x
>
0
{\displaystyle \int {1 \over x}\,dx={\begin{cases}\ln \left|x\right|+C^{-}&x<0\\\ln \left|x\right|+C^{+}&x>0\end{cases}}}
∫
c
a
x
+
b
d
x
=
c
a
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {c}{ax+b}}\,dx={\frac {c}{a}}\ln \left|ax+b\right|+C}
Exponential functions
edit
Logarithms
edit
∫
ln
x
d
x
=
x
ln
x
−
x
+
C
=
x
(
ln
x
−
1
)
+
C
{\displaystyle \int \ln x\,dx=x\ln x-x+C=x(\ln x-1)+C}
∫
log
a
x
d
x
=
x
log
a
x
−
x
ln
a
+
C
=
x
ln
a
(
ln
x
−
1
)
+
C
{\displaystyle \int \log _{a}x\,dx=x\log _{a}x-{\frac {x}{\ln a}}+C={\frac {x}{\ln a}}(\ln x-1)+C}
Trigonometric functions
edit
∫
sin
x
d
x
=
−
cos
x
+
C
{\displaystyle \int \sin {x}\,dx=-\cos {x}+C}
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos {x}\,dx=\sin {x}+C}
∫
tan
x
d
x
=
ln
|
sec
x
|
+
C
=
−
ln
|
cos
x
|
+
C
{\displaystyle \int \tan {x}\,dx=\ln {\left|\sec {x}\right|}+C=-\ln {\left|\cos {x}\right|}+C}
∫
cot
x
d
x
=
−
ln
|
csc
x
|
+
C
=
ln
|
sin
x
|
+
C
{\displaystyle \int \cot {x}\,dx=-\ln {\left|\csc {x}\right|}+C=\ln {\left|\sin {x}\right|}+C}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
C
=
ln
|
tan
(
x
2
+
π
4
)
|
+
C
{\displaystyle \int \sec {x}\,dx=\ln {\left|\sec {x}+\tan {x}\right|}+C=\ln \left|\tan \left({\dfrac {x}{2}}+{\dfrac {\pi }{4}}\right)\right|+C}
∫
csc
x
d
x
=
−
ln
|
csc
x
+
cot
x
|
+
C
=
ln
|
csc
x
−
cot
x
|
+
C
=
ln
|
tan
x
2
|
+
C
{\displaystyle \int \csc {x}\,dx=-\ln {\left|\csc {x}+\cot {x}\right|}+C=\ln {\left|\csc {x}-\cot {x}\right|}+C=\ln {\left|\tan {\frac {x}{2}}\right|}+C}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}x\,dx=\tan x+C}
∫
csc
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int \csc ^{2}x\,dx=-\cot x+C}
∫
sec
x
tan
x
d
x
=
sec
x
+
C
{\displaystyle \int \sec {x}\,\tan {x}\,dx=\sec {x}+C}
∫
csc
x
cot
x
d
x
=
−
csc
x
+
C
{\displaystyle \int \csc {x}\,\cot {x}\,dx=-\csc {x}+C}
∫
sin
2
x
d
x
=
1
2
(
x
−
sin
2
x
2
)
+
C
=
1
2
(
x
−
sin
x
cos
x
)
+
C
{\displaystyle \int \sin ^{2}x\,dx={\frac {1}{2}}\left(x-{\frac {\sin 2x}{2}}\right)+C={\frac {1}{2}}(x-\sin x\cos x)+C}
∫
cos
2
x
d
x
=
1
2
(
x
+
sin
2
x
2
)
+
C
=
1
2
(
x
+
sin
x
cos
x
)
+
C
{\displaystyle \int \cos ^{2}x\,dx={\frac {1}{2}}\left(x+{\frac {\sin 2x}{2}}\right)+C={\frac {1}{2}}(x+\sin x\cos x)+C}
∫
tan
2
x
d
x
=
tan
x
−
x
+
C
{\displaystyle \int \tan ^{2}x\,dx=\tan x-x+C}
∫
cot
2
x
d
x
=
−
cot
x
−
x
+
C
{\displaystyle \int \cot ^{2}x\,dx=-\cot x-x+C}
∫
sec
3
x
d
x
=
1
2
(
sec
x
tan
x
+
ln
|
sec
x
+
tan
x
|
)
+
C
{\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}(\sec x\tan x+\ln |\sec x+\tan x|)+C}
∫
csc
3
x
d
x
=
1
2
(
−
csc
x
cot
x
+
ln
|
csc
x
−
cot
x
|
)
+
C
=
1
2
(
ln
|
tan
x
2
|
−
csc
x
cot
x
)
+
C
{\displaystyle \int \csc ^{3}x\,dx={\frac {1}{2}}(-\csc x\cot x+\ln |\csc x-\cot x|)+C={\frac {1}{2}}\left(\ln \left|\tan {\frac {x}{2}}\right|-\csc x\cot x\right)+C}
∫
sin
n
x
d
x
=
−
sin
n
−
1
x
cos
x
n
+
n
−
1
n
∫
sin
n
−
2
x
d
x
{\displaystyle \int \sin ^{n}x\,dx=-{\frac {\sin ^{n-1}{x}\cos {x}}{n}}+{\frac {n-1}{n}}\int \sin ^{n-2}{x}\,dx}
∫
cos
n
x
d
x
=
cos
n
−
1
x
sin
x
n
+
n
−
1
n
∫
cos
n
−
2
x
d
x
{\displaystyle \int \cos ^{n}x\,dx={\frac {\cos ^{n-1}{x}\sin {x}}{n}}+{\frac {n-1}{n}}\int \cos ^{n-2}{x}\,dx}
Inverse trigonometric functions
edit
∫
arcsin
x
d
x
=
x
arcsin
x
+
1
−
x
2
+
C
,
for
|
x
|
≤
1
{\displaystyle \int \arcsin {x}\,dx=x\arcsin {x}+{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq 1}
∫
arccos
x
d
x
=
x
arccos
x
−
1
−
x
2
+
C
,
for
|
x
|
≤
1
{\displaystyle \int \arccos {x}\,dx=x\arccos {x}-{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq 1}
∫
arctan
x
d
x
=
x
arctan
x
−
1
2
ln
|
1
+
x
2
|
+
C
,
for all real
x
{\displaystyle \int \arctan {x}\,dx=x\arctan {x}-{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C,{\text{ for all real }}x}
∫
arccot
x
d
x
=
x
arccot
x
+
1
2
ln
|
1
+
x
2
|
+
C
,
for all real
x
{\displaystyle \int \operatorname {arccot} {x}\,dx=x\operatorname {arccot} {x}+{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C,{\text{ for all real }}x}
∫
arcsec
x
d
x
=
x
arcsec
x
−
ln
|
x
(
1
+
1
−
x
−
2
)
|
+
C
,
for
|
x
|
≥
1
{\displaystyle \int \operatorname {arcsec} {x}\,dx=x\operatorname {arcsec} {x}-\ln \left\vert x\,\left(1+{\sqrt {1-x^{-2}}}\,\right)\right\vert +C,{\text{ for }}\vert x\vert \geq 1}
∫
arccsc
x
d
x
=
x
arccsc
x
+
ln
|
x
(
1
+
1
−
x
−
2
)
|
+
C
,
for
|
x
|
≥
1
{\displaystyle \int \operatorname {arccsc} {x}\,dx=x\operatorname {arccsc} {x}+\ln \left\vert x\,\left(1+{\sqrt {1-x^{-2}}}\,\right)\right\vert +C,{\text{ for }}\vert x\vert \geq 1}
Hyperbolic functions
edit
∫
sinh
x
d
x
=
cosh
x
+
C
{\displaystyle \int \sinh x\,dx=\cosh x+C}
∫
cosh
x
d
x
=
sinh
x
+
C
{\displaystyle \int \cosh x\,dx=\sinh x+C}
∫
tanh
x
d
x
=
ln
(
cosh
x
)
+
C
{\displaystyle \int \tanh x\,dx=\ln \,(\cosh x)+C}
∫
coth
x
d
x
=
ln
|
sinh
x
|
+
C
,
for
x
≠
0
{\displaystyle \int \coth x\,dx=\ln |\sinh x|+C,{\text{ for }}x\neq 0}
∫
sech
x
d
x
=
arctan
(
sinh
x
)
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=\arctan \,(\sinh x)+C}
∫
csch
x
d
x
=
ln
|
coth
x
−
csch
x
|
+
C
=
ln
|
tanh
x
2
|
+
C
,
for
x
≠
0
{\displaystyle \int \operatorname {csch} \,x\,dx=\ln |\operatorname {coth} x-\operatorname {csch} x|+C=\ln \left|\tanh {x \over 2}\right|+C,{\text{ for }}x\neq 0}
∫
sech
2
x
d
x
=
tanh
x
+
C
{\displaystyle \int \operatorname {sech} ^{2}x\,dx=\tanh x+C}
∫
csch
2
x
d
x
=
−
coth
x
+
C
{\displaystyle \int \operatorname {csch} ^{2}x\,dx=-\operatorname {coth} x+C}
∫
sech
x
tanh
x
d
x
=
−
sech
x
+
C
{\displaystyle \int \operatorname {sech} {x}\,\operatorname {tanh} {x}\,dx=-\operatorname {sech} {x}+C}
∫
csch
x
coth
x
d
x
=
−
csch
x
+
C
{\displaystyle \int \operatorname {csch} {x}\,\operatorname {coth} {x}\,dx=-\operatorname {csch} {x}+C}
Inverse hyperbolic functions
edit
∫
arcsinh
x
d
x
=
x
arcsinh
x
−
x
2
+
1
+
C
,
for all real
x
{\displaystyle \int \operatorname {arcsinh} \,x\,dx=x\,\operatorname {arcsinh} \,x-{\sqrt {x^{2}+1}}+C,{\text{ for all real }}x}
∫
arccosh
x
d
x
=
x
arccosh
x
−
x
2
−
1
+
C
,
for
x
≥
1
{\displaystyle \int \operatorname {arccosh} \,x\,dx=x\,\operatorname {arccosh} \,x-{\sqrt {x^{2}-1}}+C,{\text{ for }}x\geq 1}
∫
arctanh
x
d
x
=
x
arctanh
x
+
ln
(
1
−
x
2
)
2
+
C
,
for
|
x
|
<
1
{\displaystyle \int \operatorname {arctanh} \,x\,dx=x\,\operatorname {arctanh} \,x+{\frac {\ln \left(\,1-x^{2}\right)}{2}}+C,{\text{ for }}\vert x\vert <1}
∫
arccoth
x
d
x
=
x
arccoth
x
+
ln
(
x
2
−
1
)
2
+
C
,
for
|
x
|
>
1
{\displaystyle \int \operatorname {arccoth} \,x\,dx=x\,\operatorname {arccoth} \,x+{\frac {\ln \left(x^{2}-1\right)}{2}}+C,{\text{ for }}\vert x\vert >1}
∫
arcsech
x
d
x
=
x
arcsech
x
+
arcsin
x
+
C
,
for
0
<
x
≤
1
{\displaystyle \int \operatorname {arcsech} \,x\,dx=x\,\operatorname {arcsech} \,x+\arcsin x+C,{\text{ for }}0<x\leq 1}
∫
arccsch
x
d
x
=
x
arccsch
x
+
|
arcsinh
x
|
+
C
,
for
x
≠
0
{\displaystyle \int \operatorname {arccsch} \,x\,dx=x\,\operatorname {arccsch} \,x+\vert \operatorname {arcsinh} \,x\vert +C,{\text{ for }}x\neq 0}
Products of functions proportional to their second derivatives
edit
∫
cos
a
x
e
b
x
d
x
=
e
b
x
a
2
+
b
2
(
a
sin
a
x
+
b
cos
a
x
)
+
C
{\displaystyle \int \cos ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(a\sin ax+b\cos ax\right)+C}
∫
sin
a
x
e
b
x
d
x
=
e
b
x
a
2
+
b
2
(
b
sin
a
x
−
a
cos
a
x
)
+
C
{\displaystyle \int \sin ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(b\sin ax-a\cos ax\right)+C}
∫
cos
a
x
cosh
b
x
d
x
=
1
a
2
+
b
2
(
a
sin
a
x
cosh
b
x
+
b
cos
a
x
sinh
b
x
)
+
C
{\displaystyle \int \cos ax\,\cosh bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(a\sin ax\,\cosh bx+b\cos ax\,\sinh bx\right)+C}
∫
sin
a
x
cosh
b
x
d
x
=
1
a
2
+
b
2
(
b
sin
a
x
sinh
b
x
−
a
cos
a
x
cosh
b
x
)
+
C
{\displaystyle \int \sin ax\,\cosh bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(b\sin ax\,\sinh bx-a\cos ax\,\cosh bx\right)+C}
Absolute-value functions
edit
Let f continuous function , that has at most one zero . If f g f f g f
∫
|
f
(
x
)
|
d
x
=
sgn
(
f
(
x
)
)
g
(
x
)
+
C
,
{\displaystyle \int \left|f(x)\right|\,dx=\operatorname {sgn}(f(x))g(x)+C,}
sgn(x ) is the sign function , which takes the values −1, 0, 1 when x This can be proved by computing the derivative of the right-hand side of the formula, taking into account that the condition on g
This gives the following formulas (where a ≠ 0f C must be replaced by a piecewise constant function):
∫
|
(
a
x
+
b
)
n
|
d
x
=
sgn
(
a
x
+
b
)
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
+
C
{\displaystyle \int \left|(ax+b)^{n}\right|\,dx=\operatorname {sgn}(ax+b){(ax+b)^{n+1} \over a(n+1)}+C}
when n
n
≠
−
1
{\displaystyle n\neq -1}
∫
|
tan
a
x
|
d
x
=
−
1
a
sgn
(
tan
a
x
)
ln
(
|
cos
a
x
|
)
+
C
{\displaystyle \int \left|\tan {ax}\right|\,dx=-{\frac {1}{a}}\operatorname {sgn}(\tan {ax})\ln(\left|\cos {ax}\right|)+C}
when
a
x
∈
(
n
π
−
π
2
,
n
π
+
π
2
)
{\textstyle ax\in \left(n\pi -{\frac {\pi }{2}},n\pi +{\frac {\pi }{2}}\right)}
n
∫
|
csc
a
x
|
d
x
=
−
1
a
sgn
(
csc
a
x
)
ln
(
|
csc
a
x
+
cot
a
x
|
)
+
C
{\displaystyle \int \left|\csc {ax}\right|\,dx=-{\frac {1}{a}}\operatorname {sgn}(\csc {ax})\ln(\left|\csc {ax}+\cot {ax}\right|)+C}
when
a
x
∈
(
n
π
,
n
π
+
π
)
{\displaystyle ax\in \left(n\pi ,n\pi +\pi \right)}
n
∫
|
sec
a
x
|
d
x
=
1
a
sgn
(
sec
a
x
)
ln
(
|
sec
a
x
+
tan
a
x
|
)
+
C
{\displaystyle \int \left|\sec {ax}\right|\,dx={\frac {1}{a}}\operatorname {sgn}(\sec {ax})\ln(\left|\sec {ax}+\tan {ax}\right|)+C}
when
a
x
∈
(
n
π
−
π
2
,
n
π
+
π
2
)
{\textstyle ax\in \left(n\pi -{\frac {\pi }{2}},n\pi +{\frac {\pi }{2}}\right)}
n
∫
|
cot
a
x
|
d
x
=
1
a
sgn
(
cot
a
x
)
ln
(
|
sin
a
x
|
)
+
C
{\displaystyle \int \left|\cot {ax}\right|\,dx={\frac {1}{a}}\operatorname {sgn}(\cot {ax})\ln(\left|\sin {ax}\right|)+C}
when
a
x
∈
(
n
π
,
n
π
+
π
)
{\displaystyle ax\in \left(n\pi ,n\pi +\pi \right)}
n If the function f f sgn(f (x )) ∫ f (x ) dx is an antiderivative of f interval on which f f (x ) = 0step function . If we also use the fact that the absolute values of sine and cosine are periodic with period π , then we get:
∫
|
sin
a
x
|
d
x
=
2
a
⌊
a
x
π
⌋
−
1
a
cos
(
a
x
−
⌊
a
x
π
⌋
π
)
+
C
{\displaystyle \int \left|\sin {ax}\right|\,dx={2 \over a}\left\lfloor {\frac {ax}{\pi }}\right\rfloor -{1 \over a}\cos {\left(ax-\left\lfloor {\frac {ax}{\pi }}\right\rfloor \pi \right)}+C}
[citation needed
∫
|
cos
a
x
|
d
x
=
2
a
⌊
a
x
π
+
1
2
⌋
+
1
a
sin
(
a
x
−
⌊
a
x
π
+
1
2
⌋
π
)
+
C
{\displaystyle \int \left|\cos {ax}\right|\,dx={2 \over a}\left\lfloor {\frac {ax}{\pi }}+{\frac {1}{2}}\right\rfloor +{1 \over a}\sin {\left(ax-\left\lfloor {\frac {ax}{\pi }}+{\frac {1}{2}}\right\rfloor \pi \right)}+C}
[citation needed Special functions
edit
Ci , Si : Trigonometric integrals , Ei : Exponential integral , li : Logarithmic integral function , erf : Error function
∫
Ci
(
x
)
d
x
=
x
Ci
(
x
)
−
sin
x
{\displaystyle \int \operatorname {Ci} (x)\,dx=x\operatorname {Ci} (x)-\sin x}
∫
Si
(
x
)
d
x
=
x
Si
(
x
)
+
cos
x
{\displaystyle \int \operatorname {Si} (x)\,dx=x\operatorname {Si} (x)+\cos x}
∫
Ei
(
x
)
d
x
=
x
Ei
(
x
)
−
e
x
{\displaystyle \int \operatorname {Ei} (x)\,dx=x\operatorname {Ei} (x)-e^{x}}
∫
li
(
x
)
d
x
=
x
li
(
x
)
−
Ei
(
2
ln
x
)
{\displaystyle \int \operatorname {li} (x)\,dx=x\operatorname {li} (x)-\operatorname {Ei} (2\ln x)}
∫
li
(
x
)
x
d
x
=
ln
x
li
(
x
)
−
x
{\displaystyle \int {\frac {\operatorname {li} (x)}{x}}\,dx=\ln x\,\operatorname {li} (x)-x}
∫
erf
(
x
)
d
x
=
e
−
x
2
π
+
x
erf
(
x
)
{\displaystyle \int \operatorname {erf} (x)\,dx={\frac {e^{-x^{2}}}{\sqrt {\pi }}}+x\operatorname {erf} (x)}
Tangent Half-angle Substitution
edit
The tangent half-angle substitution relates an angle to the slope of a line. Introducing a new variable
t
=
tan
x
2
,
{\textstyle t=\tan {\tfrac {x}{2}},}
rational functions of
t
,
{\displaystyle t,}
d
x
{\displaystyle dx}
d
t
{\displaystyle dt}
t
,
{\displaystyle t,}
sin
x
=
2
t
1
+
t
2
,
cos
x
=
1
−
t
2
1
+
t
2
,
and
d
x
=
2
1
+
t
2
d
t
.
{\displaystyle \sin x={\frac {2t}{1+t^{2}}},\quad \cos x={\frac {1-t^{2}}{1+t^{2}}},\quad {\text{and}}\quad dx={\frac {2}{1+t^{2}}}\,dt.}
Similar expressions can be written for tan x , cot x , sec x , and csc x .
Derivation
edit
Using the double-angle formulas
sin
x
=
2
sin
x
2
cos
x
2
{\displaystyle \sin x=2\sin {\tfrac {x}{2}}\cos {\tfrac {x}{2}}}
cos
x
=
cos
2
x
2
−
sin
2
x
2
{\displaystyle \cos x=\cos ^{2}{\tfrac {x}{2}}-\sin ^{2}{\tfrac {x}{2}}}
Pythagorean identity
1
=
cos
2
x
2
+
sin
2
x
2
{\displaystyle 1=\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}}
sin
x
=
2
sin
x
2
cos
x
2
cos
2
x
2
+
sin
2
x
2
=
2
tan
x
2
1
+
tan
2
x
2
=
2
t
1
+
t
2
,
cos
x
=
cos
2
x
2
−
sin
2
x
2
cos
2
x
2
+
sin
2
x
2
=
1
−
tan
2
x
2
1
+
tan
2
x
2
=
1
−
t
2
1
+
t
2
.
{\displaystyle {\begin{aligned}\sin x&={\frac {2\sin {\tfrac {x}{2}}\,\cos {\tfrac {x}{2}}}{\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}}}={\frac {2\tan {\tfrac {x}{2}}}{1+\tan ^{2}{\tfrac {x}{2}}}}={\frac {2t}{1+t^{2}}},\\[18mu]\cos x&={\frac {\cos ^{2}{\tfrac {x}{2}}-\sin ^{2}{\tfrac {x}{2}}}{\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}}}={\frac {1-\tan ^{2}{\tfrac {x}{2}}}{1+\tan ^{2}{\tfrac {x}{2}}}}={\frac {1-t^{2}}{1+t^{2}}}.\end{aligned}}}
Finally, since
t
=
tan
x
2
{\textstyle t=\tan {\tfrac {x}{2}}}
differentiation rules imply
d
t
=
1
2
(
1
+
tan
2
x
2
)
d
x
=
1
+
t
2
2
d
x
,
{\displaystyle dt={\tfrac {1}{2}}\left(1+\tan ^{2}{\tfrac {x}{2}}\right)dx={\frac {1+t^{2}}{2}}\,dx,}
d
x
=
2
1
+
t
2
d
t
.
{\displaystyle dx={\frac {2}{1+t^{2}}}\,dt.}
References
edit
^ "Table of equivalent infinitesimals" (PDF) . Vaxa Software .^ Calculus (5th edition) , F. Ayres, E. Mendelson, Schaum's Outline Series, 2009, ISBN 978-0-07-150861-2 .^ Advanced Calculus (3rd edition) , R. Wrede, M.R. Spiegel, Schaum's Outline Series, 2010, ISBN 978-0-07-162366-7 .^ Complex Variables , M.R. Spiegel, S. Lipschutz, J.J. Schiller, D. Spellman, Schaum's Outlines Series, McGraw Hill (USA), 2009, ISBN 978-0-07-161569-3 ^ "Differentiation Rules" . University of Waterloo – CEMC Open Courseware . Retrieved 3 May 2022 .^ a b ^ Serge Lang . A First Course in Calculus , 5th edition, p. 290^ "Reader Survey: log|x | + C ", Tom Leinster, The n-category Café , March 19, 2012
![{\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {1}{x^{3}}}\left[\left({\frac {2+\cos x}{3}}\right)^{x}-1\right]&=\lim _{x\to 0}{\frac {e^{x\ln {\frac {2+\cos x}{3}}}-1}{x^{3}}}\\&=\lim _{x\to 0}{\frac {1}{x^{2}}}\ln {\frac {2+\cos x}{3}}\\&=\lim _{x\to 0}{\frac {1}{x^{2}}}\ln \left({\frac {\cos x-1}{3}}+1\right)\\&=\lim _{x\to 0}{\frac {\cos x-1}{3x^{2}}}\\&=\lim _{x\to 0}-{\frac {x^{2}}{6x^{2}}}\\&=-{\frac {1}{6}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c77dd026051c68e8a3ae139a8a19259a5ba4ae15)
![{\displaystyle [{\text{D}}(f\circ g)]_{x}=[{\text{D}}f]_{g(x)}\cdot [{\text{D}}g]_{x}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/624f2d85795ec0b7c5d54e5dd6ed8fddcdcfca36)
![{\displaystyle {\frac {d}{dx}}\left(f_{1}(x)^{f_{2}(x)^{\left(...\right)^{f_{n}(x)}}}\right)=\left[\sum \limits _{k=1}^{n}{\frac {\partial }{\partial x_{k}}}\left(f_{1}(x_{1})^{f_{2}(x_{2})^{\left(...\right)^{f_{n}(x_{n})}}}\right)\right]{\biggr \vert }_{x_{1}=x_{2}=...=x_{n}=x},{\text{ if }}f_{i<n}(x)>0{\text{ and }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e52623821c09e999ec428d4b2264ed2123e6c5c)
![{\displaystyle [0,\pi ]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb175ba2045ce5c320e0942973d0168dee16551a)
![{\displaystyle \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c631b663bb72b84a42765b6c31c1de35cc75dc2c)
![{\displaystyle [-\pi ,\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb064fd6c55820cfa660eabeeda0f6e3c4935ae6)
![{\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}&&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots &&{\text{for all }}x\\[6pt]\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}&&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots &&{\text{for all }}x\\[6pt]\tan x&=\sum _{n=1}^{\infty }{\frac {B_{2n}(-4)^{n}\left(1-4^{n}\right)}{(2n)!}}x^{2n-1}&&=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\sec x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}&&=1+{\frac {x^{2}}{2}}+{\frac {5x^{4}}{24}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\arcsin x&=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+\cdots &&{\text{for }}|x|\leq 1\\[6pt]\arccos x&={\frac {\pi }{2}}-\arcsin x\\&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&={\frac {\pi }{2}}-x-{\frac {x^{3}}{6}}-{\frac {3x^{5}}{40}}-\cdots &&{\text{for }}|x|\leq 1\\[6pt]\arctan x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}&&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots &&{\text{for }}|x|\leq 1,\ x\neq \pm i\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/158a0ae14d1c9e0d1dc21c268f7e2169b9066dc7)
![{\displaystyle {\begin{aligned}\sinh x&=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{(2n+1)!}}&&=x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+\cdots &&{\text{for all }}x\\[6pt]\cosh x&=\sum _{n=0}^{\infty }{\frac {x^{2n}}{(2n)!}}&&=1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+\cdots &&{\text{for all }}x\\[6pt]\tanh x&=\sum _{n=1}^{\infty }{\frac {B_{2n}4^{n}\left(4^{n}-1\right)}{(2n)!}}x^{2n-1}&&=x-{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}-{\frac {17x^{7}}{315}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\operatorname {arsinh} x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&=x-{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}-\cdots &&{\text{for }}|x|\leq 1\\[6pt]\operatorname {artanh} x&=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{2n+1}}&&=x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+\cdots &&{\text{for }}|x|\leq 1,\ x\neq \pm 1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cda808c97562eca785bd172eb7739711b338730a)
![{\displaystyle {\begin{aligned}{\frac {2}{\pi }}K(x)&=\sum _{n=0}^{\infty }{\frac {[(2n)!]^{2}}{16^{n}(n!)^{4}}}x^{2n}\\{\frac {2}{\pi }}E(x)&=\sum _{n=0}^{\infty }{\frac {[(2n)!]^{2}}{(1-2n)16^{n}(n!)^{4}}}x^{2n}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41ef2ac15ffd8927e6037af64226d9fe96dcc4dc)
![{\displaystyle \vartheta _{00}(x)^{-1/6}\vartheta _{01}(x)^{-2/3}{\biggl [}{\frac {\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}}{16\,x}}{\biggr ]}^{-1/24}=\sum _{n=0}^{\infty }P(n)x^{n}=\prod _{k=1}^{\infty }{\frac {1}{1-x^{k}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c898219336ceb853f972d83f60a14049d671523)
![{\displaystyle \vartheta _{00}(x)^{1/6}\vartheta _{01}(x)^{-1/3}{\biggl [}{\frac {\vartheta _{00}(x)^{4}-\vartheta _{01}(x)^{4}}{16\,x}}{\biggr ]}^{1/24}=\sum _{n=0}^{\infty }Q(n)x^{n}=\prod _{k=1}^{\infty }{\frac {1}{1-x^{2k-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa5dd403c225903926ad0c45788a5f1a0a5fb67c)
![{\displaystyle {\begin{aligned}\sin x&={\frac {2\sin {\tfrac {x}{2}}\,\cos {\tfrac {x}{2}}}{\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}}}={\frac {2\tan {\tfrac {x}{2}}}{1+\tan ^{2}{\tfrac {x}{2}}}}={\frac {2t}{1+t^{2}}},\\[18mu]\cos x&={\frac {\cos ^{2}{\tfrac {x}{2}}-\sin ^{2}{\tfrac {x}{2}}}{\cos ^{2}{\tfrac {x}{2}}+\sin ^{2}{\tfrac {x}{2}}}}={\frac {1-\tan ^{2}{\tfrac {x}{2}}}{1+\tan ^{2}{\tfrac {x}{2}}}}={\frac {1-t^{2}}{1+t^{2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0e0db3590824f5019d21786ee97ec792c129906)
