nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of Hyperbolic Sine of a x by Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sinh a x \cosh a x} = \frac 1 a \ln \size {\tanh a x} + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x $ and $\cosh a x$: $14.595$
* : $\S 17$: Tables of Special Indefinite Integrals: $(29)$ Integrals Involving $\sinh a x $ and $\cosh a x$: $17.29.4.$
Category:Primitives involving Hyperbolic Sine Function
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Hyperbolic Sine of a x by Hyperbolic Cosine of a x minus 1

== Theorem ==
:$\ds \int \frac {\d x} {\sinh a x \paren {\cosh a x - 1} } = \frac {-1} {2 a} \ln \size {\tanh \frac {a x} 2} - \frac 1 {2 a \paren {\cosh a x - 1} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x $ and $\cosh a x$: $14.603$
Category:Primitives involving Hyperbolic Sine Function
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Hyperbolic Sine of a x by Hyperbolic Cosine of a x plus 1

== Theorem ==
:$\ds \int \frac {\d x} {\sinh a x \paren {\cosh a x + 1} } = \frac 1 {2 a} \ln \size {\tanh \frac {a x} 2} + \frac 1 {2 a \paren {\cosh a x + 1} } + C$
== Proof ==
Let:
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x $ and $\cosh a x$: $14.602$
Category:Primitives involving Hyperbolic Sine Function
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Hyperbolic Sine of a x by Square of Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sinh a x \cosh^2 a x} = \frac 1 a \ln \size {\tanh \frac {a x} 2} + \frac {\sech a x} a + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x $ and $\cosh a x$: $14.597$
Category:Primitives involving Hyperbolic Sine Function
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Hyperbolic Tangent of a x

== Theorem ==
:$\ds \int \frac {\d x} {\tanh a x} = \frac {\ln \size {\sinh a x} } a + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {\sinh a x}$
* Primitive of $\dfrac 1 {\cosh a x}$
* Primitive of $\dfrac 1 {\coth a x}$
* Primitive of $\dfrac 1 {\sech a x}$
* Primitive of $\dfrac 1 {\csch a x}$
== Sources ==
* : $\S 14$: Integrals involving $\tanh a x$: $14.609$
Category:Primitives involving Hyperbolic Tangent Function
Primitive of Reciprocal of Logarithm of x

== Theorem ==
For $x > 1$:
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $\dfrac 1 {\ln x}$ has no Solution in Elementary Functions
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.533$
Category:Primitives involving Logarithm Function
Category:Primitive of Reciprocal of Logarithm of x
Primitive of Reciprocal of Logarithm of x/Proof 2

== Theorem ==
== Proof ==
From Primitive of $\dfrac {x^m} {\ln x}$:
:$\ds \int \frac {x^m \rd x} {\ln x} = \map \ln {\ln x} + \paren {m + 1} \ln x + \sum_{k \mathop \ge 2}^n \frac {\paren {m + 1}^k \paren {\ln x}^k} {k \times k!} + C$
The result follows by setting $m = 0$.
Category:Primitive of Reciprocal of Logarithm of x
Primitive of Reciprocal of Logarithm of x has no Solution in Elementary Functions

== Theorem ==
The primitive:
:$\ds \int \frac {\d x} {\ln x}$
cannot be expressed in terms of a finite number of elementary functions.
== Proof ==
== Also see ==
* Primitive of $\dfrac {\d x} {\ln x}$
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.29$: Liouville ()
Category:Primitives involving Logarithm Function
Primitive of Reciprocal of Logarithm of x has no Solution in Elementary Functions/Historical Note

== Historical Note on Primitive of $\dfrac 1 {\ln x}$ has no Solution in Elementary Functions ==
The proof that $\ds \int \dfrac {\d x} {\ln x}$ cannot be expressed with a finite number of elementary functions was first proved by .
== Sources ==
* : Chapter $\text {A}.29$: Liouville ()
Category:Historical Notes
Primitive of Reciprocal of One plus Fourth Power of x

== Theorem ==
:$\ds \int \frac 1 {1 + x^4} \rd x = \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } } + C$
== Proof 1 ==
== Proof 2 ==
Category:Primitives involving Reciprocals
Category:Primitives involving x to the fourth plus or minus a to the fourth
Category:Primitive of Reciprocal of One plus Fourth Power of x
Primitive of Reciprocal of One plus Fourth Power of x/Proof 1

== Theorem ==
== Proof ==
From Primitive of $\dfrac {1 + x^2} {1 + x^4}$, we have:
:$\ds \int \frac {x^2 + 1} {x^4 + 1} \rd x = \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C$
From Primitive of $\dfrac {-1 + x^2} {1 + x^4}$, we have:
:$\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x = \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C$
We therefore have: 
Category:Primitive of Reciprocal of One plus Fourth Power of x
Primitive of Reciprocal of Power of Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\cos^n a x} = \frac {\sin a x} {a \paren {n - 1} \cos^{n - 1} a x} + \frac {n - 2} {n - 1} \int \frac {\d x} {\cos^{n - 2} a x}$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $\dfrac 1 {\sin^n a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.397$
Category:Primitives involving Cosine Function
Primitive of Reciprocal of Power of Cosine of a x by Power of Sine of a x/Reduction of Power of Cosine

== Theorem ==
:$\ds \int \frac {\d x} {\sin^m a x \cos^n a x} = \frac 1 {a \paren {n - 1} \sin^{m - 1} a x \cos^{n - 1} a x} + \frac {m + n - 2} {n - 1} \int \frac {\d x} {\sin^m a x \cos^{n - 2} a x}$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {\sin^m a x \cos^n a x}$ : Reduction of Power of Sine
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.428$
* : $\S 17$: Tables of Special Indefinite Integrals: $(19)$ Integrals Involving $\sin a x$ and $\cos a x$: $17.19.30.$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of Power of Cosine of a x by Power of Sine of a x/Reduction of Power of Sine

== Theorem ==
:$\ds \int \frac {\d x} {\sin^m a x \cos^n a x} = \frac {-1} {a \paren {n - 1} \sin^{m - 1} a x \cos^{n - 1} a x} + \frac {m + n - 2} {m - 1} \int \frac {\d x} {\sin^{m - 2} a x \cos^n a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {\sin^m a x \cos^n a x}$ : Reduction of Power of Cosine
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.428$
* : $\S 17$: Tables of Special Indefinite Integrals: $(19)$ Integrals Involving $\sin a x$ and $\cos a x$: $17.19.30.$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of Power of Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\cosh^n a x} = \frac {\sinh a x} {a \paren {n - 1} \cosh^{n - 1} a x} + \frac {n - 2} {n - 1} \int \frac {\d x} {\cosh^{n - 2} a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac {\d x} {\sinh^n a x}$
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.588$
* : $\S 17$: Tables of Special Indefinite Integrals: $(28)$ Integrals Involving $\cosh a x$: $17.28.15.$
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Power of Hyperbolic Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sinh^n a x} = \frac {-\cosh a x} {a \paren {n - 1} \sinh^{n - 1} a x} - \frac {n - 2} {n - 1} \int \frac {\d x} {\sinh^{n - 2} a x}$
for $n \ne 1$.
== Proof ==
We note that when $n = 1$, $\dfrac {n - 2} {n - 1}$ is undefined, rendering this derivation invalid.
== Also see ==
* Primitive of $\dfrac 1 {\sinh a x}$ for the case where $n = -1$
* Primitive of $\dfrac 1 {\cosh^n a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x$: $14.560$
Category:Primitives involving Hyperbolic Sine Function
Primitive of Reciprocal of Power of Root of 2 a x minus x squared

== Theorem ==
:$\ds \int \dfrac {\d x} {\paren {\sqrt {2 a x - x^2} }^n} = \frac {\paren {x - a} \paren {\sqrt {2 a x - x^2} }^{2 - n} } {\paren {n - 2} a^2} + \frac {n - 3} {\paren {n - 2} a^2} \int \dfrac {\d x} {\paren {\sqrt {2 a x - x^2} }^{n - 2} }$
== Proof ==
from which the result follows immediately.
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $50$.
Category:Primitives of Roots of Quadratic Functions
Primitive of Reciprocal of Power of Root of x squared minus a squared

== Theorem ==
:$\ds \int \dfrac {\d x} {\paren {\sqrt {x^2 - a^2} }^n} = \dfrac {x \paren {\sqrt {x^2 - a^2} }^{2 - n} } {\paren {2 - n} a^2} - \dfrac {n - 3} {\paren {n - 2} a^2} \int \dfrac {\d x} {\paren {\sqrt {x^2 - a^2} }^{n - 2} }$
for $n \ne 2$.
== Proof ==
Let:
Also:
and:
Thus:
== Also see ==
For $n = -2$, use Primitive of $\dfrac 1 {x^2 - a^2}$.
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $39$.
Category:Primitives involving Root of x squared minus a squared
Primitive of Reciprocal of Power of Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sin^n a x} = \frac {-\cos a x} {a \paren {n - 1} \sin^{n - 1} a x} + \frac {n - 2} {n - 1} \int \frac {\d x} {\sin^{n - 2} a x}$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $\dfrac 1 {\cos^n a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.367$
Category:Primitives involving Sine Function
Primitive of Reciprocal of Power of a squared minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {\paren {a^2 - x^2}^n} = \frac x {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {a^2 - x^2}^{n - 1} }$
for $x^2 > a^2$.
== Proof ==
Aiming for an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$
in order to use the technique of Integration by Parts, let:
Thus:
Then:
== Also see ==
* Primitive of $\dfrac 1 {\paren {x^2 - a^2}^n}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.177$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of Power of a x + b by Power of p x + q

== Theorem ==
:$\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$
== Proof ==
From Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - a \paren {m + n + 2} \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$
Setting $m := -m$ and $n := -n$:
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.110$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.6.$
Category:Primitives involving a x + b and p x + q
Primitive of Reciprocal of Power of p x + q by Root of a x + b

== Theorem ==
:$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$
== Proof ==
From Primitive of Reciprocal of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$:
:$\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$
Setting $m := \dfrac 1 2$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.117$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.5.$
Category:Primitives involving Root of a x + b and p x + q
Primitive of Reciprocal of Power of x by Power of Power of x minus Power of a

== Theorem ==
:$\ds \int \frac {\d x} {x^m \ \paren {x^n - a^n}^r} = \frac 1 {a^n} \int \frac {\d x} {x^{m - n} \paren {x^n - a^n}^r} - \frac 1 {a^n} \int \frac {\d x} {x^m \paren {x^n - a^n}^{r - 1} }$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.333$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Reciprocal of Power of x by Power of Power of x plus Power of a

== Theorem ==
:$\ds \int \frac {\d x} {x^m \paren {x^n + a^n}^r} = \frac 1 {a^n} \int \frac {\d x} {x^m \paren {x^n + a^n}^{r - 1} } - \frac 1 {a^n} \int \frac {\d x} {x^{m - n} \paren {x^n + a^n}^r}$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.328$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Reciprocal of Power of x by Power of a squared minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {x^m \paren {a^2 - x^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {a^2 - x^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {a^2 - x^2}^n}$
for $x^2 < a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^m \paren {x^2 - a^2}^n}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.181$
* : $\S 17$: Tables of Special Indefinite Integrals: $(8)$ Integrals Involving $x^2 - a^2$, $x^2 < a^2$: $17.8.19.$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of Power of x by Power of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
== Proof ==
First:
Next, with a view to obtaining an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
Thus:
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.279$
* : $\S 17$: Tables of Special Indefinite Integrals: $(12)$ Integrals Involving $a x^2 + b x + c$: $17.12.15.$
Category:Primitives involving a x squared plus b x plus c
Primitive of Reciprocal of Power of x by Power of x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^m \paren {x^2 - a^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^m \paren {a^2 - x^2}^n}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.162$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.19.$
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of Power of x by Power of x squared plus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^m \paren {x^2 + a^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.143$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.19.$
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of Power of x by Root of a x + b

== Theorem ==
:$\ds \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Putting $n := -\dfrac 1 2$ and $m := -m$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt{a x + b}$: $14.95$
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.12.$
Category:Primitives involving Root of a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of Power of x by a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^n \paren {a x^2 + b x + c} } = \frac {-1} {\paren {n - 1} c x^{n - 1} } - \frac b c \int \frac {\d x} {x^{n - 1} \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.271$
Category:Primitives involving a x squared plus b x plus c
Primitive of Reciprocal of Power of x by x cubed plus a cubed

== Theorem ==
:$\ds \int \frac {\d x} {x^n \paren {x^3 + a^3} } = \frac {-1} {a^3 \paren {n - 1} x^{n - 1} } - \frac 1 {a^3} \int \frac {\d x} {x^{n - 3} \paren {x^3 + a^3} }$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.310$
* : $\S 17$: Tables of Special Indefinite Integrals: $(14)$ Integrals Involving $x^3 + a^3$: $17.14.12.$
Category:Primitives involving x cubed plus a cubed
Primitive of Reciprocal of Power of x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {\paren {x^2 - a^2}^n} = \frac {-x} {2 \paren {n - 1} a^2 \paren {x^2 - a^2}^{n - 1} } - \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {x^2 - a^2}^{n - 1} }$
for $x^2 > a^2$.
== Proof ==
Aiming for an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$
in order to use the technique of Integration by Parts, let:
Thus:
Then:
== Also see ==
* Primitive of $\dfrac 1 {\paren {a^2 - x^2}^n}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.158$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.15.$
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of Power of x squared plus a squared

== Theorem ==
:$\ds \int \frac {\d x} {\paren {x^2 + a^2}^n} = \frac x {2 \paren {n - 1} a^2 \paren {x^2 + a^2}^{n - 1} } + \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {x^2 + a^2}^{n - 1} }$
== Proof ==
Aiming for an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
in order to use the technique of Integration by Parts, let:
Thus:
Then:
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.139$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.15.$
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of Root of 1 minus x squared/Arcsine Form

== Theorem ==
:$\ds \int \frac {\d x} {\sqrt {1 - x^2} } = \arcsin x + C$
where $C$ is an arbitrary constant.
== Proof 1 ==
== Proof 2 ==
Category:Primitive of Reciprocal of Root of a squared minus x squared
Primitive of Reciprocal of Root of 1 minus x squared/Arcsine Form/Proof 1

== Corollary to Primitive of $\frac 1 {\sqrt {a^2 - x^2} }$: Arcsine Form ==
== Proof ==
From Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: Arcsine Form:
:$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$
The result follows by setting $a = 1$.
Category:Primitive of Reciprocal of Root of a squared minus x squared
Primitive of Reciprocal of Root of 1 minus x squared/Arcsine Form/Proof 2

== Theorem ==
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Integration
* : Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms
Category:Primitive of Reciprocal of Root of a squared minus x squared
Primitive of Reciprocal of Root of 2 a x minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {\sqrt {2 a x - x^2} } = \arcsin \dfrac {x - a} a + C$
where $C$ is an arbitrary constant.
== Proof ==
Let $u := x - a$.
Then:
:$\dfrac {\d u} {\d x} = 1$
and:
:$x = u + a$
and we have:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.47$
* : Front endpapers: A Brief Table of Integrals: $47$.
Category:Primitives of Roots of Quadratic Functions
Primitive of Reciprocal of Root of a minus x by Cube of Root of x minus b

== Theorem ==
:$\ds \int \dfrac {\d x} {\paren {a - x}^{1/2} \paren {x - b}^{3/2} } = \dfrac 2 {b - a} \sqrt {\dfrac {a - x} {x - b} } + C$
== Proof 1 ==
== Proof 2 ==
Category:Primitive of Reciprocal of Root of a minus x by Cube of Root of x minus b
Category:Primitives involving Root of a x + b and Root of p x + q
Primitive of Reciprocal of Root of a minus x by Cube of Root of x minus b/Proof 1

== Theorem ==
== Proof ==
Let:
Then we have:
and:
Hence:
== Sources ==
* : $\text {II}$. Calculus: Integration: Useful substitutions: Example
Category:Primitive of Reciprocal of Root of a minus x by Cube of Root of x minus b
Primitive of Reciprocal of Root of a squared minus x squared

== Theorem ==
=== Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: $\arcsin$ form ===
=== Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: $\arccos$ form ===
== Examples ==
== Also see ==
* Primitive of Reciprocal of $\sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $\sqrt {x^2 - a^2}$
Category:Primitive of Reciprocal of Root of a squared minus x squared
Category:Expressions whose Primitives are Inverse Trigonometric Functions
Category:Primitives involving Root of a squared minus x squared
Category:Primitives involving Reciprocals
Primitive of Reciprocal of Root of a squared minus x squared/Arccosine Form

== Theorem ==
:$\ds \int \frac 1 {\sqrt {a^2 - x^2} } \rd x = -\arccos \frac x a + C$
where $a$ is a strictly positive constant and $a^2 > x^2$.
== Proof ==
Substitute:
:$\cos \theta = \dfrac x a \iff x = a \cos \theta$
for $\theta \in \openint 0 \pi$.
From Real Cosine Function is Bounded and Shape of Cosine Function, this substitution is valid for all $x / a \in \openint {-1} 1$.
By hypothesis:
so this substitution will not change the domain of the integrand.
Then:
We have defined $\theta$ to be in the open interval $\openint 0 \pi$.
From Sine and Cosine are Periodic on Reals, $\sin \theta > 0$ for the entire interval. Therefore the absolute value is unnecessary, and the integral simplifies to:
As $\theta$ was stipulated to be in the open interval $\openint 0 \pi$:
:$\cos \theta = \dfrac x a \iff \theta = \arccos \dfrac x a$
The answer in terms of $x$, then, is:
:$\ds \int \frac 1 {\sqrt {a^2 - x^2} } \rd x = -\arccos \frac x a + C$
== Also see ==
* Primitive of Reciprocal of $\sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $\sqrt {x^2 - a^2}$
* Derivative of Arccosine Function
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $12$.
* : Chapter $5$. Integrals: Integrals of Special Functions: $25$
Category:Arccosine Function
Category:Primitive of Reciprocal of Root of a squared minus x squared