nyuuzyou/wikis huggingface dataset:
Primitive of Inverse Hyperbolic Sine of x over a over x

== Theorem ==
== Proof ==
For $\size x < a$:
For $x > a$:
For $x < a$:
== Also see ==
* Primitive of $\dfrac 1 x \arcosh \dfrac x a$
* Primitive of $\dfrac 1 x \artanh \dfrac x a$
* Primitive of $\dfrac 1 x \arcoth \dfrac x a$
* Primitive of $\dfrac 1 x \arsech \dfrac x a$
* Primitive of $\dfrac 1 x \arcsch \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.649$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.3.$
Category:Primitives involving Inverse Hyperbolic Sine Function
Primitive of Inverse Hyperbolic Sine of x over a over x squared

== Theorem ==
:$\ds \int \frac 1 {x^2} \arsinh \dfrac x a \rd x = -\frac 1 x \arsinh \dfrac x a - \frac 1 a \map \ln {\frac a x + \frac {\sqrt {a^2 + x^2} } {\size x} }$
== 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 $\dfrac 1 {x^2} \arcosh \dfrac x a$
* Primitive of $\dfrac 1 {x^2} \artanh \dfrac x a$
* Primitive of $\dfrac 1 {x^2} \arcoth \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.650$
Category:Primitives involving Inverse Hyperbolic Sine Function
Primitive of Inverse Hyperbolic Tangent Function

== Theorem ==
:$\ds \int \artanh x \rd x = x \artanh x + \frac {\map \ln {1 - x^2} } 2 + C$
for $x^2 < 1$.
== Proof ==
== Also presented as ==
== Also see ==
* Primitive of $\arsinh x$
* Primitive of $\arcosh x$
* Primitive of $\arcoth x$
== Sources ==
* 
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitive of Inverse Hyperbolic Tangent Function
Category:Primitives involving Inverse Hyperbolic Tangent Function
Category:Inverse Hyperbolic Tangent
Primitive of Inverse Hyperbolic Tangent Function/Also presented as

== Primitive of Inverse Hyperbolic Tangent Function: Also presented as ==
This result can also be presented as:
:$\ds \int \artanh x \rd x = x \artanh x + \ln \sqrt {1 - x^2} + C$
== Proof ==
From Primitive of $\artanh x$:
:$\ds \int \artanh x \rd x = x \artanh x + \frac {\map \ln {1 - x^2} } 2 + C$
The result follows by observing:
:$\dfrac 1 2 \log x = \log \sqrt x$
from Logarithm of Power.
== Sources ==
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
Category:Primitive of Inverse Hyperbolic Tangent Function
Primitive of Inverse Hyperbolic Tangent Function/Proof 1

== Theorem ==
== 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:
Category:Primitive of Inverse Hyperbolic Tangent Function
Primitive of Inverse Hyperbolic Tangent of x over a

== Theorem ==
:$\ds \int \artanh \frac x a \rd x = x \artanh \dfrac x a + \frac {a \map \ln {a^2 - x^2} } 2 + C$
for $x^2 < a^2$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $\arsinh \dfrac x a$
* Primitive of $\arcosh \dfrac x a$
* Primitive of $\arcoth \dfrac x a$
* Primitive of $\arsech \dfrac x a$
* Primitive of $\arcsch \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.656$
* : Appendix $2$: Table of derivatives and integrals of common functions: Inverse hyperbolic functions
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.7.$
Category:Primitive of Inverse Hyperbolic Tangent of x over a
Category:Primitive of Inverse Hyperbolic Tangent Function
Category:Primitives involving Inverse Hyperbolic Tangent Function
Primitive of Inverse Hyperbolic Tangent of x over a/Proof 1

== Theorem ==
== 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:
Category:Primitive of Inverse Hyperbolic Tangent of x over a
Primitive of Inverse Hyperbolic Tangent of x over a over x

== Theorem ==
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 x \arsinh \dfrac x a$
* Primitive of $\dfrac 1 x \arcosh \dfrac x a$
* Primitive of $\dfrac 1 x \arcoth \dfrac x a$
* Primitive of $\dfrac 1 x \arsech \dfrac x a$
* Primitive of $\dfrac 1 x \arcsch \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.659$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.9.$
Category:Primitives involving Inverse Hyperbolic Tangent Function
Primitive of Inverse Hyperbolic Tangent of x over a over x squared

== Theorem ==
:$\ds \int \frac 1 {x^2} \artanh \dfrac x a \rd x = -\frac 1 x \artanh \dfrac x a + \frac 1 {2 a} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
== 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 $\dfrac 1 {x^2} \arsinh \frac x a$
* Primitive of $\dfrac 1 {x^2} \arcosh \frac x a$
* Primitive of $\dfrac 1 {x^2} \arcoth \frac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.660$
Category:Primitives involving Inverse Hyperbolic Tangent Function
Primitive of Logarithm of a x over x

== Theorem ==
:$\ds \int \frac {\ln a x} x \rd x = \frac {\paren {\ln a x}^2} 2 + C$
for $a \ne 0$ and $x \ne 0$.
== Proof 1==
Let $z = a x$.
== Proof 2==
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $111$.
Category:Primitives involving Logarithm Function
Primitive of Logarithm of x

== Theorem ==
:$\ds \int \ln x \rd x = x \ln x - x + C$
=== Corollary ===
== Proof 1 ==
 
== Proof 2 ==
== Also presented as ==
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.525$
* : Appendix $\text I$: Table of Indefinite Integrals $10$.
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
* : $\S 17$: Tables of Special Indefinite Integrals: $(26)$ Integrals Involving $\ln x$: $17.26.1.$
* : Appendix $7$: Integrals
* : Appendix $8$: Integrals
Category:Primitive of Logarithm of x
Category:Primitives involving Logarithm Function
Primitive of Logarithm of x/Also presented as

== Primitive of $\ln x$: Also presented as ==
Some sources present this result as:
:$\ds \int \ln x \rd x = x \paren {\ln x - 1} + C$
== Sources ==
* : Appendix $2$: Table of derivatives and integrals of common functions
Category:Primitive of Logarithm of x
Primitive of Logarithm of x/Corollary

== Corollary to Primitive of $\map \ln x$ ==
:$\ds \int \map \ln {1 - x} \rd x = \paren {x - 1} \map \ln {1 - x} - x + C$
== Proof ==
Let $z = \paren {1 - x}$.
Category:Primitive of Logarithm of x
Category:Primitives involving Logarithm Function
Primitive of Logarithm of x/Proof 1

== Theorem ==
== 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:
== Sources ==
* : $\text {II}$. Calculus: Integration: Integration by Parts: Example
Category:Primitive of Logarithm of x
Primitive of Logarithm of x/Proof 2

== Theorem ==
== Proof ==
Note that we have: 
We therefore have: 
giving: 
:$\ds \frac 1 a \int_0^a \ln u \rd u = \ln a - 1$
so: 
:$\ds \int_0^a \ln u \rd u = a \ln a - a$
for all real $a > 0$. 
By Fundamental Theorem of Calculus: First Part, we have that: 
:$x \ln x - x$ is a primitive of $\ln x$ on $x > 0$.
We therefore conclude that, by Primitives which Differ by Constant:
:$\ds \int \ln x \rd x = x \ln x - x + C$
for $x > 0$.
Category:Primitive of Logarithm of x
Primitive of Logarithm of x over x

== Theorem ==
:$\ds \int \frac {\ln x} x \rd x = \frac {\ln^2 x} 2 + C$
=== Corollary ===
== 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:
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.528$
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitives involving Logarithm Function
Primitive of Logarithm of x over x squared

== Theorem ==
:$\ds \int \frac {\ln x} {x^2} \rd x = \frac {-\ln x} x - \frac 1 x + C$
== Proof ==
From Primitive of $x^m \ln x$:
:$\ds \int x^m \ln x \rd x = \frac {x^{m + 1} } {m + 1} \paren {\ln x - \frac 1 {m + 1} } + C$
Thus:
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.529$
Category:Primitives involving Logarithm Function
Primitive of Logarithm of x squared minus a squared

== Theorem ==
:$\ds \int \map \ln {x^2 - a^2} \rd x = x \map \ln {x^2 - a^2} - 2 x + a \map \ln {\frac {x + a} {x - a} } + C$
for $x^2 > a^2$.
== 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 $\map \ln {x^2 + a^2}$
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.538$
Category:Primitives involving Logarithm Function
Category:Primitives involving x squared minus a squared
Primitive of Logarithm of x squared plus a squared

== Theorem ==
:$\ds \int \map \ln {x^2 + a^2} \rd x = x \map \ln {x^2 + a^2} - 2 x + 2 a \arctan \frac x a + C$
== 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 $\map \ln {x^2 - a^2}$
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.537$
Category:Primitives involving Logarithm Function
Category:Primitives involving x squared plus a squared
Primitive of Minus One plus x Squared over One plus Fourth Power of x

== Theorem ==
:$\ds \int \frac {x^2 - 1} {x^4 + 1} \rd x = \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C$
== 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 Minus One plus x Squared over One plus Fourth Power of x
Primitive of Minus One plus x Squared over One plus Fourth Power of x/Proof 1

== Theorem ==
== Proof ==
We have:
Note that, by Derivative of Power: 
:$\dfrac \d {\d x} \paren {x + \dfrac 1 x} = 1 - \dfrac 1 {x^2}$
So, we have:
Category:Primitive of Minus One plus x Squared over One plus Fourth Power of x
Primitive of Odd Power of x over Power of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x^{2 n - 1} \rd x} {\paren {a x^2 + b x + c}^n} = \frac 1 a \int \frac {x^{2 n - 3} \rd x} {\paren {a x^2 + b x + c}^{n - 1} } - \frac c a \int \frac {x^{2 n - 3} \rd x} {\paren {a x^2 + b x + c}^n} - \frac b a \int \frac {x^{2 n - 2} \rd x} {\paren {a x^2 + b x + c}^n}$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.276$
Category:Primitives involving a x squared plus b x plus c
Primitive of One plus x Squared over One plus Fourth Power of x

== Theorem ==
:$\ds \int \frac {x^2 + 1} {x^4 + 1} \rd x = \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C$
== Proof ==
We have:
Note that, by Derivative of Power: 
:$\dfrac \d {\d x} \paren {x - \dfrac 1 x} = 1 + \dfrac 1 {x^2}$
So, we have:
Category:Primitives involving Reciprocals
Primitive of Periodic Real Function

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $F$ be a primitive of $f$ that is bounded on all of $\R$.
Let $f$ be periodic with period $L$.
Then $F$ is also periodic with period $L$.
== Proof ==
Let $f$ be periodic with period $L$.
Let $f$ have a primitive $F$ that is bounded on all of $\R$.
By definition of a periodic function, it is seen that:
:$\map f x = \map f {x + L}$.
Then:
:$\ds \int \map f x \rd x$
and:
:$\ds \int \map f {x + L} \rd x$
are both primitives of the same function.
So by Primitives which Differ by Constant:
 $C \ne 0$.
Then it is to be shown that for all $k \in \N_{> 0}$:
:$\map F x + k C = \map F {x + k L}$
The case for $k = 1$ has already been proven, so this will be proven by induction.
Suppose that for some $n \in \N_{> 0}$ we have:
:$\map F x + n C = \map F {x + n L}$
Then
Holding $x$ fixed yields:
:$\ds \lim_{k \mathop \to \infty} \size {\map F {x + k L} } = \lim_{k \mathop \to \infty} \size {\map F x + k C} = \infty$
and so $\map F x$ is unbounded.
But we had previously established that $\map F x$ was bounded.
This is a contradiction.
Therefore our assumption that $C \ne 0$ was false.
Hence $C = 0$ and so:
:$\map F x = \map F {x + L}$
It has been shown that $F$ is periodic.
Let $L'$ be the period of $F$.
Then:
But it was previously established that $L$ was the period of $f$.
Hence it follows that $L' = L$.
Hence the result.
== Also see ==
* Derivative of Periodic Real Function
Category:Primitives
Category:Periodic Functions
Primitive of Pointwise Sum of Functions

== Theorem ==
Let $f_1, f_2, \ldots, f_n$ be real functions which are integrable.
Then:
:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Sources ==
* : $\S 14$: General Rules of Integration: $14.3$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.3.$
Category:Primitive of Pointwise Sum of Functions
Category:Primitives
Primitive of Pointwise Sum of Functions/Examples

== Examples of Use of Primitive of Pointwise Sum of Functions ==
=== Primitive of $f + g$ ===
=== Primitive of $u + v + w$ ===
=== Primitive of $u + v - w$ ===
=== Primitive of $3 x \paren {2 x^2 + 5 x - 4}$ ===
Category:Primitive of Pointwise Sum of Functions
Primitive of Pointwise Sum of Functions/Examples/f+g

== Examples of Use of Primitive of Pointwise Sum of Functions ==
Let $f$ and $g$ be real functions of $x$ which are integrable.
Then:
:$\ds \int \paren {\map f x + \map g x} \rd x = \int \map f x \rd x + \int \map g x \rd x$
== Proof ==
This is an instance of Primitive of Pointwise Sum of Functions:
:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
where $f = f_1$ and $g = f_2$.
== Sources ==
* : Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms
* : Appendix $\text I$: Table of Indefinite Integrals $2$.
* : Chapter $25$: Fundamental Integration Formulas: $2$.
* 
Category:Primitive of Pointwise Sum of Functions
Primitive of Pointwise Sum of Functions/Examples/u+v+w

== Examples of Use of Primitive of Pointwise Sum of Functions ==
Let $u$, $v$ and $w$ be real functions of $x$ which are integrable.
Then:
:$\ds \int \paren {u + v + w} \rd x = \int u \rd x + \int v \rd x + \int w \rd x$
== Proof ==
This is an instance of Primitive of Pointwise Sum of Functions:
:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
where $u = f_1$, $v = f_2$ and $w = f_3$.
== Sources ==
* : Chapter $\text {III}$: Integration: Three rules for integration: $\text {II}$
Category:Primitive of Pointwise Sum of Functions
Primitive of Pointwise Sum of Functions/Examples/u+v-w

== Examples of Use of Primitive of Pointwise Sum of Functions ==
Let $u$, $v$ and $w$ be real functions of $x$ which are integrable.
Then:
:$\ds \int \paren {u + v - w} \rd x = \int u \rd x + \int v \rd x - \int w \rd x$
== Proof ==
This is an instance of Primitive of Pointwise Sum of Functions:
:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
where $u = f_1$, $v = f_2$ and $w = f_3$.
== Sources ==
* : Chapter $10$: Integration: $10.4$. Standard integrals: General Rules: $\text I$.
Category:Primitive of Pointwise Sum of Functions
Primitive of Pointwise Sum of Functions/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
$\map P 1$ is true, as this just says:
:$\ds \int \map {f_1} x \rd x = \int \map {f_1} x \rd x$
=== Basis for the Induction ===
$\map P 2$ is the case:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x$
Then we need to show:
:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_{k + 1} } x \rd x$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: \ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
Category:Primitive of Pointwise Sum of Functions
Primitive of Pointwise Sum of Functions/Proof 2

== Theorem ==
== Proof ==
From Sum Rule for Derivatives:
:$\ds \map {\dfrac \d {\d x} } {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {\dfrac \d {\d x} } {\map {f_i} x}$
The result follows by definition of primitive.
== Sources ==
* : Chapter $\text {III}$: Integration: Three rules for integration: $\text {II}$
Category:Primitive of Pointwise Sum of Functions
Primitive of Power

== Theorem ==
Let $n \in \R: n \ne -1$.
Then:
:$\ds \int x^n \rd x = \frac {x^{n + 1} } {n + 1} + C$
where $C$ is an arbitrary constant.
That is:
:$\dfrac {x^{n + 1} } {n + 1}$ is a primitive of $x^n$.
== Proof ==
== Also known as ==
== Examples ==
== Also see ==
* Primitive of Reciprocal for the case where $n = -1$
== Sources ==
* : Chapter $\text {III}$: Integration: Integration
* : $\text {II}$. Calculus: Integration
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(i) (a)}$
* : $\S 14$: General Rules of Integration: $14.7$
* : Appendix $\text I$: Table of Indefinite Integrals $5$.
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
* : Appendix $2$: Table of derivatives and integrals of common functions
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.7.$
* : Appendix $7$: Integrals
* : Appendix $8$: Integrals
Category:Primitive of Power
Category:Primitives
Primitive of Power/Proof

== Theorem ==
== Proof ==
When $n = -1$ we have $n + 1 = 0$, and $\dfrac {x^{n + 1} } {n + 1} = \dfrac {x^0} 0$ is undefined.
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $1$.
* : Chapter $\text {III}$: Integration: Integration
* : $\text {II}$. Calculus: Integration
* : Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms
* : Chapter $25$: Fundamental Integration Formulas: $4$.
* : Chapter $5$. Integrals: Integrals of Special Functions: $1$
* : $9$ Sums and Integrals: $9.8$ Standard Integrals
* 
Category:Primitive of Power
Primitive of Power of Cosecant of a x

== Theorem ==
:$\ds \int \csc^n a x \rd x = \frac{-\csc^{n - 2} a x \cot a x} {a \paren {n - 1} } + \frac {n - 2} {n - 1} \int \csc^{n - 2} a x \rd x$
where $n \ne -1$.
== 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 $\csc a x$ for the case where $n = 1$
* Primitive of $\sin^n a x$
* Primitive of $\cos^n a x$
* Primitive of $\tan^n a x$
* Primitive of $\cot^n a x$
* Primitive of $\sec^n a x$
== Sources ==
* : $\S 14$: Integrals involving $\csc a x$: $14.470$
* : Back endpapers: A Brief Table of Integrals: $93$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(23)$ Integrals Involving $\csc a x$: $17.23.10.$
Category:Primitives involving Cosecant Function
Primitive of Power of Cosecant of a x by Cotangent of a x

== Theorem ==
:$\ds \int \csc^n a x \cot a x \rd x = \frac {-\csc^n a x} {n a} + C$
for $n \ne 0$.
== Proof ==
== Also see ==
* Primitive of $\sec^n a x \tan a x$
== Sources ==
* : $\S 14$: Integrals involving $\csc a x$: $14.464$
* : Back endpapers: A Brief Table of Integrals: $95$.
Category:Primitives involving Cosecant Function
Category:Primitives involving Cotangent Function
Primitive of Power of Cosine of a x by Sine of a x

== Theorem ==
:$\ds \int \cos^n a x \sin a x \rd x = \frac {-\cos^{n + 1} a x} {\paren {n + 1} a} + C$
for $n \ne -1$.
== Proof ==
== Also see ==
* Primitive of $\sin^n a x \cos a x$
For $n = -1$, use Primitive of $\dfrac {\sin a x} {\cos a x}$.
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.402$
* : Front endpapers: A Brief Table of Integrals: $66$.
Category:Primitives involving Sine Function and Cosine Function
Primitive of Power of Cosine of a x over Power of Sine of a x/Reduction of Both Powers

== Theorem ==
:$\ds \int \frac {\cos^m a x} {\sin^n a x} \rd x = \frac {-\cos^{m - 1} a x} {a \paren {n - 1} \sin^{n - 1} a x} - \frac {m - 1} {n - 1} \int \frac {\cos^{m - 2} a x} {\sin^{n - 2} a x} \rd x + C$
== 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 $\dfrac {\sin^m a x} {\cos^n a x}$: Reduction of Both Powers
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.427$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Power of Cosine of a x over Power of Sine of a x/Reduction of Power of Cosine

== Theorem ==
:$\ds \int \frac {\cos^m a x} {\sin^n a x} \rd x = \frac {\cos^{m - 1} a x} {a \paren {m - n} \sin^{n - 1} a x} + \frac {m - 1} {m - n} \int \frac {\cos^{m - 2} a x} {\sin^n a x} \rd x + C$
== 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:
Thus:
and so:
Then let:
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.427$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Power of Cosine of a x over Power of Sine of a x/Reduction of Power of Sine

== Theorem ==
:$\ds \int \frac {\cos^m a x} {\sin^n a x} \rd x = \frac {-\cos^{m + 1} a x} {a \paren {n - 1} \sin^{n - 1} a x} - \frac {m - n + 2} {n - 1} \int \frac {\cos^m a x} {\sin^{n - 2} a x} \rd x + C$
== Proof ==
Hence, rearranging:
== Also see ==
* Primitive of $\dfrac {\sin^m a x} {\cos^n a x}$ : Reduction of $\cos^n$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.427$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Power of Cotangent of a x

== Theorem ==
:$\ds \int \cot^n a x \rd x = \frac {-\cot^{n - 1} a x} {\paren {n - 1} a} - \int \cot^{n - 2} a x \rd x$
for $n \ne 1$.
== Proof ==
== Also see ==
* Primitive of $\cot a x$ for $n = 1$
* Primitive of $\sin^n a x$
* Primitive of $\cos^n a x$
* Primitive of $\tan^n a x$
* Primitive of $\sec^n a x$
* Primitive of $\csc^n a x$
== Sources ==
* : $\S 14$: Integrals involving $\cot a x$: $14.450$
* : Back endpapers: A Brief Table of Integrals: $87$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(21)$ Integrals Involving $\cot a x$: $17.21.11.$
Category:Primitives involving Cotangent Function
Primitive of Power of Cotangent of a x by Square of Cosecant of a x

== Theorem ==
:$\ds \int \cot^n a x \csc^2 a x \rd x = \frac {-\cot^{n + 1} a x} {\paren {n + 1} a} + C$
== Proof ==
== Also see ==
* Primitive of $\tan^n a x \sec^2 a x$
== Sources ==
* : $\S 14$: Integrals involving $\cot a x$: $14.443$
Category:Primitives involving Cotangent Function
Category:Primitives involving Cosecant Function