nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of p plus q by Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p + q \sin a x} = \begin{cases} 
\ds \frac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\frac {p \tan \dfrac {a x} 2 + q} {\sqrt {p^2 - q^2} } } + C & : q^2 - p^2 < 0 \\
\ds \frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\frac {p \tan \dfrac {a x} 2 + q - \sqrt {p^2 - q^2} } {p \tan \dfrac {a x} 2 + q + \sqrt {p^2 - q^2} } } + C & : q^2 - p^2 > 0 \\
\end{cases}$
== Proof ==
The discriminant of $p u^2 + 2 q u + p$ is $4 q^2 - 4 p^2$.
Thus:
== Examples ==
== Also presented as ==
== Also see ==
* Primitive of $\dfrac 1 {p + q \cos a x}$
* Primitive of $\dfrac 1 {p + q \tan a x}$
* Primitive of $\dfrac 1 {p + q \cot a x}$
* Primitive of $\dfrac 1 {q + p \sec a x}$
* Primitive of $\dfrac 1 {q + p \csc a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.360$
Category:Primitive of Reciprocal of p plus q by Sine of a x
Category:Primitives involving Sine Function
Primitive of Reciprocal of p plus q by Sine of a x/Also presented as

== Primitive of $\frac 1 {p + q \sin a x}$: Also presented as ==
This result can be seen presented in different forms.
=== $p^2 > q^2$: Also presented as ===
=== $p^2 < q^2$: Also presented as ===
Category:Primitive of Reciprocal of p plus q by Sine of a x
Primitive of Reciprocal of p plus q by Sine of a x/Examples/3 + 2 sin x

== Example of Use of Primitive of $\dfrac 1 {p + q \sin a x}$ ==
:$\ds \int \dfrac {\d x} {3 + 2 \sin x} = \frac 2 {\sqrt 5} \map \arctan {\frac {3 \tan \dfrac x 2 + 2} {\sqrt 5} } + C$
== Proof ==
From Primitive of $\dfrac 1 {p + q \sin a x}$:
The result follows on setting $a = 1$, $p \gets 3$ and $q \gets 2$, as follows.
We have that $2^2 - 3^2 < 0$, and so:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $29$.
Category:Examples of Use of Primitive of Reciprocal of p plus q by Sine of a x
Primitive of Reciprocal of p plus q by Sine of a x/Weierstrass Substitution

== Lemma for Primitive of Reciprocal of $p + q \sin a x$ ==
The Weierstrass Substitution of $\ds \int \frac {\d x} {p + q \sin a x}$ is:
:$\ds \frac 2 a \int \frac {\d u} {p u^2 + 2 q u + p}$
where $u = \tan \dfrac {a x} 2$.
== Proof ==
Category:Primitives involving Sine Function
Category:Weierstrass Substitutions
Primitive of Reciprocal of p plus q by Sine of a x/p^2 greater than q^2/Also presented as

== Primitive of $\frac 1 {p + q \sin a x}$ for $p^2 > q^2$: Also presented as ==
The result for $p^2 > q^2$ is also seen presented in the form:
:$\ds \int \frac {\d x} {p + q \sin a x} = -\frac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\sqrt {\dfrac {p - q} {p + q} } \map \tan {\dfrac \pi 4 - \dfrac {p x} 2} } + C$
== Proof ==
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $70$.
Category:Primitive of Reciprocal of p plus q by Sine of a x
Primitive of Reciprocal of p plus q by Sine of a x/p^2 less than q^2/Also presented as

== Primitive of $\frac 1 {p + q \sin a x}$ for $p^2 < q^2$: Also presented as ==
The result for $p^2 < q^2$ is also seen presented in the form:
:$\ds \int \frac {\d x} {p + q \sin a x} = -\frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\dfrac {q + p \sin a x + \sqrt {q^2 - p^2} \cos a x} {p + q \sin a x} } + C$
== Proof ==
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $71$.
Category:Primitive of Reciprocal of p plus q by Sine of a x
Primitive of Reciprocal of p plus q by Tangent of a x

== Theorem ==
:$\ds \int \frac {\d x} {p + q \tan a x} = \frac {p x} {p^2 + q^2} + \frac q {a \paren {p^2 + q^2} } \ln \size {q \sin a x + p \cos a x} + C$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $\dfrac 1 {p + q \sin a x}$
* Primitive of $\dfrac 1 {p + q \cos a x}$
* Primitive of $\dfrac 1 {p + q \cot a x}$
* Primitive of $\dfrac 1 {q + p \sec a x}$
* Primitive of $\dfrac 1 {q + p \csc a x}$
== Sources ==
* : $\S 14$: Integrals involving $\tan a x$: $14.438$
Category:Primitives involving Tangent Function
Category:Primitive of Reciprocal of p plus q by Tangent of a x
Primitive of Reciprocal of p plus q by Tangent of a x/Proof 2

== Theorem ==
== Proof ==
We have:
:$\dfrac \d {\d x} \paren {q \sin a x + p \cos a x} = a q \cos a x - a p \sin a x$
Thus:
Category:Primitive of Reciprocal of p plus q by Tangent of a x
Primitive of Reciprocal of p squared by square of Sine of a x minus q squared by square of Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p^2 \sin^2 a x - q^2 \cos^2 a x} = \frac 1 {2 a p q} \ln \size {\frac {p \tan a x - q} {p \tan a x + q} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.424$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p squared by square of Sine of a x plus q squared by square of Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x} = \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.423$
Category:Primitives involving Sine Function and Cosine Function
Category:Primitive of Reciprocal of p squared by square of Sine of a x plus q squared by square of Cosine of a x
Primitive of Reciprocal of p squared by square of Sine of a x plus q squared by square of Cosine of a x/Proof 1

== Theorem ==
== Proof ==
Let $u = p^2 + q^2$ and $v = q^2 - p^2$.
Then:
Also:
Therefore:
Category:Primitive of Reciprocal of p squared by square of Sine of a x plus q squared by square of Cosine of a x
Primitive of Reciprocal of p squared minus Square of q by Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p^2 - q^2 \cosh^2 a x} = \begin {cases}
\dfrac 1 {2 a p \sqrt {p^2 - q^2} } \ln \size {\dfrac {p \tanh a x + \sqrt {p^2 - q^2} } {p \tanh a x - \sqrt {p^2 - q^2} } } + C & : p^2 > q^2 \\
\dfrac 1 {a p \sqrt {q^2 - p^2} } \arctan \dfrac {p \tanh a x} {\sqrt {q^2 - p^2} } + C & : p^2 < q^2 \\
\end {cases}$
== Proof ==
Let:
There are two cases to address.
First, suppose $p^2 > q^2$.
Then we have that $p^2 - q^2 > 0$, and so:
Now suppose $p^2 < q^2$.
Then we have that $p^2 - q^2 < 0$, and so:
== Also see ==
* Primitive of $\dfrac 1 {p^2 - q^2 \sinh^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.583$
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of p squared minus Square of q by Hyperbolic Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p^2 - q^2 \sinh^2 a x} = \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {p + \sqrt {p^2 + q^2} \tanh a x} {p - \sqrt {p^2 + q^2} \tanh a x} } + C$
== Proof ==
Then:
== Also see ==
* Primitive of $\dfrac 1 {p^2 - q^2 \cosh^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x$: $14.556$
Category:Primitives involving Hyperbolic Sine Function
Primitive of Reciprocal of p squared minus square of q by Cosine of a x

== Theorem ==
:$\ds \int \frac {\rd x} {p^2 - q^2 \cos^2 a x} = \begin {cases}
\dfrac 1 {a p \sqrt {p^2 - q^2} } \arctan \dfrac {p \tan a x} {\sqrt {p^2 - q^2} } & : p^2 > q^2 \\
\dfrac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\dfrac {p \tan a x - \sqrt {q^2 - p^2} } {p \tan a x + \sqrt {q^2 - p^2} } } & : p^2 < q^2
\end {cases}$
== Proof ==
Let:
There are two cases to address.
First, suppose $p^2 > q^2$.
Then we have that $p^2 - q^2 > 0$, and so:
Now suppose $p^2 < q^2$.
Then we have that $p^2 - q^2 < 0$, and so:
== Also see ==
* Primitive of $\dfrac 1 {p^2 - q^2 \sin^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\cos a x$: $14.393$
Category:Primitives involving Cosine Function
Primitive of Reciprocal of p squared minus square of q by Sine of a x

== Theorem ==
$\quad \ds \int \frac {\d x} {p^2 - q^2 \sin^2 a x} = \begin {cases}
\dfrac 1 {a p \sqrt {p^2 - q^2} } \arctan \dfrac {\sqrt {p^2 - q^2} \tan a x} p & : p^2 > q^2 \\
\dfrac 1 {2 a p \sqrt {q^2 - p^2} } \ln \size {\dfrac {\sqrt {q^2 - p^2} \tan a x + p} {\sqrt {q^2 - p^2} \tan a x - p} } & : p^2 < q^2
\end {cases}$
== Proof ==
Let:
There are two cases to address.
First, suppose $p^2 > q^2$.
Then we have that $p^2 - q^2 > 0$, and so:
Now suppose $p^2 < q^2$.
Then we have that $p^2 - q^2 < 0$, and so:
== Also see ==
* Primitive of $\dfrac 1 {p^2 - q^2 \cos^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.363$
Category:Primitives involving Sine Function
Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x

== Theorem ==
=== Logarithm Form ===
=== Inverse Hyperbolic Tangent Form ===
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.584$
Category:Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x/Inverse Hyperbolic Tangent Form

== Theorem ==
:$\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {a p \sqrt {p^2 + q^2} } \tanh^{-1} \dfrac {p \tanh a x} {\sqrt {p^2 + q^2} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.584$
Category:Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x
Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x/Logarithm Form

== Theorem ==
:$\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }$
== Proof ==
Let:
We have:
== Also see ==
* Primitive of $\dfrac 1 {p^2 + q^2 \sinh^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.584$
Category:Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x
Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x/Mistake

== Source Work ==
:Chapter $14$: Indefinite Integrals
::Integrals involving $\cosh a x$: $14.584$
This mistake can be seen in the edition as published by Schaum: ISBN 0-07-060224-7 (unknown printing).
== Mistake ==
:$\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \begin {cases}
\dfrac 1 {2 a p \sqrt {p^2 + q^2} } \map \ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } } \\ \\
\dfrac 1 {a p \sqrt {p^2 + q^2} } \arctan \dfrac {p \tanh a x} {\sqrt {p^2 + q^2} }
\end {cases}$
== Correction ==
As demonstrated in Primitive of $\dfrac {\d x} {p^2 + q^2 \cosh^2 a x}$, the second expression is incorrect.
It should just be:
:$\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \map \ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }$
The second expression is not completely wrong, however.
The result can also be presented as:
:$\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {a p \sqrt {p^2 + q^2} } \tanh^{-1} \dfrac {p \tanh a x} {\sqrt {p^2 + q^2} }$
but it is a genuine equivalence which is valid over the whole range.
Hence presenting it as two separate cases, each expected to valid over a subdomain of $\R$ (in this case unspecified), is misleading.
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.584$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p^2 + q^2 \sinh^2 a x} = \begin {cases}
\dfrac 1 {a p \sqrt{q^2 - p^2} } \arctan \dfrac {\sqrt {q^2 - p^2} \tanh a x} p + C & : p^2 < q^2 \\
\dfrac 1 {2 a p \sqrt{p^2 - q^2} } \ln \size {\dfrac {p + \sqrt {p^2 - q^2} \tanh a x} {p - \sqrt {p^2 - q^2} \tanh a x} } + C & : p^2 > q^2 \\
\end {cases}$
== Proof ==
Then:
There are two cases to address: $q^2 - p^2 > 0$ and $q^2 - p^2 < 0$.
First suppose that $q^2 - p^2 > 0$.
Then:
Now suppose that $q^2 - p^2 < 0$.
Then:
== Also see ==
* Primitive of $\dfrac 1 {p^2 + q^2 \cosh^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x$: $14.555$
Category:Primitives involving Hyperbolic Sine Function
Primitive of Reciprocal of p squared plus square of q by Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p^2 + q^2 \cos^2 a x} = \frac 1 {a p \sqrt{p^2 + q^2} } \arctan \frac {p \tan a x} {\sqrt {p^2 + q^2} } + C$
where $C$ is an arbitrary constant.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {p^2 + q^2 \sin^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\cos a x$: $14.392$
Category:Primitives involving Cosine Function
Primitive of Reciprocal of p squared plus square of q by Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p^2 + q^2 \sin^2 a x} = \frac 1 {a p \sqrt {p^2 + q^2} } \arctan \frac {\sqrt {p^2 + q^2} \tan a x} p + C$
where $C$ is an arbitrary constant.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {p^2 + q^2 \cos^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.362$
Category:Primitives involving Sine Function
Primitive of Reciprocal of p squared plus square of q by Sine of a x/Weierstrass Substitution

== Lemma for Primitive of Reciprocal of $\paren {p + q \sin a x}^2$ ==
The Weierstrass Substitution of $\ds \int \frac {\d x} {p^2 + q^2 \sin^2 a x}$ is:
:$\ds \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {p^2 \paren {u^2}^2 + \paren {2 p + 4 q^2} u^2 + p}$
== Proof ==
Category:Primitives involving Sine Function
Category:Weierstrass Substitutions
Primitive of Reciprocal of p x + q by Root of a x + b

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.
Then:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases}
\dfrac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C & : p \paren {b p - a q} > 0 \\ \\
\dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C & : p \paren {b p - a q} < 0 \\
\end {cases}$
== Proof ==
=== Case $1$: $p \paren {b p - a q} > 0$ ===
=== Case $2$: $p \paren {b p - a q} < 0$ ===
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.114$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.2.$
Category:Primitive of Reciprocal of p x + q by Root of a x + b
Category:Primitives involving Root of a x + b and p x + q
Primitive of Reciprocal of p x + q by Root of a x + b/Lemma

== Lemma for Primitive of $\frac 1 {\paren {p x + q} \sqrt {a x + b} }$ ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.
Then:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \dfrac 2 p \int \frac {\d u} {u^2 - \paren {\dfrac {b p - a q} p} }$
where:
:$u = \sqrt {a x + b}$
== Proof ==
Let:
and:
Then:
Category:Primitive of Reciprocal of p x + q by Root of a x + b
Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) greater than 0

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.
Let $p \paren {b p - a q} > 0$.
Then:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \frac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$
== Proof ==
=== Lemma ===
We have that:
:$p \paren {b p - a q} > 0$
which means:
:$\dfrac {b p - a q} p > 0$
Hence let:
:$d^2 = \dfrac {b p - a q} p$
Thus:
== Also presented as ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.30$
Category:Primitive of Reciprocal of p x + q by Root of a x + b
Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) greater than 0/Also presented as

== Primitive of $\frac 1 {\paren {p x + q} \sqrt {a x + b} }$ where $p \paren {b p - a q} > 0$: Also presented as ==
This result can also be seen presented as:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \frac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$
but this presupposes both that $p > 0$ and $b p - a q > 0$.
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.114$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.2.$
Category:Primitive of Reciprocal of p x + q by Root of a x + b
Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) less than 0

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.
Let $p \paren {b p - a q} > 0$.
Then:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C$
== Proof ==
=== Lemma ===
We have that:
:$p \paren {b p - a q} < 0$
which means:
:$\dfrac {b p - a q} p < 0$
Hence let:
:$d^2 = -\dfrac {b p - a q} p$
Thus:
== Also presented as ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.29$
Category:Primitive of Reciprocal of p x + q by Root of a x + b
Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) less than 0/Also presented as

== Primitive of $\frac 1 {\paren {p x + q} \sqrt {a x + b} }$ where $p \paren {b p - a q} < 0$: Also presented as ==
This result can also be seen presented as:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C$
but this presupposes both that $p > 0$ and $b p - a q < 0$.
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.114$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.2.$
Category:Primitive of Reciprocal of p x + q by Root of a x + b
Primitive of Reciprocal of p x + q by Root of a x + b by Root of p x + q

== Theorem ==
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } } = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C$
== Proof ==
From Primitive of $\paren {p x + q}^n \sqrt {a x + b}$:
:$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} }$
Putting $n = \dfrac 3 2$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $14.124$
* : $\S 17$: Tables of Special Indefinite Integrals: $(5)$ Integrals Involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $17.5.5.$
Category:Primitives involving Root of a x + b and Root of p x + q
Primitive of Reciprocal of q plus p by Cosecant of a x

== Theorem ==
:$\ds \int \frac {\d x} {q + p \csc a x} = \frac x q - \frac p q \int \frac {\d x} {p + q \sin a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {p + q \sin a x}$
* Primitive of $\dfrac 1 {p + q \cos a x}$
* Primitive of $\dfrac 1 {p + q \tan a x}$
* Primitive of $\dfrac 1 {p + q \cot a x}$
* Primitive of $\dfrac 1 {q + p \sec a x}$
== Sources ==
* : $\S 14$: Integrals involving $\csc a x$: $14.469$
Category:Primitives involving Cosecant Function
Primitive of Reciprocal of q plus p by Hyperbolic Cosecant of a x

== Theorem ==
:$\ds \int \frac {\d x} {q + p \csch a x} = \frac x q - \frac p q \int \frac {\d x} {p + q \sinh a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {p + q \sinh a x}$
* Primitive of $\dfrac 1 {p + q \cosh a x}$
* Primitive of $\dfrac 1 {p + q \tanh a x}$
* Primitive of $\dfrac 1 {p + q \coth a x}$
* Primitive of $\dfrac 1 {q + p \sech a x}$
== Sources ==
* : $\S 14$: Integrals involving $\csch a x$: $14.644$
Category:Primitives involving Hyperbolic Cosecant Function
Primitive of Reciprocal of q plus p by Hyperbolic Secant of a x

== Theorem ==
:$\ds \int \frac {\d x} {q + p \sech a x} = \frac x q - \frac p q \int \frac {\d x} {p + q \cosh a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {p + q \sinh a x}$
* Primitive of $\dfrac 1 {p + q \cosh a x}$
* Primitive of $\dfrac 1 {p + q \tanh a x}$
* Primitive of $\dfrac 1 {p + q \coth a x}$
* Primitive of $\dfrac 1 {q + p \csch a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sech a x$: $14.634$
Category:Primitives involving Hyperbolic Secant Function
Primitive of Reciprocal of q plus p by Secant of a x

== Theorem ==
:$\ds \int \frac {\d x} {q + p \sec a x} = \frac x q - \frac p q \int \frac {\d x} {p + q \cos a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {p + q \sin a x}$
* Primitive of $\dfrac 1 {p + q \cos a x}$
* Primitive of $\dfrac 1 {p + q \tan a x}$
* Primitive of $\dfrac 1 {p + q \cot a x}$
* Primitive of $\dfrac 1 {q + p \csc a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sec a x$: $14.459$
Category:Primitives involving Secant Function
Primitive of Reciprocal of square of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^2} = \frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$
== Proof ==
Let:
Then:
Let $u = z^2$.
Let:
Recall the result Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$:
:$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$
Let:
Then:
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.272$
Category:Primitives involving a x squared plus b x plus c
Primitive of Reciprocal of square of p plus q by Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\paren {p + q \cos a x}^2} = \frac {q \sin a x} {a \paren {q^2 - p^2} \paren {p + q \cos a x} } - \frac p {q^2 - p^2} \int \frac {\d x} {p + q \cos a x}$
== Proof ==
Hence the result.
== Also see ==
* Primitive of $\dfrac 1 {\paren {p + q \sin a x}^2}$
== Sources ==
* : $\S 14$: Integrals involving $\cos a x$: $14.391$
Category:Primitives involving Cosine Function
Primitive of Reciprocal of square of p plus q by Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\paren {p + q \sin a x}^2} = \frac {q \cos a x} {a \paren {p^2 - q^2} \paren {p + q \sin a x} } + \frac p {p^2 - q^2} \int \frac {\d x} {p + q \sin a x}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $\dfrac 1 {\paren {p + q \cos a x}^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.361$
Category:Primitives involving Sine Function
Category:Primitive of Reciprocal of square of p plus q by Sine of a x
Primitive of Reciprocal of square of p plus q by Sine of a x/Lemma

== Lemma for Primitive of $\dfrac 1 {\paren {p + q \sin a x}^2}$ ==
Let $u = \tan \dfrac \theta 2$.
Then:
:$\dfrac 1 {p + q \sin a x} = \dfrac {u^2 + 1} {p u^2 + 2 q u + p}$
== Proof ==
From Tangent Half-Angle Substitution for Sine:
Category:Primitive of Reciprocal of square of p plus q by Sine of a x
Primitive of Reciprocal of square of p plus q by Sine of a x/Weierstrass Substitution

== Lemma for Primitive of Reciprocal of $\paren {p + q \sin a x}^2$ ==
The Weierstrass Substitution for $\ds \int \frac {\d x} {\paren {p + q \sin a x}^2}$ yields:
:$\ds \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}$
where $u = \tan \dfrac {a x} 2$.
== Proof ==
Category:Primitive of Reciprocal of square of p plus q by Sine of a x
Category:Weierstrass Substitutions
Primitive of Reciprocal of x by Cube of Root of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3} = \frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.293$
Category:Primitives involving Root of a x squared plus b x plus c