nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of a x squared + b by Root of c x squared + d/b c greater than a d

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers such that $a d \ne b c$.
Let $a d > b c$.
Then:
:$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac 1 {2 \sqrt {b \paren {a d - b c} } } \ln \size {\dfrac {\sqrt {b \paren {c x^2 + d} } + x \sqrt {b c - a d} } {\sqrt {b \paren {c x^2 + d} } - x \sqrt {b c - a d} } } + C$
== Proof ==
=== Lemma ===
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.50$
Category:Primitive of Reciprocal of a x squared + b by Root of c x squared + d
Primitive of Reciprocal of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:
$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end {cases}$
=== $a$ equal to Zero ===
=== $b$ equal to Zero ===
=== $c$ equal to Zero ===
== Proof ==
First:
Put:
Let $D = b^2 - 4 a c$.
Thus:
=== Negative Discriminant ===
=== Positive Discriminant ===
=== Zero Discriminant ===
== Also presented as ==
== Examples ==
== Also see ==
* Primitive of $\dfrac 1 {a x + b}$ for the case where $a = 0$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.265$
* : $\S 17$: Tables of Special Indefinite Integrals: $(12)$ Integrals Involving $a x^2 + b x + c$: $17.12.1.$
Category:Primitive of Reciprocal of a x squared plus b x plus c
Category:Primitives involving a x squared plus b x plus c
Category:Primitives involving Reciprocals
Primitive of Reciprocal of a x squared plus b x plus c/Also presented as

== Primitive of $\dfrac 1 {a x^2 + b x + c}$: Also presented as ==
In some older works, this result can also be seen presented as:
:$\ds \int \frac {\d x} {a x^2 + 2 b x + c}$
where the solution is then developed via the form:
:$\ds \dfrac 1 a \int \frac {\d x} {\paren {x + \frac b a}^2 + \paren {\frac c a - \frac {b^2} {a^2} } }$
== Sources ==
* : $\text {II}$. Calculus: Integration: Algebraic Integration
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/Examples

== Examples of Use of Primitive of $\dfrac 1 {a x^2 + b x + c}$ ==
=== Primitive of $\dfrac 1 {3 x^2 + 4 x + 2}$ ===
=== Primitive of $\dfrac 1 {x^2 + 4 x + 5}$ ===
=== Primitive of $\dfrac 1 {x^2 + 2 a x + b}$ ===
Category:Examples of Use of Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2

== Example of Use of Primitive of $\dfrac 1 {a x^2 + b x + c}$ ==
:$\ds \int \frac {\d x} {3 x^2 + 4 x + 2} = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C$
== Proof 1 ==
== Proof 2 ==
Category:Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2
Category:Examples of Use of Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2/Proof 1

== Example of Use of Primitive of $\dfrac 1 {a x^2 + b x + c}$ ==
== Proof ==
We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:
We note that:
Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
Substituting for $a$, $b$ and $c$ and simplifying:
Category:Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2
Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2/Proof 2

== Example of Use of Primitive of $\dfrac 1 {a x^2 + b x + c}$ ==
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Integration: Algebraic Integration
Category:Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2
Primitive of Reciprocal of a x squared plus b x plus c/Examples/x^2 + 2 a x + b

== Example of Use of Primitive of $\dfrac 1 {x^2 + 2 a x + b}$ ==
:$\ds \int \frac {\d x} {x^2 + 2 a x + b} = \dfrac 1 {\sqrt {b - a^2} } \map \arctan {\dfrac {x + a} {\sqrt {b - a^2} } } + C$
where $b > a^2$.
== Proof ==
We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:
We note that:
Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
Substituting for $a$, $b$ and $c$ :
== Sources ==
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
Category:Examples of Use of Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/Examples/x^2 + 4 x + 5

== Example of Use of Primitive of $\dfrac 1 {a x^2 + b x + c}$ ==
:$\ds \int \dfrac {\d x} {x^2 + 4 x + 5} = \map \arctan {x + 2} + C$
== Proof ==
We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:
We note that:
Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
Substituting for $a$, $b$ and $c$ and simplifying:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $15$.
Category:Examples of Use of Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/Negative Discriminant

== Theorem ==
Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c < 0$.
Then:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
== Proof ==
First:
Put:
Let $D = b^2 - 4 a c$.
Thus:
Let $b^2 - 4 a c < 0$.
Then:
Thus:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.16$
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/Positive Discriminant

== Theorem ==
Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c > 0$.
Then:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C$
== Proof ==
First:
Put:
Let $D = b^2 - 4 a c$.
Thus:
Let $b^2 - 4 a c > 0$.
Then:
Thus:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.17$
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/Zero Discriminant

== Theorem ==
Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c = 0$.
Then:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac {-2} {2 a x + b} + C$
== Proof ==
First:
Let $b^2 - 4 a c = 0$.
Then:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.18$
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.265$
* : $\S 17$: Tables of Special Indefinite Integrals: $(12)$ Integrals Involving $a x^2 + b x + c$: $17.12.1.$
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/a equal to 0

== Theorem ==
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 b \ln \size {b x + c} + C$
when $a = 0$.
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + bx + c$: $14.265$
* : $\S 17$: Tables of Special Indefinite Integrals: $(12)$ Integrals Involving $a x^2 + b x + c$: $17.12.1.$
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/b equal to 0

== Theorem ==
Let $a \in \R_{\ne 0}$.
Let $b = 0$.
Then:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C & : a c > 0 \\ \\
\dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C & : a c < 0 \\ \\
\dfrac {-1} {a x} + C & : c = 0
\end {cases}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.265$
* : $\S 17$: Tables of Special Indefinite Integrals: $(12)$ Integrals Involving $a x^2 + b x + c$: $17.12.1.$
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/b equal to 0/Proof 1

== Theorem ==
Let $a \in \R_{\ne 0}$.
== Proof ==
First:
Let $a c > 0$.
Then $\dfrac c a > 0$ and:
Thus:
Let $a c < 0$.
Then $\dfrac c a > 0$ and:
Thus:
Let $c = 0$.
Then:
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/b equal to 0/Proof 2

== Theorem ==
Let $a \in \R_{\ne 0}$.
== Proof ==
Let $b = 0$.
From Primitive of Reciprocal of a x squared plus b x plus c, we have:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end {cases}$
Let $a c > 0$.
Then $b^2 - 4 a c = 0 - 4 a c < 0$ and so:
Let $a c < 0$.
Then $b^2 - 4 a c = 0 - 4 a c > 0$ and so:
Let $c = 0$.
Then $b^2 - 4 a c = 0 - 0 = 0$ and so:
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/c equal to 0

== Theorem ==
Let $a, b \in \R_{\ne 0}$.
Let $c = 0$.
Then:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 b \ln \size {\frac x {a x + b} } + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + bx + c$: $14.265$
* : $\S 17$: Tables of Special Indefinite Integrals: $(12)$ Integrals Involving $a x^2 + b x + c$: $17.12.1.$
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of a x squared plus b x plus c/c equal to 0/Proof 2

== Theorem ==
== Proof ==
Let $c = 0$.
From Primitive of $\dfrac 1 {a x^2 + b x + c}$, we have:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end{cases}$
As $b \ne 0$ it follows that $b^2 - 4 a c = b^2 > 0$.
Thus:
Category:Primitive of Reciprocal of a x squared plus b x plus c
Primitive of Reciprocal of p by Exponential of a x plus q by Exponential of -a x

== Theorem ==
:$\ds \int \frac {\d x} {p e^{a x} + q e^{-a x} } = \begin{cases}
\dfrac 1 {a \sqrt {p q} } \map \arctan {\sqrt {\dfrac p q} e^{a x} } & : \sqrt {p q} > 0 \\
\dfrac 1 {2 a \sqrt {-p q} } \ln \size {\dfrac {e^{a x} - \sqrt {-\dfrac q p} } {e^{a x} + \sqrt {-\dfrac q p} } } & : \sqrt {p q} < 0 \\
\end{cases}$
== Proof ==
Let $\dfrac q p > 0$.
Then:
Let $\dfrac q p < 0$.
Then let $d^2 = -\dfrac q p$ and so:
== Sources ==
* : $\S 14$: Integrals involving $e^{a x}$: $14.517$
Category:Primitives involving Exponential Function
Primitive of Reciprocal of p by Sine of a x plus q by 1 plus Cosine of a x

== Theorem ==
:$\ds \int \frac {\rd x} {p \sin a x + q \paren {1 + \cos a x} } = \frac 1 {a p} \ln \size {q + p \tan \frac {a x} 2} + C$
== Proof ==
Let $z = a x$.
Then $\d x = \dfrac {\d z} a$ and so:
$(1): \quad \ds \int \frac {\rd x} {p \sin a x + q \paren {1 + \cos a x} } = \dfrac 1 a \int \frac {\rd z} {p \sin z + q \paren {1 + \cos z} }$
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.421$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p \sin a x + q \cos a x} = \frac 1 {a \sqrt {p^2 + q^2} } \ln \tan \size {\frac {a x + \arctan \dfrac q p} 2} + C$
== Proof ==
=== Lemma ===
== Also see ==
* Primitive of $\dfrac 1 {p \sin a x + q \cos a x + r}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.419$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x/Lemma

== Lemma for Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x ==
:$\ds \frac 1 2 \map \arctan {\dfrac {-p} q} + \frac \pi 4 = \frac {\arctan \dfrac q p} 2$
== Proof ==
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x minus Root of p squared plus q squared

== Theorem ==
:$\ds \int \frac {\d x} {p \sin a x + q \cos a x - \sqrt {p^2 + q^2} } = \frac {-1} {a \sqrt {p^2 + q^2} } \map \tan {\frac \pi 4 + \frac {a x + \arctan \frac q p} 2} + C$
== Proof ==
Let $\theta = \arctan \dfrac p q$.
Then by the definitions of sine, cosine and tangent:
Now consider:
Then let $\theta' = \arctan \dfrac q p$.
From Arctangent of Reciprocal equals Arccotangent:
:$\theta' = \arccot \dfrac p q$.
Hence:
Thus:
== Also see ==
* Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x plus r: the general case for where $r^2 \ne p^2 + q^2$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.422$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x plus Root of p squared plus q squared

== Theorem ==
:$\ds \int \frac {\d x} {p \sin a x + q \cos a x + \sqrt {p^2 + q^2} } = \frac {-1} {a \sqrt {p^2 + q^2} } \map \tan {\frac \pi 4 - \frac {a x + \arctan \frac q p} 2} + C$
== Proof ==
Let $\theta = \arctan \dfrac p q$.
Then by the definitions of sine, cosine and tangent:
Now consider:
Then let $\theta' = \arctan \dfrac q p$.
From Arctangent of Reciprocal equals Arccotangent:
:$\theta' = \arccot \dfrac p q$.
Hence:
Thus:
== Also see ==
* Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x plus r: the general case for where $r^2 \ne p^2 + q^2$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.422$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x plus or minus Root of p squared plus q squared

== Theorem ==
:$\ds \int \frac {\d x} {p \sin a x + q \cos a x \pm \sqrt {p^2 + q^2} } = \frac {-1} {a \sqrt {p^2 + q^2} } \map \tan {\frac \pi 4 \mp \frac {a x + \arctan \frac q p} 2} + C$
=== Primitive of $\dfrac 1 {p \sin a x + q \cos a x + \sqrt {p^2 + q^2}}$ ===
=== Primitive of $\dfrac 1 {p \sin a x + q \cos a x - \sqrt {p^2 + q^2}}$ ===
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.422$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x plus r

== Theorem ==
:$\ds \int \frac {\d x} {p \sin a x + q \cos a x + r} = \begin{cases}
\ds \frac 2 {a \sqrt {r^2 - p^2 - q^2} } \map \arctan {\frac {p + \paren {r - q} \tan \dfrac {a x} 2} {\sqrt {r^2 - p^2 - q^2} } } + C & : p^2 + q^2 < r^2 \\
\ds \frac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \size {\frac {p - \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} {p + \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} } + C & : p^2 + q^2 > r^2
\end{cases}$
== Proof ==
Let $u = \tan \dfrac {a x} 2$.
Then we have:
Hence:
There are $2$ cases: $r^2 > p^2 + q^2$ and $r^2 < p^2 + q^2$.
First suppose $r^2 > p^2 + q^2$.
Then:
:$r^2 - \paren {p^2 + q^2} > 0$
and so $(1)$ may be written as:
Next suppose $r^2 < p^2 + q^2$.
Then:
:$r^2 - \paren {p^2 + q^2} < 0$
and so $(1)$ may be written as:
== Also see ==
* Primitive of $\dfrac 1 {p \sin a x + q \cos a x}$ for the case where $r = 0$
* Primitive of $\dfrac 1 {p \sin a x + q \paren {1 + \cos a x} }$ for the case where $r = q$
* Primitive of $\dfrac 1 {p \sin a x + q \cos a x + \sqrt {p^2 + q^2} }$ for the case where $r = \sqrt {p^2 + q^2}$
* Primitive of $\dfrac 1 {p \sin a x + q \cos a x - \sqrt {p^2 + q^2} }$ for the case where $r = -\sqrt {p^2 + q^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.420$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of p plus q by Cosine of a x

== Theorem ==
:$\ds \int \frac {\rd x} {p + q \cos a x} = \begin {cases} 
\dfrac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\sqrt {\dfrac {p - q} {p + q} } \tan \dfrac {a x} 2} + C & : p^2 > q^2 \\
\dfrac 1 {a \sqrt {q^2 - p^2} } \ln \size {\dfrac {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } } + C & : p^2 < q^2 \\
\end {cases}$
for $p \ne q$.
== Proof ==
== Examples ==
== Also presented as ==
== Also see ==
* For $p = q$: Primitive of $\dfrac 1 {1 + \cos a x}$
* For $p = -q$: Primitive of $\dfrac 1 {1 - \cos a x}$
* Primitive of $\dfrac 1 {p + q \sin a x}$
* 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 $\cos a x$: $14.390$
* : Back endpapers: A Brief Table of Integrals: $74$.
Category:Primitive of Reciprocal of p plus q by Cosine of a x
Category:Primitives involving Cosine Function
Primitive of Reciprocal of p plus q by Cosine of a x/Examples/5 + 3 cos x

== Example of Use of Primitive of $\dfrac 1 {p + q \cos a x}$ ==
:$\ds \int \dfrac {\d x} {5 + 3 \cos x} = \dfrac 1 2 \map \arctan {\dfrac 1 2 \tan \dfrac x 2} + C$
== Proof ==
From Primitive of $\dfrac 1 {p + q \cos a x}$:
The result follows on setting $a \gets 1$, $p \gets 5$ and $q \gets 3$, as follows.
We have that $5^2 > 3^2$, and so:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $28$.
Category:Examples of Use of Primitive of Reciprocal of p plus q by Cosine of a x
Primitive of Reciprocal of p plus q by Cosine of a x/Proof 1

== Theorem ==
== Proof ==
Let $p^2 > q^2$.
Then, by Sign of Quotient of Factors of Difference of Squares:
:$\dfrac {p + q} {p - q} > 0$
Thus, let $\dfrac {p + q} {p - q} = d^2$.
Then:
Now let $p^2 < q^2$.
Then, by Sign of Quotient of Factors of Difference of Squares:
:$\dfrac {p + q} {p - q} < 0$
Thus, let:
:$-\dfrac {p + q} {p - q} = d^2$
or:
:$\dfrac {q + p} {q - p} = d^2$
Then:
Category:Primitive of Reciprocal of p plus q by Cosine of a x
Primitive of Reciprocal of p plus q by Cosine of a x/Weierstrass Substitution

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

== Primitive of $\frac 1 {p + q \cos a x}$ for $p^2 < q^2$: Also presented as ==
The result for $p^2 < q^2$ is also seen presented in the form:
:$\ds \int \frac {\d x} {p + q \cos a x} = \frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\dfrac {q + p \cos a x + \sqrt {q^2 - p^2} \sin a x} {p + q \cos a x} } + C$
== Proof ==
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $75$.
Category:Primitive of Reciprocal of p plus q by Cosine of a x
Primitive of Reciprocal of p plus q by Cotangent of a x

== Theorem ==
:$\ds \int \frac {\d x} {p + q \cot a x} = \frac {p x} {p^2 + q^2} - \frac q {a \paren {p^2 + q^2} } \ln \size {p \sin a x + q \cos a x} + C$
== Proof ==
We have:
:$\dfrac \d {\d x} \paren {p \sin a x + q \cos a x} = a p \cos a x - a q \sin a x$
Thus:
== 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 $\dfrac 1 {q + p \sec a x}$
* Primitive of $\dfrac 1 {q + p \csc a x}$
== Sources ==
* : $\S 14$: Integrals involving $\cot a x$: $14.449$
Category:Primitives involving Cotangent Function
Primitive of Reciprocal of p plus q by Exponential of a x

== Theorem ==
:$\ds \int \frac {\d x} {p + q e^{a x} } = \frac x p - \frac 1 {a p} \ln \size {p + q e^{a x} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $e^{a x}$: $14.515$
Category:Primitives involving Exponential Function
Primitive of Reciprocal of p plus q by Hyperbolic Cosine of a x

== Theorem ==
For $p \ne q$:
:$\ds \int \frac {\d x} {p + q \cosh a x} = \begin {cases}
\dfrac 2 {a \sqrt {q^2 - p^2} } \arctan \dfrac {q e^{a x} + p} {\sqrt {q^2 - p^2} } + C & : p^2 < q^2 \\
\dfrac 1 {a \sqrt {p^2 - q^2} } \ln \size {\dfrac {q e^{a x} + p - \sqrt {p^2 - q^2} } {q e^{a x} + p + \sqrt {p^2 - q^2} } } + C & : p^2 > q^2
\end {cases}$
== Proof ==
Let:
Hence:
The discriminant of $q u^2 + 2 p u + q$ is given by:
Hence the result depends on the sign of $p^2 - q^2$.
Let $p^2 < q^2$.
Then:
Let $p^2 > q^2$.
Then:
== Also see ==
* Primitive of $\dfrac 1 {\cosh a x + 1}$ for the case where $p = q$
* Primitive of $\dfrac 1 {p + q \sinh a x}$
* Primitive of $\dfrac 1 {p + q \tanh a x}$
* Primitive of $\dfrac 1 {p + q \coth a x}$
* Primitive of $\dfrac 1 {q + p \sech a x}$
* Primitive of $\dfrac 1 {q + p \csch a x}$
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.581$
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of p plus q by Hyperbolic Cotangent of a x

== Theorem ==
:$\ds \int \frac {\d x} {p + q \coth a x} = \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {p \sinh a x + q \cosh a x} + C$
== Proof ==
We have:
:$\dfrac \d {\d x} \paren {p \sinh a x + q \cosh a x} = a p \cosh a x + a q \sinh a x$
Thus:
== 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 $\dfrac 1 {q + p \sech a x}$
* Primitive of $\dfrac 1 {q + p \csch a x}$
== Sources ==
* : $\S 14$: Integrals involving $\coth a x$: $14.624$
* : $\S 17$: Tables of Special Indefinite Integrals: $(31)$ Integrals Involving $\coth a x$: $17.31.7.$
Category:Primitives involving Hyperbolic Cotangent Function
Primitive of Reciprocal of p plus q by Hyperbolic Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {p + q \sinh a x} = \frac 1 {a \sqrt{p^2 + q^2} } \ln \size {\frac {q e^{a x} + p - \sqrt {p^2 + q^2} } {q e^{a x} + p + \sqrt {p^2 + q^2} } } + C$
== Proof ==
Let:
Hence:
The discriminant of $q u^2 + 2 p u - q$ is given by:
Hence the sign of $\map {\operatorname {Disc} } {q u^2 + 2 p u - q}$ is always positive.
So:
== Also see ==
* 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 $\dfrac 1 {q + p \sech a x}$
* Primitive of $\dfrac 1 {q + p \csch a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x$: $14.553$
Category:Primitives involving Hyperbolic Sine Function
Primitive of Reciprocal of p plus q by Hyperbolic Tangent of a x

== Theorem ==
:$\ds \int \frac {\d x} {p + q \tanh a x} = \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {q \sinh a x + p \cosh a x} + C$
== Proof ==
We have:
:$\dfrac \d {\d x} \paren {q \sinh a x + p \cosh a x} = a q \cosh a x + a p \sinh a x$
Thus:
== Also see ==
* Primitive of $\dfrac 1 {p + q \sinh a x}$
* Primitive of $\dfrac 1 {p + q \cosh a x}$
* Primitive of $\dfrac 1 {p + q \coth a x}$
* Primitive of $\dfrac 1 {q + p \sech a x}$
* Primitive of $\dfrac 1 {q + p \csch a x}$
== Sources ==
* : $\S 14$: Integrals involving $\tanh a x$: $14.613$
* : $\S 17$: Tables of Special Indefinite Integrals: $(30)$ Integrals Involving $\tanh a x$: $17.30.7.$
Category:Primitives involving Hyperbolic Tangent Function