hcju/proofwikips huggingface dataset:
Primitive of Reciprocal of x cubed plus a cubed/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $x^3 + a^3$ == :$\dfrac 1 {x^3 + a^3} = \dfrac 1 {3 a^2 \paren {x + a} } - \dfrac {x - 2 a} {3 a^2 \paren {x^2 - a x + a^2} }$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {x^3 + a^3}       | o = \equiv       | r = \frac 1 {\paren {x + a} \paren {x^2 - a x + a^2} }       | c = Sum of Two Cubes }} {{eqn | o = \equiv       | r = \frac A {x + a} + \frac {B x + C} {x^2 - a x + a^2}       | c =  }} {{eqn | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {x^2 - a x + a^2} + \paren {B x + C} \paren {x + a}       | c = multiplying through by $x^3 + a^3$ }} {{eqn | n = 1       | o = \equiv       | r = A x^2 - A a x + A a^2 + B x^2 + B a x + C x + C a       | c = multiplying out }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = A + B       | r = 0       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = -A       | r = B       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = -A a + B a + C       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = -A a + -A a + C       | r = 0       | c = from $(2)$ }} {{eqn | n = 3       | ll= \leadsto       | l = 2 A a       | r = C }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^2 + C a       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A a^2 + \paren {2 A a} a       | r = 1       | c = from $(3)$ }} {{eqn | ll= \leadsto       | l = A a^2 + \paren {2 A a} a       | r = \frac 1 {3 a^2}       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = 1 - \frac a {3 a^2}       | c = from $(3)$ }} {{eqn | r = \frac {2 a} {3 a^2}       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\frac {-1} {3 a^2}       | c = from $(2)$ }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {3 a^2} }} {{eqn | l = B       | r = -\frac 1 {3 a^2} }} {{eqn | l = C       | r = \frac {2 a} {3 a^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x fourth plus a fourth/Lemma 1
== Lemma for Primitive of Reciprocal of $x^4 + a^4$ == :$\displaystyle \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2} = \frac {\sqrt 2} a \, \map \arctan {1 + \frac {x \sqrt 2} a}$ == Proof == The discriminant of $x^2 + a x \sqrt 2 + a^2$ is: {{begin-eqn}} {{eqn | l = \map {\operatorname {Disc} } {x^2 + a x \sqrt 2 + a^2}       | r = \paren {a \sqrt 2}^2 - 4 \times 1 \times a^2       | c =  }} {{eqn | r = 2 a^2 - 4 a^2       | c =  }} {{eqn | r = - 2 a^2       | c =  }} {{eqn | r = < 0       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2}       | r = \frac 2 {\sqrt {4 a^2 - \paren {a \sqrt 2}^2} } \map \arctan {\frac {2 x + a \sqrt 2} {\sqrt {4 a^2 - \paren {a \sqrt 2}^2} } }       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ }} {{eqn | r = \frac 2 {a \sqrt 2} \, \map \arctan {\frac {2 x + a \sqrt 2} {a \sqrt 2} }       | c = simplifying }} {{eqn | r = \frac {\sqrt 2} a \, \map \arctan {1 + \frac {x \sqrt 2} a}       | c = simplifying }} {{end-eqn}}
Primitive of Reciprocal of x fourth plus a fourth/Lemma 2
== Lemma for Primitive of Reciprocal of $x^4 + a^4$ == :$\displaystyle \int \frac {\mathrm d x} {x^2 - a x \sqrt 2 + a^2} = \frac {-\sqrt 2} a \arctan \left({1 - \frac {x \sqrt 2} a}\right)$ == Proof == The discriminant of $x^2 - a x \sqrt 2 + a^2$ is: {{begin-eqn}} {{eqn | l = \operatorname{Disc} \left({x^2 - a x \sqrt 2 + a^2}\right)       | r = \left({-a \sqrt 2}\right)^2 - 4 \times 1 \times a^2       | c =  }} {{eqn | r = 2 a^2 - 4 a^2       | c =  }} {{eqn | r = - 2 a^2       | c =  }} {{eqn | r = < 0       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 - a x \sqrt 2 + a^2}       | r = \frac 2 {\sqrt {4 a^2 - \left({-a \sqrt 2}\right)^2} } \arctan \left({\frac {2 x - a \sqrt 2} {\sqrt {4 a^2 - \left({-a \sqrt 2}\right)^2} } }\right)       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ }} {{eqn | r = \frac 2 {a \sqrt 2} \arctan \left({\frac {2 x - a \sqrt 2} {a \sqrt 2} }\right)       | c = simplifying }} {{eqn | r = \frac {\sqrt 2} a \arctan \left({\frac {x \sqrt 2} a - 1}\right)       | c = simplifying }} {{eqn | r = \frac {-\sqrt 2} a \arctan \left({1 - \frac {x \sqrt 2} a}\right)       | c = Arctangent is Odd Function }} {{end-eqn}}
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 $\displaystyle \int \frac {\mathrm d x} {p + q \sin a x}$ is: :$\displaystyle \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u + p}$ where $u = \tan \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {p + q \sin a x}       | r = \frac 1 a \int \frac {\mathrm d z} {p + q \sin z}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac 1 {p + q \frac {2 u} {u^2 + 1} } \frac {2 \ \mathrm d u} {u^2 + 1}       | c = Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \ \mathrm d u} {\left({u^2 + 1}\right) \frac {p \left({u^2 + 1}\right) + 2 q u} {u^2 + 1} }       | c = common denominator }} {{eqn | r = \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u + p}       | c = simplifying }} {{end-eqn}}
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 of $\displaystyle \int \frac {\d x} {\paren {p + q \sin a x}^2}$ is: :$\displaystyle \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}$ where $u = \tan \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\paren {p + q \sin a x}^2}       | r = \frac 1 a \int \frac {\d z} {\paren {p + q \sin z}^2}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac 1 {\paren {p + q \frac {2 u} {u^2 + 1} }^2} \frac {2 \rd u} {u^2 + 1}       | c = Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \rd u} {\paren {u^2 + 1} \paren {\frac {p \paren {u^2 + 1} + 2 q u} {u^2 + 1} }^2}       | c = common denominator }} {{eqn | r = \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}       | c = simplifying }} {{end-eqn}}
Primitive of Exponential of a x by Sine of b x/Lemma
== Lemma for Primitive of $e^{a x} \sin b x$ ==  :$\displaystyle \int e^{a x} \sin b x \rd x = \frac {e^{a x} \sin b x} a - \frac b a \int e^{a x} \cos b x \rd x$  == Proof == With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \sin b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = b \cos b x       | c = Derivative of $\sin a x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\ d x}       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} } a       | c = Primitive of $e^{a x}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int e^{a x} \sin b x \rd x       | r = \sin b x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \paren {b \cos b x} \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {e^{a x} \sin b x} a - \frac b a \int e^{a x} \cos b x \rd x       | c = Primitive of Constant Multiple of Function }} {{end-eqn}}
Primitive of Exponential of a x by Cosine of b x/Lemma
== Lemma for Primitive of $e^{a x} \cos b x$ ==  :$\displaystyle \int e^{a x} \cos b x \rd x = \frac {e^{a x} \cos b x} a + \frac b a \int e^{a x} \sin b x \rd x$  == Proof == With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v}{\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \cos b x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = -b \sin b x       | c = Derivative of $\cos a x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} } a       | c = Primitive of $e^{a x}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int e^{a x} \cos b x \rd x       | r = \cos b x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \paren {-b \sin b x} \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {e^{a x} \cos b x} a + \frac b a \int e^{a x} \sin b x \rd x       | c = Primitive of Constant Multiple of Function }} {{end-eqn}}
Primitive of Exponential of a x by Power of Sine of b x/Lemma 2
== Lemma for Primitive of $e^{a x} \sin^n b x$ == :$\dfrac {a^2 + n b^2} a e^{a x} \sin^n b x - \dfrac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right) = e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)$ == Proof == {{begin-eqn}} {{eqn | r = \frac {a^2 + n b^2} a e^{a x} \sin^n b x - \frac {n b} a e^{a x} \sin^{n - 1} b x \left({a \cos b x + b \sin b x}\right)       | c =  }} {{eqn | r = \frac {a^2 + n b^2} a e^{a x} \sin^{n - 1} b x \sin b x - \frac {n b} a e^{a x} \sin^{n - 1} b x a \cos b x - \frac {n b} a e^{a x} \sin^{n - 1} b x b \sin b x       | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \left({\frac {a^2 + n b^2} a \sin b x - \frac {n b} a a \cos b x - \frac {n b} a b \sin b x}\right)       | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \left({a \sin b x + \frac {n b^2} a \sin b x - n b \cos b x - \frac {n b^2} a \sin b x}\right)       | c =  }} {{eqn | r = e^{a x} \sin^{n - 1} b x \left({a \sin b x - n b \cos b x}\right)       | c =  }} {{end-eqn}}
Primitive of Exponential of a x by Power of Cosine of b x/Lemma 2
== Lemma for Primitive of $e^{a x} \cos^n b x$ == :$\dfrac {a^2 + n b^2} a e^{a x} \cos^n b x + \dfrac {n b} a e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x} = e^{a x} \cos^{n - 1} b x \paren {a \cos b x + n b \sin b x}$ == Proof == {{begin-eqn}} {{eqn | r = \dfrac {a^2 + n b^2} a e^{a x} \cos^n b x + \dfrac {n b} a e^{a x} \cos^{n - 1} b x \paren {a \sin b x - b \cos b x}       | c =  }} {{eqn | r = \frac {a^2 + n b^2} a e^{a x} \cos^{n - 1} b x \cos b x + \frac {n b} a e^{a x} \cos^{n - 1} b x a \sin b x - \frac {n b} a e^{a x} \cos^{n - 1} b x b \cos b x       | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {\frac {a^2 + n b^2} a \cos b x + \frac {n b} a a \sin b x - \frac {n b} a b \cos b x}       | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {a \cos b x + \frac {n b^2} a \cos b x + n b \sin b x - \frac {n b^2} a \cos b x}       | c =  }} {{eqn | r = e^{a x} \cos^{n - 1} b x \paren {a \cos b x + n b \sin b x}       | c =  }} {{end-eqn}}
Primitive of Power of Hyperbolic Tangent of a x
Primitive of Power of Hyperbolic Tangent of a x 0 34887 191348 2014-08-12T07:05:34Z Prime.mover 59 Created page with "== Theorem ==  :$\displaystyle \int \tanh^n a x \ \mathrm d x = \frac {-\tanh^{n - 1} a x} {a \left({n - 1}\right)} + \int \tanh^{n - 2} a x \ \mathrm d x + C$ </..." wikitext text/x-wiki == Theorem ==  :$\displaystyle \int \tanh^n a x \ \mathrm d x = \frac {-\tanh^{n - 1} a x} {a \left({n - 1}\right)} + \int \tanh^{n - 2} a x \ \mathrm d x + C$  == Proof == {{begin-eqn}} {{eqn | l = \int \tanh^n a x \ \mathrm d x       | r = \int \tanh^{n - 2} a x \tanh^2 a x \ \mathrm d x       | c =  }} {{eqn | r = \int \tanh^{n - 2} a x \left({1 - \operatorname{sech}^2 a x}\right) \ \mathrm d x       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{eqn | r = -\int \tanh^{n - 2} a x \operatorname{sech}^2 a x \ \mathrm d x + \int \tanh^{n - 2} \ \mathrm d x       | c = Linear Combination of Integrals }} {{eqn | r = \frac {-\tanh^{n - 1} a x} {a \left({n - 1}\right)} + \int \tanh^{n - 2} a x \ \mathrm d x + C       | c = Primitive of $\tanh^n a x \operatorname{sech}^2 a x$ }} {{end-eqn}}  == Also see == * Primitive of $\sinh^n a x$ * Primitive of $\cosh^n a x$ * Primitive of $\coth^n a x$ * Primitive of $\operatorname{sech}^n a x$ * Primitive of $\operatorname{csch}^n a x$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of Reciprocal of p plus q by Hyperbolic Tangent of a x|next=Primitive of Hyperbolic Cotangent of a x}}: $\S 14$: Integrals involving $\tanh a x$: $14.614$
Derivative of Tangent Function/Corollary
== Corollary to Derivative of Tangent Function ==  :$\map {\dfrac \d {\d x} } {\tan a x} = a \sec^2 a x$  == Proof == {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\tan x}       | r = \sec^2 x       | c = Derivative of $\tan x$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\tan a x}       | r = a \sec^2 a x       | c = Derivative of Function of Constant Multiple }} {{end-eqn}}  == Also see == * Derivative of $\sin a x$ * Derivative of $\cos a x$ * Derivative of $\cot a x$ * Derivative of $\sec a x$ * Derivative of $\csc a x$
Derivative of Cotangent Function/Corollary
== Corollary to Derivative of Cotangent Function ==  :$\map {\dfrac \d {\d x} } {\cot a x} = -a \csc^2 a x$  == Proof == {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\cot x}       | r = -\csc^2 x       | c = Derivative of $\cot x$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\cot a x}       | r = -a \csc^2 a x       | c = Derivative of Function of Constant Multiple }} {{end-eqn}}  == Also see == * Derivative of $\sin a x$ * Derivative of $\cos a x$ * Derivative of $\tan a x$ * Derivative of $\sec a x$ * Derivative of $\csc a x$
Hyperbolic Tangent Half-Angle Substitution for Cosine
== Corollary to Double Angle Formula for Hyperbolic Cosine ==  :$\cosh 2 x = \dfrac {1 + \tanh^2 x}{1 - \tanh^2 x}$  where $\cosh$ and $\tanh$ denote hyperbolic cosine and hyperbolic tangent respectively. == Proof == {{begin-eqn}} {{eqn | l = \cosh 2 x       | r = \cosh^2 x + \sinh^2 x       | c = Double Angle Formula for Hyperbolic Cosine }} {{eqn | r = \paren {\cosh^2 x + \sinh^2 x} \frac {\cosh^2 x} {\cosh^2 x}       | c =  }} {{eqn | r = \paren {1 + \tanh^2 x} \cosh^2 x       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {1 + \tanh^2 x} {\sech^2 x}       | c = {{Defof|Hyperbolic Secant|index = 2}} }} {{eqn | r = \frac {1 + \tanh^2 x} {1 - \tanh^2 x}       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{end-eqn}}
Hyperbolic Tangent Half-Angle Substitution for Sine
== Corollary to Double Angle Formula for Hyperbolic Sine ==  :$\sinh 2 x = \dfrac {2 \tanh x} {1 - \tanh^2 x}$  where $\sin$ and $\tan$ denote hyperbolic sine and hyperbolic tangent respectively. == Proof == {{begin-eqn}} {{eqn | l = \sinh 2 x       | r = 2 \sinh x \cosh x       | c = Double Angle Formula for Hyperbolic Sine }} {{eqn | r = 2 \sinh x \cosh x \frac {\cosh x} {\cosh x}       | c =  }} {{eqn | r = 2 \tanh x \cosh^2 x       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {2 \tanh x} {\operatorname{sech}^2 x}       | c = {{Defof|Hyperbolic Secant|index = 2}} }} {{eqn | r = \frac {2 \tanh x} {1 - \tanh^2 x}       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{end-eqn}}
Square Modulo 3/Corollary 1
== Corollary to Square Modulo 3 ==  Let $x, y \in \Z$ be integers. Then: :$3 \divides \paren {x^2 + y^2} \iff 3 \divides x \land 3 \divides y$ where $3 \divides x$ denotes that $3$ divides $x$.  == Proof == === Sufficient Condition === Let $3 \divides x \land 3 \divides y$. Then by definition of divisibility: :$x \equiv 0 \pmod 3$ and :$y \equiv 0 \pmod 3$ From Square Modulo 3: :$x^2 \equiv 0 \pmod 3$ and :$y^2 \equiv 0 \pmod 3$ Then from Modulo Addition is Well-Defined: :$\paren {x^2 + y^2} \equiv 0 \pmod 3$ Thus: :$3 \divides \paren {x^2 + y^2}$ {{qed|lemma}} === Necessary Condition === Now suppose $3 \divides \paren {x^2 + y^2}$. Then by definition of divisibility: :$\paren {x^2 + y^2} \pmod 3$ From Square Modulo 3: : $x^2 \equiv 0 \pmod 3$ or $x^2 \equiv 1 \pmod 3$ and : $y^2 \equiv 0 \pmod 3$ or $y^2 \equiv 1 \pmod 3$ Thus the only way $\paren {x^2 + y^2} \equiv 0 \pmod 3$ is for: : $x^2 \equiv 0 \pmod 3$ and: : $y^2 \equiv 0 \pmod 3$  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Square Modulo 3|next = Square Modulo 3/Corollary 2}}: Chapter $2$: Some Properties of $\Z$: Exercise $2.9$
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 $\displaystyle \int \frac {\d x} {p + q \cos a x}$ is: :$\displaystyle \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + \dfrac {p + q} {p - q} }$ where $u = \tan \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\d x} {p + q \cos a x}       | r = \frac 1 a \int \frac {\d z} {p + q \cos z}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac 1 {p + q \paren {\frac {1 - u^2} {1 + u^2} } } \frac {2 \rd u} {1 + u^2}       | c = Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \rd u} {\paren {1 + u^2} \paren {\frac {p \paren {1 + u^2} + q \paren {1 - u^2} } {1 + u^2} } }       | c = common denominator }} {{eqn | r = \frac 2 a \int \frac {\d u} {\paren {p - q} u^2 + \paren {p + q} }       | c = simplifying }} {{eqn | r = \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + \dfrac {p + q} {p - q} }       | c = simplifying further }} {{end-eqn}}
Primitive of Reciprocal of p plus q by Hyperbolic Sine of a x/Hyperbolic Tangent Half-Angle Substitution
Primitive of Reciprocal of p plus q by Hyperbolic Sine of a x/Hyperbolic Tangent Half-Angle Substitution 0 35075 191968 2014-08-23T15:33:05Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $p + q \sinh a x$ ==  The Hyperbolic Tangent Half-Angle Substitutio..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $p + q \sinh a x$ == The Hyperbolic Tangent Half-Angle Substitution of $\displaystyle \int \frac {\mathrm d x} {p + q \sinh a x}$ is: :$\displaystyle \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u - p}$ where $u = \tanh \dfrac {a x} 2$. == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {p + q \sinh a x}       | r = \frac 1 a \int \frac {\mathrm d z} {p + q \sinh z}       | c = Primitive of Function of Constant Multiple: $z = a x$ }} {{eqn | r = \frac 1 a \int \frac {\dfrac {2 \ \mathrm d u} {u^2 - 1} } {p + q \dfrac {2 u} {u^2 - 1} }       | c = Hyperbolic Tangent Half-Angle Substitution: $u = \tanh \dfrac z 2 = \tanh \dfrac {a x} 2$ }} {{eqn | r = \frac 1 a \int \frac {2 \ \mathrm d u} {p \left({u^2 - 1}\right) + 2 q u}       | c = multiplying top and bottom by $u^2 - 1$ }} {{eqn | r = \frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u - p}       | c = simplifying }} {{end-eqn}}
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 == :$\displaystyle \frac 1 2 \left({\arctan \dfrac {-p} q}\right) + \frac \pi 4 = \frac {\arctan \dfrac q p} 2$ == Proof == {{begin-eqn}} {{eqn | l = y       | r = \frac 1 2 \left({\arctan \dfrac {-p} q}\right) + \frac \pi 4       | c = making a definition }} {{eqn | ll= \implies       | l = 2 y - \frac \pi 2       | r = \arctan \dfrac {-p} q       | c = rearranging }} {{eqn | ll= \implies       | l = \tan \left({2 y - \frac \pi 2}\right)       | r = \dfrac {-p} q       | c = Definition of Arctangent }} {{eqn | ll= \implies       | l = \tan \left({2 y + \frac {3 \pi} 2}\right)       | r = \dfrac {-p} q       | c = Tangent Function is Periodic on Reals }} {{eqn | ll= \implies       | l = -\cot \left({2 y}\right)       | r = \dfrac {-p} q       | c = Tangent of Angle plus Three Right Angles }} {{eqn | ll= \implies       | l = \tan \left({2 y}\right)       | r = \dfrac q p       | c = Cotangent is Reciprocal of Tangent }} {{eqn | ll= \implies       | l = y       | r = \frac {\arctan \dfrac q p} 2       | c = Definition of Arctangent }} {{end-eqn}}
Cartesian Product of Countable Sets is Countable/Corollary/Proof 2
== Corollary to Cartesian Product of Countable Sets is Countable == {{:Cartesian Product of Countable Sets is Countable/Corollary}} == Proof ==  Proof by induction: === Basis for the Induction === When $k = 2$, the case is the same as Cartesian Product of Countable Sets is Countable. So shown for basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$ where $f_k$ is an injection. Now we need to show that for $n = k + 1$: :$\exists f_{k+1}: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$ where $f_{k+1}$ is an injection. === Induction Step === This is our induction step: By the induction hypothesis: :$\exists f_k: S_1 \times S_2 \times \cdots \times S_k \to \N$ where $f_k$ is an injection. Thus by definition, $S_1 \times S_2 \times \cdots \times S_k$ is countable. By hypothesis $S_{k + 1}$ is countable. So by the basis for the induction: :$\exists g: \left({S_1 \times S_2 \times \cdots \times S_k}\right) \times S_{k+1} \to \N \times \N$ where $g$ is an injection. By Cartesian Product of Countable Sets is Countable, :$\exists r: \N \times \N \to \N$ where $r$ is an injection. Therefore, by Composite of Injections is Injection: :$f_{k+1} = r \circ g: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$ is an injection. The result follows by induction.   == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Cartesian Product of Countable Sets is Countable|next = }}: $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Theorem $7.6$
Finding Center of Circle/Porism
== Porism to Finding Center of Circle ==  {{:Euclid:Proposition/III/1/Porism}}  == Proof == See Perpendicular Bisector of Chord Passes Through Center, where the proof has been extracted from the proof given by Euclid. == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Finding Center of Circle/Proof 1|next=Perpendicular Bisector of Chord Passes Through Center}}: Book $\text{III}$. Propositions
Ratio of Areas of Similar Triangles/Porism
== Porism to Ratio of Areas of Similar Triangles == {{:Euclid:Proposition/VI/19/Porism}} == Proof == Follows immediately from Ratio of Areas of Similar Triangles.  {{Euclid Note|19|VI}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Ratio of Areas of Similar Triangles|next=Similar Polygons are composed of Similar Triangles}}: Book $\text{VI}$. Propositions
Euclidean Algorithm/Euclid's Proof/Porism
== Porism to Euclidean Algorithm == {{:Euclid:Proposition/VII/2/Porism}} == Proof == An algebraic proof of this is given in Common Divisor Divides GCD.  {{Euclid Note|2|VII}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Euclidean Algorithm/Euclid's Proof|next = Greatest Common Divisor of Three Numbers}}: Book $\text{VII}$. Propositions
Axiom:Euclid's Common Notion 1
== Common Notion ==  {{EuclidCommonNotionStatement|1|Things which are equal to the same thing are also equal to each other.}}  == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Axiom:Euclid's Fifth Postulate|next = Axiom:Euclid's Common Notion 2}}: Book $\text{I}$. Common Notions * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Axiom:Euclid's Postulates|next = Axiom:Euclid's Fourth Postulate}}: Chapter $2$: The Logic of Shape: Euclid
Axiom:Euclid's Common Notion 4
== Common Notion ==  {{EuclidCommonNotionStatement|4|Things which coincide with one another are equal to one another.}}  == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 1|1926|ed=2nd|edpage=Second Edition|Sir Thomas L. Heath|prev=Axiom:Euclid's Common Notion 3|next=Axiom:Euclid's Common Notion 5}}: Book $\text{I}$. Common Notions