hcju/proofwikips huggingface dataset:
Primitive of Reciprocal of a squared minus x squared/Logarithm Form/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {a^2 - x^2}$ == :$\dfrac 1 {a^2 - x^2} \equiv \dfrac 1 {2 a \paren {a + x} } + \dfrac 1 {2 a \paren {a - x} }$ == Proof == {{begin-eqn}} {{eqn | l = \frac 1 {\paren {a^2 - x^2} }       | o = \equiv       | r = \frac A {a - x} + \frac B {a + x}       | c = Difference of Two Squares }} {{eqn | n = 1       | ll= \leadsto       | l = 1       | o = \equiv       | r = A \paren {a + x} + B \paren {a - x}       | c = multiplying through by $\paren {a^2 - x^2}$ }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 2 a + B \cdot 0       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \frac 1 {2 a}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = A \cdot 0 + B \cdot \paren {2 a}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = \frac 1 {2 a}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {2 a} }} {{eqn | l = B       | r = \frac 1 {2 a} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal/Corollary
== Corollary to Primitive of Reciprocal ==  :$\displaystyle \int \frac {\d x} x = \ln x + C$ for $x > 0$.  == Proof == From Primitive of Reciprocal: :$\displaystyle \int \frac {\d x} x = \ln \size x + C$ for $x \ne 0$. By definition of absolute value: :$\forall x \in \R_{>0}: \size x = x$ Hence the result.
Primitive of Reciprocal of x by a squared minus x squared/Partial Fraction Expansion
Primitive of Reciprocal of x by a squared minus x squared/Partial Fraction Expansion 0 34052 188687 2014-06-06T18:57:30Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x \left({a^2 - x^2}\right)$ ==  :$\dfrac 1 {x \left({a^2 - x^2}\right)} ..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x \left({a^2 - x^2}\right)$ == :$\dfrac 1 {x \left({a^2 - x^2}\right)} \equiv \dfrac 1 {a^2 x} + \dfrac x {a^2 \left({a^2 - x^2}\right)}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x \left({a^2 - x^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac {B x + C} {a^2 - x^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a^2 - x^2}\right) + B x^2 + C x       | c = multiplying through by $x \left({a^2 - x^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = A a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = A       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A + B       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {a^2} }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of x by a squared minus x squared
Primitive of Reciprocal of x by a squared minus x squared 0 34053 188690 2014-06-06T19:03:04Z Prime.mover 59 Created page with " :$\displaystyle \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)} = \frac 1 {2 a^2} \ln \left({\frac {x^2} {a^2 - x^2} }\right) + C$ for $x^2 < a^2$. </onlyi..." wikitext text/x-wiki  :$\displaystyle \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)} = \frac 1 {2 a^2} \ln \left({\frac {x^2} {a^2 - x^2} }\right) + C$ for $x^2 < a^2$.  == Proof == {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)}       | r = \int \left({\frac 1 {a^2 x} + \frac x {a^2 \left({a^2 - x^2}\right)} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} x + \frac 1 {a^2} \int \frac {x \ \mathrm d x} {a^2 - x^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \ln \left\vert{x}\right\vert + \frac 1 {a^2} \int \frac {x \ \mathrm d x} {a^2 - x^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^2} \ln \left\vert{x}\right\vert + \frac 1 {a^2} \left({-\frac 1 2 \ln \left ({a^2 - x^2}\right) }\right) + C       | c = Primitive of $\dfrac x {a^2 - x^2}$ }} {{eqn | r = \frac 1 {2 a^2} \ln \left\vert{x^2}\right\vert - \frac 1 {2 a^2} \ln \left ({a^2 - x^2}\right) + C       | c = Logarithm of Power }} {{eqn | r = \frac 1 {2 a^2} \ln \left({x^2}\right) - \frac 1 {2 a^2} \ln \left ({a^2 - x^2}\right) + C       | c = as $x^2 > 0$ }} {{eqn | r = \frac 1 {2 a^2} \ln \left({\frac {x^2} {x^2 - a^2} }\right) + C       | c = Difference of Logarithms }} {{end-eqn}}  == Also see == * Primitive of $\dfrac 1 {x \left({x^2 - a^2}\right)}$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of x cubed over a squared minus x squared|next=Primitive of Reciprocal of x squared by a squared minus x squared}}: $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.167$
Primitive of Reciprocal of x squared by a squared minus x squared/Partial Fraction Expansion
Primitive of Reciprocal of x squared by a squared minus x squared/Partial Fraction Expansion 0 34054 188699 2014-06-06T19:42:33Z Prime.mover 59 Created page with "== Lemma for Primitive of Reciprocal of $x^2 \left({a^2 - x^2}\right)$ ==  :$\dfrac 1 {x^2 \left({a^2 - x..." wikitext text/x-wiki == Lemma for Primitive of Reciprocal of $x^2 \left({a^2 - x^2}\right)$ == :$\dfrac 1 {x^2 \left({a^2 - x^2}\right)} \equiv \dfrac 1 {a^2 \left({a^2 - x^2}\right)} + \dfrac 1 {a^2 x^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^2 \left({a^2 - x^2}\right)}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac {C x + D} {a^2 - x^2}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x \left({a^2 - x^2}\right) + B \left({a^2 - x^2}\right) + C x^3 + D x^2       | c = multiplying through by $x^2 \left({a^2 - x^2}\right)$ }} {{end-eqn}} Setting $x = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = B a^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^2       | c =  }} {{eqn | ll= \implies       | l = A       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -A + C       | c =  }} {{eqn | ll= \implies       | l = C       | r = 0       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -B + D       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {a^2}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = 0 }} {{eqn | l = B       | r = \frac 1 {a^2} }} {{eqn | l = C       | r = 0 }} {{eqn | l = D       | r = \frac 1 {a^2} }} {{end-eqn}} Hence the result.
Primitive of Reciprocal of a squared minus x squared squared/Partial Fraction Expansion
== Lemma for Primitive of Reciprocal of $\left({a^2 - x^2}\right)^2$ == :$\dfrac 1 {\left({a^2 - x^2}\right)^2} \equiv \dfrac 1 {4 a^3 \left({a + x}\right)} + \dfrac 1 {4 a^3 \left({a - x}\right)} + \dfrac 1 {4 a^2 \left({a + x}\right)^2} + \dfrac 1 {4 a^2 \left({a - x}\right)^2}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({a^2 - x^2}\right)^2}       | r = \dfrac 1 {\left({a + x}\right)^2 \left({a - x}\right)^2}       | c = Difference of Two Squares }} {{eqn | o = \equiv       | r = \dfrac A {a + x} + \dfrac B {\left({a + x}\right)^2} + \dfrac C {a - x} + \dfrac D {\left({a - x}\right)^2}       | c =  }} {{eqn | ll= \implies       | l = 1       | o = \equiv       | r = A \left({a^2 - x^2}\right) \left({a - x}\right) + B \left({a - x}\right)^2 + C \left({a^2 - x^2}\right) \left({a + x}\right) + D \left({a + x}\right)^2       | c = multiplying through by $\left({a^2 - x^2}\right)^2$ }} {{eqn | n = 1       | o = \equiv       | r = A a^3 - A a^2 x - A a x^2 + A x^3 + B a^2 - 2 B a x + B x^2       | c = multiplying out }} {{eqn | o =        | ro= +       | r = C a^3 + C a^2 x - C a x^2 - C x^3 + D a^2 + 2 D a x + D x^2       | c =  }} {{end-eqn}} Setting $x = a$ in $(1)$: {{begin-eqn}} {{eqn | l = D \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = D       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Setting $x = -a$ in $(1)$: {{begin-eqn}} {{eqn | l = B \left({2 a}\right)^2       | r = 1       | c =  }} {{eqn | ll= \implies       | l = B       | r = \frac 1 {4 a^2}       | c =  }} {{end-eqn}} Equating coefficients of $x^3$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A - C       | c =  }} {{eqn | ll= \implies       | l = A       | r = C       | c =  }} {{end-eqn}} Equating coefficients of $x^2$ in $(1)$: {{begin-eqn}} {{eqn | l = - A a - C a + B + D       | r = 0       | c =  }} {{eqn | ll= \implies       | l = - A a - C a + \frac 1 {4 a^2} + \frac 1 {4 a^2}       | r = 0       | c =  }} {{eqn | ll= \implies       | l = -C a - C a + \frac 1 {2 a^2}       | r = 0       | c = as $A = C$ }} {{eqn | ll= \implies       | l = 2 C        | r = \frac 1 {2 a^3}       | c =  }} {{eqn | ll= \implies       | l = C        | r = \frac 1 {4 a^3}       | c =  }} {{eqn | ll= \implies       | l = A        | r = \frac 1 {4 a^3}       | c =  }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac 1 {4 a^3} }} {{eqn | l = B       | r = \frac 1 {4 a^2} }} {{eqn | l = C       | r = \frac 1 {4 a^3} }} {{eqn | l = D       | r = \frac 1 {4 a^2} }} {{end-eqn}} Hence the result.
Primitive of x squared over Root of a squared minus x squared
Primitive of x squared over Root of a squared minus x squared 0 34139 188942 2014-06-12T20:57:02Z Prime.mover 59 Created page with "== Theorem ==  :$\displaystyle \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} } = \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C$ </only..." wikitext text/x-wiki == Theorem ==  :$\displaystyle \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} } = \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C$  == Proof == With a view to expressing the problem in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u}{\mathrm d x}       | r = 1       | c = Power Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v}{\mathrm d x}       | r = \frac x {\sqrt {a^2 - x^2} }       | c =  }} {{eqn | ll= \implies       | l = v       | r = -\sqrt {a^2 - x^2}       | c = Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} }       | r = \int x \frac {x \ \mathrm d x} {\sqrt {a^2 - x^2} }       | c =  }} {{eqn | r = -x \sqrt {a^2 - x^2} - \int \left({-\sqrt {a^2 - x^2} }\right) \ \mathrm d x       | c = Integration by Parts }} {{eqn | r = -x \sqrt {a^2 - x^2} + \int \left({\sqrt {a^2 - x^2} }\right) \ \mathrm d x       | c = simplifying }} {{eqn | r = -x \sqrt {a^2 - x^2} + \left({\frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a}\right) + C       | c = Primitive of $\sqrt {a^2 - x^2}$ }} {{eqn | r = \frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C       | c =  }} {{end-eqn}}  == Also see == * Primitive of $\dfrac {x^2} {\sqrt {x^2 + a^2} }$ * Primitive of $\dfrac {x^2} {\sqrt {x^2 - a^2} }$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of x over Root of a squared minus x squared|next=Primitive of x cubed over Root of a squared minus x squared}}: $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.239$
Sum of Infinite Geometric Sequence/Corollary 2
== Corollary to Sum of Infinite Geometric Sequence == Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$. Let $z \in S$. Let $\size z < 1$, where $\size z$ denotes: :the absolute value of $z$, for real and rational $z$ :the complex modulus of $z$ for complex $z$. Then:  :$\displaystyle \sum_{n \mathop = 0}^\infty a z^n = \frac a {1 - z}$  == Proof == {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty a z^n       | r = a \sum_{n \mathop = 0}^\infty z^n       | c = }} {{eqn | r = a \frac 1 {1 - z}       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \frac a {1 - z}       | c = }} {{end-eqn}}  == Sources == * {{BookReference|Handbook of Mathematical Functions|1964|Milton Abramowitz|author2 = Irene A. Stegun|prev = Sum of Geometric Sequence/Corollary 1|next = Definition:Arithmetic Mean}}: $3.1.10$: Sum of Geometric Progression to $n$ Terms * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Sum of Geometric Sequence/Corollary 1|next = Definition:Arithmetic-Geometric Series}}: $\S 19$: Geometric Series: $19.5$ * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Sum of Geometric Sequence/Corollary 1|next = Sum of Infinite Geometric Sequence/Corollary 2/Mistake|entry = geometric series}} * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Geometric Series|next = Definition:Geometry|entry = geometric series}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Geometric Series|next = Definition:Geometry|entry = geometric series}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Sum of Geometric Sequence/Corollary 1|next = Definition:Geometry|entry = geometric series}}
Perpendicular in Right-Angled Triangle makes two Similar Triangles/Porism
== Porism to Perpendicular in Right-Angled Triangle makes two Similar Triangles == {{:Euclid:Proposition/VI/8/Porism}} == Proof == Follows directly from Perpendicular in Right-Angled Triangle makes two Similar Triangles.  {{Euclid Note|8|VI}} == Sources == * {{BookReference|Euclid: The Thirteen Books of The Elements: Volume 2|1926|ed = 2nd|edpage = Second Edition|Sir Thomas L. Heath|prev = Perpendicular in Right-Angled Triangle makes two Similar Triangles|next = Construction of Part of Line}}: Book $\text{VI}$. Propositions
Primitive of x over Root of a x squared plus b x plus c
Primitive of x over Root of a x squared plus b x plus c 0 34340 189802 2014-07-06T05:51:32Z Prime.mover 59 Created page with "== Theorem ==  Let $a \in \R_{\ne 0}$.  Then: :$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt {a x^2 + b x + c} } = \frac {\sqrt {a x^2 + b x + c} } a - \frac..." wikitext text/x-wiki == Theorem ==  Let $a \in \R_{\ne 0}$. Then: :$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt {a x^2 + b x + c} } = \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$  == Proof == First: {{begin-eqn}} {{eqn | l = z       | r = a x^2 + b x + c       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a x + b       | c = Derivative of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\sqrt {a x^2 + b x + c} }       | r = \frac 1 {2 a} \int \frac {2 a x \ \mathrm d x} {\sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b - b}\right) \ \mathrm d x} {\sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b}\right) \ \mathrm d x} {\sqrt {a x^2 + b x + c} } - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }        | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {2 a} \int \frac {\mathrm d z} {\sqrt z} - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }        | c = Integration by Substitution }} {{eqn | r = \frac 1 {2 a} 2 \sqrt z - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | c = Primitive of Power }} {{eqn | r = \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | c = substituting for $z$ }} {{end-eqn}}  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev=Primitive of Reciprocal of Root of a x squared plus b x plus c|next=Primitive of x squared over Root of a x squared plus b x plus c}}: $\S 14$: Integrals involving $\sqrt {a x^2 + bx + c}$: $14.281$
Primitive of Sine Function/Corollary
== Corollary to Primitive of Sine Function ==  :$\displaystyle \int \sin a x \rd x = - \frac {\cos a x} a + C$  where $a$ is a non-zero constant. == Proof == {{begin-eqn}} {{eqn | l = \int \sin x \rd x       | r = -\cos x + C       | c = Primitive of $\sin x$ }} {{eqn | ll= \leadsto       | l = \int \sin a x \rd x       | r = \frac 1 a \paren {-\cos a x} + C       | c = Primitive of Function of Constant Multiple }} {{eqn | r = -\frac {\cos a x} a + C       | c = simplifying }} {{end-eqn}}  == Also see == * Primitive of $\cos a x$ * Primitive of $\tan a x$ * Primitive of $\cot a x$ * Primitive of $\sec a x$ * Primitive of $\csc a x$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Primitive of Power of x over Odd Power of x minus Odd Power of a|next = Primitive of x by Sine of a x}}: $\S 14$: Integrals involving $\sin a x$: $14.339$ * {{BookReference|Calculus|1983|K.G. Binmore|prev = Primitive of Cosine of a x|next = Primitive of Reciprocal of x squared plus a squared/Arctangent Form}}: $9$ Sums and Integrals: $9.8$ Standard Integrals
Primitive of Cosine Function/Corollary
== Corollary to Primitive of Cosine Function ==  :$\displaystyle \int \cos a x \rd x = \frac {\sin a x} a + C$  where $a$ is a non-zero constant. == Proof == {{begin-eqn}} {{eqn | l = \int \cos x \rd x       | r = \sin x + C       | c = Primitive of $\cos x$ }} {{eqn | ll= \leadsto       | l = \int \cos a x \rd x       | r = \frac 1 a \paren {\sin a x} + C       | c = Primitive of Function of Constant Multiple }} {{eqn | r = \frac {\sin a x} a + C       | c = simplifying }} {{end-eqn}}  == Also see == * Primitive of $\sin a x$ * Primitive of $\tan a x$ * Primitive of $\cot a x$ * Primitive of $\sec a x$ * Primitive of $\csc a x$ == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Primitive of x over Power of Sine of a x|next = Primitive of x by Cosine of a x}}: $\S 14$: Integrals involving $\cos a x$: $14.369$ * {{BookReference|Calculus|1983|K.G. Binmore|prev = Primitive of Exponential of a x|next = Primitive of Sine of a x}}: $9$ Sums and Integrals: $9.8$ Standard Integrals
Weierstrass Substitution
 == Proof Technique == The '''Weierstrass Substitution''' is an application of Integration by Substitution. The substitution is: :$u \leftrightarrow \tan \dfrac \theta 2$ for $-\pi < \theta < \pi$, $u \in \R$. It yields: {{begin-eqn}} {{eqn | l = \sin \theta       | r = \frac {2 u} {1 + u^2} }} {{eqn | l = \cos \theta       | r = \frac {1 - u^2} {1 + u^2} }} {{eqn | l = \frac {\d \theta} {\d u}       | r = \frac 2 {1 + u^2} }} {{end-eqn}} This can be stated: :$\displaystyle \int \map F {\sin \theta, \cos \theta} \rd \theta = 2 \int \map F {\frac {2 u} {1 + u^2}, \frac {1 - u^2} {1 + u^2} } \frac {d u} {1 + u^2}$ where $u = \tan \dfrac \theta 2$. == Proof == Let $u = \tan \dfrac \theta 2$ for $-\pi < \theta < \pi$. From Shape of Tangent Function, this substitution is valid for all real $u$. Then: {{begin-eqn}} {{eqn | l = u       | r = \tan \dfrac \theta 2       | c =  }} {{eqn | ll= \leadsto       | l = \theta       | r = 2 \tan^{-1} u       | c = Definition of Inverse Tangent }} {{eqn | ll= \leadsto       | l = \dfrac {\d \theta} {\d u}       | r = \dfrac 2 {1 + u^2}       | c = Derivative of Arctangent Function and Derivative of Constant Multiple }} {{eqn | l = \sin \theta       | r = \dfrac {2 u} {1 + u^2}       | c = Tangent Half-Angle Substitution for Sine }} {{eqn | l = \cos \theta       | r = \dfrac {1 - u^2} {1 + u^2}       | c = Tangent Half-Angle Substitution for Cosine }} {{end-eqn}} The result follows from Integration by Substitution.  == Also known as == This technique is also known as '''tangent half-angle subsitution'''. == Also see == * Hyperbolic Tangent Half-Angle Substitution {{Namedfor|Karl Theodor Wilhelm Weierstrass|cat = Weierstrass}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Primitive of Function of Arcsine|next = Primitive of Reciprocal of a x + b}}: $\S 14$: Important Transformations: $14.58$ * {{Planetmath|url = weierstrasssubstitutionformulas|title = Weierstrass substitution formulas}} * {{MathWorld|Weierstrass Substitution|WeierstrassSubstitution}}
Primitive of x by Root of a x squared plus b x plus c/Lemma
Primitive of x by Root of a x squared plus b x plus c/Lemma 0 34503 190359 2014-07-19T10:53:22Z Prime.mover 59 Created page with "== Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$ ==  Let $a \in \R_{\ne 0}$.  Then: :$\displaystyle..." wikitext text/x-wiki == Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$ ==  Let $a \in \R_{\ne 0}$. Then: :$\displaystyle \int x \sqrt {a x^2 + b x + c} \ \mathrm d x = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x$  == Proof == {{begin-eqn}} {{eqn | o =        | r = \int x \sqrt {a x^2 + b x + c} \ \mathrm d x }} {{eqn | r = \int \frac {2 a x \sqrt {a x^2 + b x + c} \ \mathrm d x} {2 a}       | c = multiplying top and bottom by $2 a$ }} {{eqn | r = \int \frac {\left({2 a x + b - b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x} {2 a} }} {{eqn | n = 1       | r = \frac 1 {2 a} \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = Linear Combination of Integrals }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = a x^2 + b x + c }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a x + b       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x       | r = \int \sqrt z \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac {2 \left({\sqrt z}\right)^3} 3       | c = Primitive of Power }} {{eqn | n = 2       | r = \frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3       | c = substituting for $z$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn | o =        | r = \int x \sqrt {a x^2 + b x + c} \ \mathrm d x }} {{eqn | r = \frac 1 {2 a} \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = from $(1)$ }} {{eqn | r = \frac 1 {2 a} \left({\frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3}\right) - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = from $(2)$ }} {{eqn | r = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = simplifying }} {{end-eqn}}
Primitive of Reciprocal of x cubed plus a cubed/Lemma
== Lemma for Primitive of Reciprocal of $x^3 + a^3$ ==  :$\displaystyle \int \frac {\d x} {x^2 - a x + a^2} = \frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} }$  == Proof == The discriminant of $x^2 - a x + a^2$ is: {{begin-eqn}} {{eqn | l = \map {\mathrm {Disc} } {x^2 - a x + a^2}       | r = \paren {-a}^2 - 4 \times 1 \times a^2       | c =  }} {{eqn | r = a^2 - 4 a^2       | c =  }} {{eqn | r = - 3 a^2       | c =  }} {{eqn | r = < 0       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 - a x + a^2}       | r = \frac 2 {\sqrt {4 a^2 - \paren {-a}^2} } \map \arctan {\frac {2 x - a} {\sqrt {4 a^2 - \paren {-a}^2} } }       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ }} {{eqn | r = \frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} }       | c = simplifying }} {{end-eqn}}