hcju/proofwikiqa huggingface dataset:
Let $u = a x + b$. Then: :$\dfrac {\d u} {\d x} = a$ Then: {{begin-eqn}} {{eqn | l = \int \paren {a x + b}^n \rd x       | r = \int \dfrac {u^n} a \rd u       | c = Integration by Substitution }} {{eqn | r = \dfrac 1 a \dfrac {u^{n + 1} } {n + 1}       | c = Primitive of Power }} {{eqn | r = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C       | c = substituting back for $u$ }} {{end-eqn}} {{qed}}
Let $\map f x = x - \sin x$. By Derivative of Sine Function: :$\map {f'} x = 1 - \cos x$ From Real Cosine Function is Bounded we know $\cos x \le 1$ for all $x$. Hence $\map f x \ge 0$ for all $x$. From Derivative of Monotone Function, $\map f x$ is increasing.  By Sine of Zero is Zero, $\map f x = 0$. It follows that $\map f x \ge 0$ for all $x \ge 0$. {{qed|lemma}} Now let $\map g x = x^2 - \sin^2 x$. {{begin-eqn}} {{eqn | l = \map {g'} x       | r = 2 x - 2 \sin x \cos x       | c = Chain Rule for Derivatives and Derivative of Sine Function }} {{eqn | r = 2 x - \sin 2 x       | c = Double Angle Formula for Sine }} {{eqn | o = \ge       | r = 0       | c = for $x \ge 0$, as shown above }} {{end-eqn}} From Derivative of Monotone Function, $\map g x$ is increasing for $x \ge 0$.  By Sine of Zero is Zero, $\map g x = 0$. It follows that $\map g x \ge 0$ for all $x \ge 0$. Observe that $\map g x$ is an even function. This implies $\map g x \ge 0$ for all $x \in \R$. Finally note that $\sin^2 x \le x^2 \iff \size {\sin x} \le \size x$. {{qed}} Category:Sine Function Category:Named Theorems lnyrzyp3giyxd41fhu5add5w5q0lmlt
The right inequality is true by Sine Inequality. The left inequality is true for $x = 0$ and $x = \dfrac {\pi} 2$, where we have equality. Now consider $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$. From Shape of Sine Function, $\sin x$ is concave on the interval $\left[{0 \,.\,.\, \pi }\right]$. Letting $x_1 = 0$, $x_2 = x$ and $x_3 = \dfrac {\pi} 2$ in the definition of Concave Real Function we obtain: :$\dfrac {\sin x} x \ge \dfrac {1 - \sin x} {\dfrac {\pi} 2 - x}$ Rearranging gives the desired inequality. {{qed}} {{namedfor|Marie Ennemond Camille Jordan}}
Let $\map f x = \cos x - \paren {1 - \dfrac {x^2} 2}$. By Derivative of Cosine Function: :$\map {f'} x = x - \sin x$ From Sine Inequality, we know $\sin x \le x$ for $x \ge 0$. Hence $\map {f'} x \ge 0$ for $x \ge 0$. From Derivative of Monotone Function, $\map f x$ is increasing for $x \ge 0$. By Cosine of Zero is One, $\map f 0 = 0$. It follows that $\map f x \ge 0$ for $x \ge 0$. By Cosine Function is Even, $\map f x$ is an even function. This implies $\map f x \ge 0$ for all $x \in \R$. {{qed}} Category:Cosine Function sxm3zaovh90d8feog19hsx3m5qogudc
{{begin-eqn}} {{eqn | l = \laptrans {\map {f_a} t}       | r = \laptrans a       | c = {{Defof|Constant Mapping}} }} {{eqn | r = a \, \laptrans 1       | c = Linear Combination of Laplace Transforms }} {{eqn | r = a \int_0^{\to +\infty} e^{-s t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \intlimits {- \frac a s e^{-st} } {t \mathop = 0} {t \mathop \to +\infty} }} {{eqn | r = 0 - \paren {- \frac a s}       | c = Complex Exponential Tends to Zero, Exponential of Zero }} {{eqn | r = \frac a s }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map {f_a} t}       | r = \laptrans a       | c = {{Defof|Constant Mapping}} }} {{eqn | r = a \, \laptrans 1       | c = Linear Combination of Laplace Transforms }} {{eqn | r = a \frac 1 s       | c = Laplace Transform of 1 }} {{eqn | r = \frac a s }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map {I_\R} t}       | r = \laptrans t       | c = {{Defof|Identity Mapping}} }} {{eqn | r = \int_0^{\to +\infty} t e^{-st} \rd t       | c = {{Defof|Laplace Transform}} }} {{end-eqn}} From Integration by Parts: :$\displaystyle \int f g' \rd t = f g - \int f'g \rd t$ Here: {{begin-eqn}} {{eqn | l = f       | r = t       | c =  }} {{eqn | ll= \leadsto       | l = f'       | r = 1       | c = Derivative of Identity Function }} {{eqn | l = g'       | r = e^{-st} }} {{eqn | ll= \leadsto       | l = g       | r = -\frac 1 s e^{-s t}       | c = Primitive of Exponential Function }} {{end-eqn}} So:  {{begin-eqn}} {{eqn | l = \int t e^{-s t} \rd t       | r = -\frac t s e^{-s t} - \frac 1 s \int e^{-s t} \rd t }} {{eqn | r = -\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t}       | c = Primitive of Exponential Function }} {{end-eqn}} Evaluating at $t = 0$ and $t \to +\infty$: {{begin-eqn}} {{eqn | l = \laptrans t       | r = \intlimits {-\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t} } {t \mathop = 0} {t \mathop \to +\infty} }} {{eqn | r = -\frac 1 s \lim_{t \mathop \to +\infty} \frac t { e^{s t} } - \paren {0 - \frac 1 {s^2} }       | c = Exponential of Zero and One, Exponent Combination Laws: Negative Power }} {{eqn | r = 0 + \frac 1 {s^2}       | c = Limit at Infinity of Polynomial over Complex Exponential }} {{eqn | r = \frac 1 {s^2} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {t^n}       | r = \int_0^{\to +\infty} t^n e^{-s t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \intlimits {\frac {e^{-s t} } {-s} \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} t^{n - k} } {\paren {-s}^k} } } {t \mathop = 0} {t \mathop \to +\infty}       | c = Primitive of $x^n e^{a x}$ }} {{eqn | r = \sum_{k \mathop = 0}^n \paren {0 - \frac 1 {-s} \paren {-1}^k \frac {n^{\underline k} 0^{n - k} } {\paren {-s}^k} }       | c = Exponential Tends to Zero and Infinity, Exponent Combination Laws: Negative Power }} {{eqn | r = \sum_{k \mathop = 0}^n \paren {\dfrac 1 {s^{k + 1} } n^{\underline k} 0^{n - k} }       | c = simplifying }} {{eqn | r = \dfrac 1 {s^{k + 1} } n^{\underline n}       | c = $0^{n - k} = 0$ for all terms except $n = k$, when $0^{n - n} = 1$ }} {{eqn | r = \dfrac {n!} {s^{k + 1} }       | c = Integer to Power of Itself Falling is Factorial }} {{end-eqn}} {{qed}}
By Integration by Substitution (with appropriate renaming of variables): :$\ds \int \map g u \rd u = \int \map g {\map f x} \map {f'} x \rd x$ Then: {{begin-eqn}} {{eqn | l = \int \frac {\map {f'} x} {\map f x} \rd x       | r = \int \frac {\d u} u       | c = putting $\map g u := \dfrac 1 u$ }} {{eqn | r = \ln \size u + C       | c = Primitive of Reciprocal }} {{eqn | r = \ln \size {\map f x} + C       | c = Definition of $u$ }} {{end-eqn}} {{qed}}
Let $u = x \ln a$. {{begin-eqn}} {{eqn | l = \int a^x \rd x       | r = \int \map \exp {x \ln a} \rd x       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \frac 1 {\ln a} \int \map \exp u \rd u       | c = Primitive of Function of Constant Multiple }} {{eqn | r = \frac {\map \exp u} {\ln a} + C       | c = Primitive of Exponential Function }} {{eqn | r = \frac {\map \exp {x \ln a} } {\ln a} + C       | c = Definition of $u$ }} {{eqn | r = \frac {a^x} {\ln a} + C       | c = {{Defof|Power to Real Number}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \tan x \rd x       | r = -\ln \size {\cos x} + C       | c = Primitive of $\tan x$: Cosine Form }} {{eqn | r = \ln \size {\frac 1 {\cos x} } + C       | c = Logarithm of Reciprocal }} {{eqn | r = \ln \size {\sec x} + C       | c = Secant is Reciprocal of Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \sec x \rd x       | r = \ln \size {\tan x + \sec x} + C       | c = Primitive of $\sec x$: Secant plus Tangent Form }} {{eqn | r = \ln \size {\map \tan {\frac x 2 + \frac \pi 4} } + C       | c = Tangent of Half Angle plus $\dfrac \pi 4$ }} {{end-eqn}} {{qed}}
For $x > - 1$: {{begin-eqn}} {{eqn | l = e^x       | r = \lim_{n \mathop \to \infty} \left({1 + \frac x n}\right)^n       | c = {{Defof|Real Exponential Function}} }} {{eqn | o = \ge       | r = \lim_{n \mathop \to \infty} \left({1 + x}\right)       | c = Bernoulli's Inequality }} {{eqn | r = 1 + x       | c =  }} {{end-eqn}} as required. For $x \le -1$, the inequality follows from the fact that $e^x$ is positive for all $x$. {{qed}} Category:Exponential Function Category:Inequalities hqlcsp872ooy00iyqpsw50x7c4n1qxm
{{begin-eqn}} {{eqn | l = \int \csc x \rd x       | r = \ln \size {\tan \frac x 2} + C       | c = Primitive of $\csc x$: Tangent Form }} {{eqn | r = \ln \size {\frac {1 - \cos x} {\sin x} } + C       | c = Half Angle Formula for Tangent: Corollary 2 }} {{eqn | r = \ln \size {\frac 1 {\sin x} - \frac {\cos x} {\sin x} } + C       | c =  }} {{eqn | r = \ln \size {\csc x - \cot x} + C       | c = {{Defof|Cosecant}} and {{Defof|Cotangent}} }} {{end-eqn}} {{qed}}
From Derivative of Tangent Function: :$\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x$ The result follows from the definition of primitive. {{Qed}}
From Derivative of Cotangent Function: :$\dfrac \d {\d x} \cot x = -\csc^2 x$ The result follows from the definition of primitive. {{Qed}}
{{begin-eqn}} {{eqn | l = \int \tan^2 x \ \mathrm d x       | r = \int \left({\sec^2 x - 1}\right) \ \mathrm d x       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \int \sec^2 x \ \mathrm d x + \int \left({-1}\right) \ \mathrm d x       | c = Linear Combination of Integrals }} {{eqn | r = \tan x + C + \int \left({-1}\right) \ \mathrm d x       | c = Primitive of Square of Secant Function }} {{eqn | r = \tan x - x + C       | c = Primitive of Constant }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn	| l = I_n 	| r = \int \tan^n\left({x}\right) \mathrm d x }} {{eqn	| r = \frac {\tan^{n - 1} \left({x}\right)} {n - 1} - I_{n - 2}  	| c = Reduction Formula for Integral of Power of Tangent }} {{eqn	| l = I_0 	| r = \int \left(\tan\left({x}\right)\right)^0 \mathrm d x }} {{eqn	| r = \int \mathrm d x 	| c =  }} {{eqn	| r = x + C 	| c = Primitive of Constant }} {{eqn	| ll= \leadsto 	| l = I_2 	| r = \tan\left({x}\right) - x + C' 	| c = setting $n = 2$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \cot^2 x \rd x       | r = \int \paren {\csc^2 x - 1} \rd x       | c = Difference of Squares of Cosecant and Cotangent }} {{eqn | r = \int \csc^2 x \rd x + \int \paren {-1} \rd x       | c = Linear Combination of Integrals }} {{eqn | r = -\cot x + C + \int \paren {-1} \rd x       | c = Primitive of Square of Cosecant Function }} {{eqn | r = -\cot x - x + C       | c = Primitive of Constant }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = \int \sin^2 x \rd x       | r = \int \paren {\frac {1 - \cos 2 x} 2} \rd x       | c = Square of Sine }} {{eqn | r = \frac 1 2 \int \d x - \frac 1 2 \int \cos 2 x \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \frac x 2 - \frac 1 2 \int \cos 2 x \rd x + C       | c = Primitive of Constant }} {{eqn | r = \frac x 2 - \frac 1 2 \paren {\frac {\sin 2 x} 2} + C       | c = Primitive of $\cos a x$ }} {{eqn | r = \frac x 2 - \frac {\sin 2 x} 4 + C       | c = rearranging }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn	| l = I_n 	| r = \int \cos^n x \rd x }} {{eqn	| r = \dfrac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n I_{n-2} 	| c = Reduction Formula for Integral of Power of Cosine }} {{eqn	| l = I_0 	| r = \int \left({\cos x}\right)^0 \rd x 	| c =  }} {{eqn	| r = \int \rd x 	| c =  }} {{eqn	| r = x + C 	| c = Primitive of Constant }} {{eqn	| ll= \leadsto 	| l  = I_2 	| r  = \frac {\cos x \sin x} 2 + \frac x 2 + \frac C 2 	| c  = setting $n = 2$ }} {{eqn	| r = \frac {\sin 2 x} 4 + \frac x 2 + C' 	| c = Double Angle Formula for Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \cos^2 x \rd x       | r = \int \paren {\frac {1 + \cos 2 x} 2} \rd x       | c = Square of Cosine }} {{eqn | r = \int \frac 1 2 \rd x + \int \paren {\frac {\cos 2 x} 2} \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \frac x 2 + C + \int \paren {\frac {\cos 2 x} 2} \rd x       | c = Primitive of Constant }} {{eqn | r = \frac x 2 + C + \frac 1 2 \int \cos 2 x \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac x 2 + \frac 1 2 \paren {\frac {\sin 2 x} 2} + C       | c = Primitive of Function of Constant Multiple and Primitive of Cosine Function }} {{eqn | r = \frac x 2 + \frac {\sin 2 x} 4 + C       | c = Primitive of Function of Constant Multiple and Primitive of Cosine Function }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = \int \cos^2 x \rd x        | r = \frac 1 4 \int \left({e^{i x} + e^{-i x} }\right)^2 \rd x       | c = Cosine Exponential Formulation }} {{eqn | r = \frac 1 4 \int \left({e^{2 i x} + 2 + e^{-2 i x} }\right) \rd x }} {{eqn | r = \frac 1 4 \left({\frac{e^{2 i x} - e^{-2 i x} } {2 i} + 2 x}\right) + C       | c = Primitive of $e^{a x}$, Primitive of Constant }} {{eqn | r = \frac {\sin 2 x} 4 + \frac x 2 + C       | c = Sine Exponential Formulation }} {{end-eqn}} {{qed}}
From Derivative of Secant Function: :$\dfrac \d {\d x} \sec x = \sec x \tan x$ The result follows from the definition of primitive. {{Qed}}
From Derivative of Cosecant Function: :$\dfrac \d {\d x} \csc x = -\csc x \cot x$ The result follows from the definition of primitive. {{Qed}}
From Derivative of Hyperbolic Cosine: :$\map {\dfrac \d {\d x} } {\cosh x} = \sinh x$ The result follows from the definition of primitive. {{qed}}
From Derivative of Hyperbolic Sine: :$\map {\dfrac \d {\d x} } {\sinh x} = \cosh x$ The result follows from the definition of primitive. {{qed}}
{{begin-eqn}} {{eqn	| l = \int \tanh x \rd x 	| r = -i \int \tan i x \rd x 	| c = Hyperbolic Tangent in terms of Tangent }} {{eqn	| r = -\int \tan i x \rd \paren {i x} 	| c = Primitive of Function of Constant Multiple }} {{eqn	| r = \ln \cmod {\cos i x} + C 	| c = Primitive of $\tan x$: Cosine Form }} {{eqn	| r = \ln \cmod {\cosh x} + C 	| c = {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \map \ln {\cosh x} + C       | c = Graph of Hyperbolic Cosine Function: $\cosh x > 0$ for all $x$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \tanh x \rd x       | r = \int \frac {\sinh x} {\cosh x} \rd x       | c = {{Defof|Hyperbolic Tangent}} }} {{eqn | r = \int \frac {\paren {\cosh x}'} {\cosh x} \rd x       | c = Derivative of Hyperbolic Cosine }} {{eqn | r = \ln \size {\cosh x} + C       | c = Primitive of Function under its Derivative }} {{eqn | r = \map \ln {\cosh x} + C       | c = Graph of Hyperbolic Cosine Function: $\cosh x > 0$ for all $x$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \coth x \rd x       | r = \int \frac {\cosh x} {\sinh x} \rd x       | c = {{Defof|Hyperbolic Cotangent}} }} {{eqn | r = \int \frac {\paren {\sinh x}'} {\sinh x} \rd x       | c = Derivative of Hyperbolic Sine }} {{eqn | r = \ln \size {\sinh x} + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn |l = \lim_{x \mathop \to 0} \frac {\tan x} x        |r = \lim_{x \mathop \to 0} \frac 1 {\cos x} \frac {\sin x} x       |c = {{Defof|Tangent Function}} }} {{eqn |r = \lim_{x \mathop \to 0} \frac 1 {\cos x} \lim_{x \mathop \to 0} \frac {\sin x} x       |c = Product Rule for Limits of Functions }} {{eqn |r = \lim_{x \mathop \to 0} \frac {\sin x} x        |c = Cosine of Zero is One }} {{eqn |r = 1       |c = Limit of Sine of X over X }} {{end-eqn}} {{qed}}
By L'Hôpital's Rule: {{begin-eqn}} {{eqn |l = \lim_{x \mathop \to 0} \frac {\tan x} x        |r = \lim_{x \mathop \to 0} \frac {\sec^2 x} 1       |c = Derivative of Tangent Function }} {{eqn |r = 1       |c = Secant of Zero }} {{end-eqn}} {{qed}}
Using De Moivre's Formula: :$\sin x = \dfrac {\left({\cos \dfrac x n + i \sin \dfrac x n}\right)^n - \left({\cos \dfrac x n - i \sin \dfrac x n}\right)^n} {2i}$ The difference between two $n$th powers can be extracted into linear factors using $n$th roots of unity. For large $n$, we can replace: : $\cos \dfrac x n$ by $1$ : $\sin \dfrac x n$ by $\dfrac x n$ {{proof wanted}}
We have that $\sin x$  has a power series representation: :$\sin x = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots$ The roots of sine are the numbers $k \pi$, where $k$ is any integer. From the Polynomial Factor Theorem, the following ''might'' be true: :$\displaystyle \sin x = A x \prod \paren {1 - \frac x {k \pi} }$ where the product is taken over all $n \in \Z \setminus \set 0$, and $A$ is some constant. The intuition is as follows. {{begin-eqn}} {{eqn | l = \sin x       | r = \ldots \paren {1 - \frac x {2 \pi} } \paren {1 - \frac x \pi} A x \paren {1 + \frac x \pi} \paren {1 + \frac x {2 \pi} } \cdots }} {{eqn | r = A x \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {2^2 \pi^2} } \paren {1 - \frac {x^2} {3^2 \pi^2} } \cdots }} {{eqn | ll= \leadsto       | l = \frac {\sin x} x       | r = A \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {2^2 \pi^2} } \paren {1 - \frac {x^2} {3^2 \pi^2} } \cdots       | c = for $x \ne 0$ }} {{end-eqn}} That $\dfrac {\sin x} x \to 1$ as $x \to 0$ is a well known limit. Letting $x$ tend to $0$ in the above equation implies that $A = 1$. We now formalize the above claims. {{proof wanted}}
{{begin-eqn}} {{eqn | l = \map {\log_b} {\dfrac 1 x}       | r = \log_b 1 - \log_b x       | c = Difference of Logarithms }} {{eqn | r = 0 - \log_b x       | c = Logarithm of 1 is 0 }} {{end-eqn}} Hence the result. {{qed}} Category:Logarithms Category:Reciprocals kgfbjw7h1n6cupqiu7qtvossewt0xp8
Let: {{begin-eqn}} {{eqn | l = \int \sech x \rd x       | r = \int \frac {\sech^2 x} {\sech x} \rd x       | c =  }} {{eqn | r = \int \frac {\sech^2 x} {\sqrt {1 - \tanh^2 x} } \rd x       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = u       | r = \tanh x       | c =  }} {{eqn | ll= \leadsto       | l = u'       | r = \sech^2 x       | c = Derivative of Hyperbolic Tangent }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \sech x \rd x       | r = \int \frac {\d u} {\sqrt {1 - u^2} }       | c = Integration by Substitution }} {{eqn | r = \map \arcsin u + C       | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: Arcsine Form }} {{eqn | r = \map \arcsin {\tanh x} + C       | c = Definition of $u$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = \int \sech x \rd x       | r = \int \frac 2 {e^x + e^{-x} } \rd x       | c = {{Defof|Hyperbolic Secant|index = 1}} }} {{eqn | r = \int \frac {2 e^x} {e^{2 x} + 1} \rd x       | c = multiplying top and bottom by $e^x$ }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = u       | r = e^x       | c =  }} {{eqn | ll= \leadsto       | l = u'       | r = e^x       | c = Derivative of Exponential Function }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \sech x \rd x       | r = \int \frac {2 \rd u} {u^2 + 1}       | c = Integration by Substitution }} {{eqn | r = 2 \arctan u + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form }} {{eqn | r = 2 \map \arctan {e^x} + C       | c = Definition of $u$ }} {{end-eqn}} {{qed}}
For $x \in \R$ and $n \in \N_{> 0}$, let: :$\map {f_n} x = \dfrac 1 2 \paren {\paren {1 + \dfrac x n}^n - \paren {1 - \dfrac x n}^n }$ Then $\map {f_n} x = 0$ {{iff}}: {{begin-eqn}} {{eqn | l = \paren {1 + \frac x n}^n       | r = \paren {1 - \frac x n}^n       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 1 + \frac x n        | r = \paren {1 - \frac x n} e^{2 \pi i \frac k n}       | c = for $k \in \Z$ }} {{eqn | ll= \leadstoandfrom       | l = x       | r = n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1}       | c =  }} {{eqn | r = n i \, \map \tan {\frac {k \pi} n }       | c = Tangent Exponential Formulation }} {{end-eqn}} Hence the roots of $\map {f_{2 m + 1} } x$ are: :$\paren {2 m + 1} i \, \map \tan {\dfrac {k \pi} {2 m + 1} }$  for $-m \le k \le m$. Observe that $\map {f_{2 m + 1} } x$ is a polynomial of degree $2 m + 1$.  Then for some constant $C$, we have: {{begin-eqn}} {{eqn | l = \map {f_{2 m + 1} } x       | r = C x \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \paren {1 - \frac x {\paren {2 m + 1} i \, \map \tan {k \pi / \paren {2 m + 1} } } }       | c = Polynomial Factor Theorem }} {{eqn | r = C x \prod_{k \mathop = 1}^m \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }       | c = Tangent Function is Odd }} {{end-eqn}} It can be seen from the Binomial Theorem that the coefficient of $x$ in $\map {f_{2 m + 1} } x$ is $1$. Hence $C = 1$, and we obtain: :$\displaystyle \map {f_{2 m + 1} } x = x \prod_{k \mathop = 1}^m \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }$ Let $l < m$. Then: :$\displaystyle x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } \le \map {f_{2 m + 1} } x$ Taking the limit as $m \to \infty$ we have: {{begin-eqn}} {{eqn | l = \lim_{m \mathop \to \infty} x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} \paren {\frac {k \pi / \paren {2 m + 1} } {\map \tan {k \pi / \paren {2 m + 1} } } }^2 }       | o = \le        | r = \frac 1 2 \paren {e^x - e^{-x} }       | c = {{Defof|Exponential Function/Real|subdef = Limit of Sequence|Exponential Function}} }} {{eqn | ll= \leadsto       | o = \le       | l = x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} }       | r = \sinh x       | c = Limit of Tan X over X and Definition of Hyperbolic Sine }} {{end-eqn}} By Tangent Inequality, we have: :$\map \tan {\dfrac {k \pi} {2 m + 1} } \ge \dfrac {k \pi} {2 m + 1}$ and hence: :$\displaystyle \map {f_{2 l + 1} } x \le x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} } \le \sinh x$ Taking the limit as $l \to \infty$, we have by the Squeeze Theorem: :$\displaystyle x \prod_{k \mathop = 1}^\infty \paren {1 + \frac {x^2} {k^2 \pi^2} } = \sinh x$ Substituting $x \mapsto i x$, we obtain: :$\displaystyle \sin x = x \prod_{k \mathop = 1}^\infty \paren {1 - \frac {x^2} {k^2 \pi^2} }$ {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \coth x + \csch x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac \d {\d x} \coth x + \frac \d {\d x} \csch x        | c = Linear Combination of Derivatives }} {{eqn | r = -\csch^2 x + \frac \d {\d x} \csch x        | c = Derivative of Hyperbolic Cotangent }} {{eqn | r = -\csch^2 x - \csch x \coth x        | c = Derivative of Hyperbolic Cosecant }} {{eqn | r = -\csch x \left({\csch x + \coth x}\right)       | c = factorising }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \csch x \rd x       | r = \int \frac {\csch x \paren {\csch x + \coth x} } {\csch x + \coth x} \rd x       | c = multiplying top and bottom by $\csch x + \coth x$ }} {{eqn | r = -\int \frac {-\csch x \paren {\csch x + \coth x} } {\csch x + \coth x} \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -\ln \size {\csch x + \coth x} + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = \int \csch x \rd x       | r = \int \frac 2 {e^x - e^{-x} } \rd x       | c = {{Defof|Hyperbolic Cosecant}} }} {{eqn | r = \int \frac {2 e^x} {e^{2 x} - 1} \rd x       | c = multiplying top and bottom by $e^x$ }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = u       | r = e^x       | c =  }} {{eqn | ll= \leadsto       | l = u'       | r = e^x       | c = Derivative of Exponential Function }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \csch x \rd x       | r = \int \frac {2 \rd u} {u^2 - 1}       | c = Integration by Substitution }} {{eqn | r = -2 \coth^{-1} u + C       | c = Primitive of Reciprocal of $x^2 - a^2$: $\coth^{-1}$ form }} {{eqn | r = -2 \map {\coth^{-1} } {e^x} + C       | c = Definition of $u$ }} {{end-eqn}} {{qed}}
From Derivative of Hyperbolic Tangent: :$\map {\dfrac \d {\d x} } {\tanh x} = \sech^2 x$ The result follows from the definition of primitive. {{Qed}}
From Derivative of Hyperbolic Cotangent: :$\map {\dfrac \d {\d x} } {\coth x} = -\csch^2 x$ The result follows from the definition of primitive. {{Qed}}
{{begin-eqn}} {{eqn | l = \int \tanh^2 x \rd x       | r = \int \paren {1 - \sech^2 x} \rd x       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{eqn | r = \int 1 \rd x - \int \sech^2 x \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \int 1 \rd x - \tanh x + C       | c = Primitive of Square of Hyperbolic Secant Function }} {{eqn | r = x - \tanh x + C       | c = Primitive of Constant }} {{end-eqn}} {{Qed}}