hcju/proofwikiqa huggingface dataset:
With a view to expressing the primitive 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 = \coth^{-1} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac {-a} {x^2 - a^2}       | c = Derivative of $\coth^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = x^2       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^3} 3       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^2 \coth^{-1} \frac x a \ \mathrm d x       | r = \frac {x^3} 3 \coth^{-1} \frac x a - \int \frac {x^3} 3 \left({\frac {-a} {x^2 - a^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^3} 3 \coth^{-1} \frac x a + \frac a 3 \int \frac {x^3 \ \mathrm d x} {x^2 - a^2} + C       | c = simplifying }} {{eqn | r = \frac {x^3} 3 \coth^{-1} \frac x a + \frac a 3 \left({\frac {x^2} 2 + \frac {a^2} 2 \ln \left({x^2 - a^2}\right) }\right) + C       | c = Primitive of $\dfrac {x^3} {x^2 - a^2}$ }} {{eqn | r = \frac {a x^2} 6 + \frac {x^3} 3 \coth^{-1} \frac x a + \frac {a^3} 6 \ln \left({x^2 - a^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive 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 = \sinh^{-1} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 {\sqrt {x^2 + a^2} }       | c = Derivative of $\sinh^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = \frac 1 {x^2}       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {-1} x       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\sinh^{-1} \dfrac x a \ \mathrm d x} {x^2}       | r = \left({\sinh^{-1} \frac x a}\right) \left({\frac {-1} x}\right) - \int \left({\frac {-1} x}\right) \left({\frac 1 {\sqrt {x^2 + a^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {-\sinh^{-1} \dfrac x a} x + \int \frac {\mathrm d x} {x \sqrt {x^2 + a^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\sinh^{-1} \dfrac x a} x - \frac 1 a \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C       | c = Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$ }} {{end-eqn}} {{qed}}
With a view to expressing the primitive 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 = \cosh^{-1} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 {\sqrt {x^2 - a^2} }       | c = Derivative of $\cosh^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = \frac 1 {x^2}       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {-1} x       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\cosh^{-1} \dfrac x a \ \mathrm d x} {x^2}       | r = \left({\cosh^{-1} \frac x a}\right) \left({\frac {-1} x}\right) - \int \left({\frac {-1} x}\right) \left({\frac 1 {\sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {-\cosh^{-1} \dfrac x a} x + \int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\cosh^{-1} \dfrac x a} x - \frac 1 a \operatorname{arcsec} \left\vert{\frac x a}\right\vert + C       | c = Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$ }} {{end-eqn}} {{qed}} {{finish|Not sure how to get there from here, I wonder if there is a problem with the required target}}
With a view to expressing the primitive in the form: :$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \tanh^{-1} \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac a {a^2 - x^2}       | c = Derivative of $\tanh^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = \frac 1 {x^2}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {-1} x       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\tanh^{-1} \dfrac x a \rd x} {x^2}       | r = \paren {\tanh^{-1} \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac a {a^2 - x^2} } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {-\tanh^{-1} \dfrac x a} x + a \int \frac {\d x} {x \paren {a^2 - x^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\tanh^{-1} \dfrac x a} x + a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2} {a^2 - x^2} } } + C       | c = Primitive of $\dfrac 1 {x \paren {a^2 - x^2} }$ }} {{eqn | r = \frac {-\tanh^{-1} \dfrac x a} x + \frac 1 {2 a} \map \ln {\frac {x^2} {a^2 - x^2} } + C       | c = simplifying }} {{end-eqn}} {{qed}}
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 = \coth^{-1} \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac {-a} {x^2 - a^2}       | c = Derivative of $\coth^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = \frac 1 {x^2}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {-1} x       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\coth^{-1} \dfrac x a \rd x} {x^2}       | r = \paren {\coth^{-1} \frac x a} \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac {-a} {x^2 - a^2} } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {-\coth^{-1} \dfrac x a} x - a \int \frac {\d x} {x \paren {x^2 - a^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\coth^{-1} \dfrac x a} x - a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } } + C       | c = Primitive of $\dfrac 1 {x \paren {x^2 - a^2} }$ }} {{eqn | r = \frac {-\coth^{-1} \dfrac x a} x - \frac 1 {2 a} \map \ln {\frac {x^2 - a^2} {x^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\coth^{-1} \dfrac x a} x + \frac 1 {2 a} \map \ln {\frac {x^2} {x^2 - a^2} } + C       | c = Logarithm of Reciprocal }} {{end-eqn}} {{qed}}
With a view to expressing the primitive 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 = \sinh^{-1} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 {\sqrt {x^2 + a^2} }       | c = Derivative of $\sinh^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = x^m       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^{m + 1} } {m + 1}       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\sinh^{-1} \dfrac x a \ \mathrm d x} {x^2}       | r = \left({\sinh^{-1} \frac x a}\right) \left({\frac {x^{m + 1} } {m + 1} }\right) - \int \left({\frac {x^{m + 1} } {m + 1} }\right) \left({\frac 1 {\sqrt {x^2 + a^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^{m + 1} } {m + 1} \sinh^{-1} \frac x a - \frac 1 {m + 1}  \int \frac {x^{m + 1} } {\sqrt {x^2 + a^2} } \ \mathrm d x + C       | c = simplifying }} {{end-eqn}} {{qed}}
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 = \coth^{-1} \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac {-a} {x^2 - a^2}       | c = Derivative of $\coth^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = x^m       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {x^{m + 1} } {m + 1}       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^m \coth^{-1} \frac x a \rd x       | r = \paren {\coth^{-1} \frac x a} \paren {\frac {x^{m + 1} } {m + 1} } - \int \paren {\frac {x^{m + 1} } {m + 1} } \paren {\frac {-a} {x^2 - a^2} } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^{m + 1} } {m + 1} \coth^{-1} \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} } {a^2 - x^2} \rd x + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \operatorname{sech}^{-1} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\rd u} {\rd x}       | r = \frac {-a} {\left\vert{x}\right\vert \sqrt{a^2 + x^2} }       | c = Derivative of $\operatorname{csch}^{-1} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\rd v} {\rd x}       | r = x^m       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^{m + 1} } {m + 1}       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\operatorname{csch}^{-1} \dfrac x a \rd x} {x^2}       | r = \left({\operatorname{csch}^{-1} \frac x a}\right) \left({\frac {x^{m + 1} } {m + 1} }\right) - \int \left({\frac {x^{m + 1} } {m + 1} }\right) \left({\frac {-a} {\left\vert{x}\right\vert \sqrt{a^2 + x^2} } }\right) \rd x + C       | c = Integration by Parts }} {{eqn | r = \begin{cases} \displaystyle \frac {x^{m + 1} } {m + 1} \operatorname{csch}^{-1} \frac x a + \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x > 0 \\ \displaystyle \frac {x^{m + 1} } {m + 1} \operatorname{csch}^{-1} \frac x a - \frac 1 {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x < 0 \\ \end{cases}       | c = Definition of Absolute Value }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = u       | r = \tanh \dfrac x 2       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = 2 \tanh^{-1} u       | c = {{Defof|Inverse Hyperbolic Tangent|subdef = Real|index = 1}} }} {{eqn | ll= \leadsto       | l = \dfrac {\d x} {\d u}       | r = \dfrac 2 {1 - u^2}       | c = Derivative of Inverse Hyperbolic Tangent and Derivative of Constant Multiple }} {{eqn | l = \sinh x       | r = \dfrac {2 u} {1 - u^2}       | c = Hyperbolic Tangent Half-Angle Substitution for Sine }} {{eqn | l = \cosh x       | r = \dfrac {1 + u^2} {1 - u^2}       | c = Hyperbolic Tangent Half-Angle Substitution for Cosine }} {{end-eqn}} The result follows from Integration by Substitution. {{qed}}
{{begin-eqn}} {{eqn | l = 1 + \cos x       | r = \cos 0 + \cos x       | c = Cosine of Zero is One }} {{eqn | r = 2 \map \cos {\dfrac {0 + x} 2} \map \cos {\dfrac {0 - x} 2}       | c = Prosthaphaeresis Formula for Cosine plus Cosine }} {{eqn | r = 2 \map \cos {\dfrac x 2} \map \cos {\dfrac {-x} 2}       | c = simplifying }} {{eqn | r = 2 \map \cos {\dfrac x 2} \map \cos {\dfrac x 2}       | c = Cosine Function is Even }} {{eqn | r = 2 \map {\cos^2} {\frac x 2}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \frac 1 {1 + \cos x}       | r = \frac 1 2 \map {\sec^2} {\frac x 2}       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos x       | r = 2 \cos^2 \frac x 2 - 1       | c = Cosine Double Angle Formula }} {{eqn | ll= \leadstoandfrom       | l = 1 + \cos x       | r = 2 \cos^2 \frac x 2       | c = adding $1$ to both sides }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 {1 + \cos x}       | r = \frac 1 2 \frac 1 {\cos^2 \frac x 2}       | c = taking the reciprocal of both sides }} {{eqn | r = \frac 1 2 \sec^2 \frac x 2       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \frac 1 {1 + \cos x}  	| r = \frac 1 {1 + \frac {1- \tan^2 \frac x 2} {1 + \tan^2 \frac x 2} } 	| c = Tangent Half-Angle Substitution for Cosine }} {{eqn	| r = \frac {1 + \tan^2 \frac x 2} 2 	| c = multiplying through $\frac {1 + \tan^2 \frac x 2} {1 + \tan^2 \frac x 2}$ }} {{eqn	| r = \frac 1 2 \sec^2 \frac x 2 	| c = Difference of Squares of Secant and Tangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 1 - \cos x       | r = \cos 0 - \cos x       | c = Cosine of Zero is One }} {{eqn | r = -2 \sin \left({\dfrac {0 + x} 2}\right) \sin \left({\dfrac {0 - x} 2}\right)       | c = Prosthaphaeresis Formula for Cosine minus Cosine }} {{eqn | r = -2 \sin \left({\dfrac x 2}\right) \sin \left({\dfrac {-x} 2}\right)       | c = simplifying }} {{eqn | r = 2 \sin \left({\dfrac x 2}\right) \sin \left({\dfrac x 2}\right)       | c = Sine Function is Odd }} {{eqn | r = 2 \sin^2 \left({\frac x 2}\right)       | c = simplifying }} {{eqn | ll= \implies       | l = \frac 1 {1 - \cos x}       | r = \frac 1 2 \csc^2 \left({\frac x 2}\right)       | c = {{Defof|Cosecant}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos x       | r = 1 - 2 \sin^2 \frac x 2       | c = Cosine Double Angle Formula }} {{eqn | ll= \iff       | l = 1 - \cos x       | r = 2 \sin^2 \frac x 2       | c = rearranging }} {{eqn | ll= \iff       | l = \frac 1 {1 - \cos x}       | r = \frac 1 2 \frac 1 {\sin^2 \frac x 2}       | c = taking the reciprocal of both sides }} {{eqn | r = \frac 1 2 \csc^2 \frac x 2       | c = Definition of Cosecant }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \frac 1 {1 - \cos x} 	| r = \frac 1 {1 - \frac {1 - \map {\tan^2} {\frac x 2} } {1 + \map {\tan^2} {\frac x 2} } } 	| c = Tangent Half-Angle Substitution for Cosine }} {{eqn	| r = \frac {1 + \map {\tan^2} {\frac x 2} } {1 + \map {\tan^2} {\frac x 2} - 1 + \map {\tan^2} {\frac x 2} } 	| c = multiplying through $\frac {1 + \map {\tan^2} {\frac x 2} } {1 + \map {\tan^2} {\frac x 2} }$ }} {{eqn	| r = \frac {\map {\sec^2} {\frac x 2} } {2 \map {\tan^2} {\frac x 2} } 	| c = Difference of Squares of Secant and Tangent }} {{eqn	| r = \frac 1 2 \cdot \frac 1 {\map {\cos^2} {\frac x 2} } \cdot \frac {\map {\cos^2} {\frac x 2} } {\map {\sin^2} {\frac x 2} } 	| c = {{Defof|Secant Function}}, {{Defof|Tangent}} }} {{eqn	| r = \frac 1 {2 \map {\sin^2} {\frac x 2} } }} {{eqn	| r = \frac 1 2 \map {\csc^2} {\frac x 2} 	| c = {{Defof|Cosecant}} }} {{end-eqn}} {{qed}}
{{AimForCont}} $\log_b a$ is rational. Then: :$\exists p, q \in \N_{>0} : \log_b a = \dfrac p q$ where $p \perp q$. Then: {{begin-eqn}} {{eqn | l = \log_b a       | r = \dfrac p q       | c =  }} {{eqn | ll= \leadsto       | l = b^{\frac p q}       | r = a       | c = {{Defof|Real General Logarithm}} }} {{eqn | ll= \leadsto       | l = \sqrt [q] {\paren {b^p} }       | r = a       | c = {{Defof|Rational Power}} }} {{eqn | ll= \leadsto       | l = b^p       | r = a^q       | c = {{Defof|Root (Analysis)|Root}} }} {{end-eqn}} which contradicts the initial assumption: :$\nexists m, n \in \N: a^m = b^n$ Hence the result by Proof by Contradiction. {{qed}} Category:Number Theory Category:Irrationality Proofs Category:Logarithms qgxchhi9q3h3ade7m4iw14nq2jb189f
{{begin-eqn}} {{eqn | l = \tan 22.5^\circ       | r = \tan \frac {45^\circ} 2       | c =  }} {{eqn | r = \frac {1 - \cos 45^\circ} {\sin 45^\circ}       | c = Half Angle Formulas/Tangent/Corollary 2 }} {{eqn | r = \frac {1 - \frac {\sqrt 2} 2} {\frac {\sqrt 2} 2}       | c = Cosine of 45 Degrees and Sine of 45 Degrees }} {{eqn | r = \sqrt 2 - 1       | c = multiplying top and bottom by $\sqrt 2$ }} {{end-eqn}} {{qed}} Category:Tangent Function 1m3wwb0zdcdp5uazkbqzc1k5uga9wd2
{{begin-eqn}} {{eqn | l = \tan 67.5 \degrees       | r = \map \tan {45 \degrees + 22.5 \degrees}       | c =  }} {{eqn | r = \frac {\tan 45 \degrees + \tan 22.5 \degrees} {1 - \tan 45 \degrees \tan 22.5 \degrees}       | c = Tangent of Sum }} {{eqn | r = \frac {1 + \paren {\sqrt 2 - 1} } {1 - 1 \times \paren {\sqrt 2 - 1} }       | c = Tangent of $45 \degrees$ and Tangent of $22.5 \degrees$ }} {{eqn | r = \frac {\sqrt 2} {2 - \sqrt 2}       | c = simplifying }} {{eqn | r = \frac {\sqrt 2 \paren {2 + \sqrt 2} } {\paren {2 - \sqrt 2} \paren {2 + \sqrt 2} }       | c = multiplying top and bottom by $2 + \sqrt 2$ }} {{eqn | r = \frac {2 \sqrt 2 + 2} {4 - 2}       | c = Difference of Two Squares }} {{eqn | r = \sqrt 2 + 1       | c = simplifying }} {{end-eqn}} {{qed}} Category:Tangent Function hul6jsgdvh6kjjq1uwbjqvmiu04qpvy
{{begin-eqn}} {{eqn | l = \sin x + \cos x       | r = \sin x + \map \sin {\frac \pi 2 - x}       | c = Sine of Complement equals Cosine }} {{eqn | r = 2 \, \map \sin {\frac {x + \paren {\frac \pi 2 - x} } 2} \map \cos {\frac {x - \paren {\frac \pi 2 - x} } 2}       | c = Prosthaphaeresis Formula for Sine plus Sine }} {{eqn | r = 2 \sin \frac \pi 4 \, \map \cos {x - \frac \pi 4}       | c = simplifying }} {{eqn | r = \sqrt 2 \, \map \cos {x - \frac \pi 4}       | c = Sine of $\dfrac \pi 4$ }} {{end-eqn}} {{qed}} Category:Sine Function Category:Cosine Function ajw6vg0ofm5gr79irlb0v9tmr1erk5t
{{begin-eqn}} {{eqn | l = \sin x - \cos x       | r = \sin x - \sin \left({\frac \pi 2 - x}\right)       | c = Sine of Complement equals Cosine }} {{eqn | r = 2 \cos \left({\frac {x + \left({\frac \pi 2 - x}\right)} 2}\right) \sin \left({\frac {x - \left({\frac \pi 2 - x}\right)} 2}\right)       | c = Prosthaphaeresis Formula for Sine minus Sine }} {{eqn | r = 2 \cos \frac \pi 4 \sin \left({x - \frac \pi 4}\right)       | c = simplifying }} {{eqn | r = \sqrt 2 \sin \left({x - \frac \pi 4}\right)       | c = Cosine of $\dfrac \pi 4$ }} {{end-eqn}} {{qed}} Category:Sine Function Category:Cosine Function mhg93r8g7ltbh2e0t9nwjzdio0bfyfm
{{begin-eqn}} {{eqn | l = \sin x - \cos x       | r = \sqrt 2 \, \map \sin {x - \dfrac \pi 4}       | c = Sine of x minus Cosine of x: Sine Form }} {{eqn | r = \sqrt 2 \, \map \cos {\frac \pi 2 - \paren {x - \dfrac \pi 4} }       | c = Cosine of Complement equals Sine }} {{eqn | r = \sqrt 2 \, \map \cos {\frac \pi 2 - x + \dfrac \pi 4}       | c = simplifying }} {{eqn | r = \sqrt 2 \, \map \cos {\frac {3 \pi} 4 - x}       | c = simplifying }} {{eqn | r = \sqrt 2 \, \map \cos {x - \frac {3 \pi} 4}       | c = Cosine Function is Even }} {{end-eqn}} {{qed}} Category:Sine Function Category:Cosine Function 4w1mna76rlqbh5e3l9cnpda8q6g5jgw
Let it be assumed that $p \sin x + q \cos x$ can be expressed in the form $M \map \cos {x + \phi}$. Then: {{begin-eqn}} {{eqn | l = p \sin x + q \cos x       | r = M \cos \map \cos {x + \phi}       | c =  }} {{eqn | ll= \leadsto       | l = \frac p M \sin x + \frac q M \cos x       | r = \map \cos {x + \phi}       | c =  }} {{eqn | r = \cos x \cos \phi - \sin x \sin \phi       | c = Cosine of Sum }} {{eqn | ll= \leadsto       | l = \sin x \sin \phi       | r = \frac {-p} M \sin x       | c = equating terms in $\sin x$ }} {{eqn | l = \cos x \cos \phi       | r = \frac q M \cos x       | c =  and $\cos x$ }} {{eqn | n = 1       | ll= \leadsto       | l = \sin \phi       | r = \frac {-p} M       | c =  }} {{eqn | n = 2       | l = \cos \phi       | r = \frac q M       | c =  }} {{eqn | ll= \leadsto       | l = \sin^2 \phi + \cos^2 \phi       | r = \paren {\frac {-p} M}^2 + \paren {\frac q M}^2       | c = squaring and adding $(1)$ and $(2)$ }} {{eqn | ll= \leadsto       | l = 1       | r = \frac {p^2 + q^2} {M^2}       | c =  }} {{eqn | ll= \leadsto       | r = M       | l = \sqrt {p^2 + q^2}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \tan \phi       | r = \frac {\sin \phi} {\cos \phi}       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {-p / M} {q / M}       | c = from $(1)$ and $(2)$ }} {{eqn | r = \frac {-p} q       | c = simplifying }} {{end-eqn}} Hence the result. {{qed}} Category:Sine Function Category:Cosine Function jr2opcgzj2mrev53xztoxdexi2w2o40
Let it be assumed that $p \sin x + q \cos x$ can be expressed in the form $M \sin \left({x + \phi}\right)$. Then: {{begin-eqn}} {{eqn | l = p \sin x + q \cos x       | r = M \sin \left({x + \phi}\right)       | c =  }} {{eqn | ll= \implies       | l = \frac p M \sin x + \frac q M \cos x       | r = \sin \left({x + \phi}\right)       | c =  }} {{eqn | r = \sin x \cos \phi + \cos x \sin \phi       | c = Sine of Sum }} {{eqn | ll= \implies       | l = \sin x \cos \phi       | r = \frac p M \sin x       | c = equating terms in $\sin x$ }} {{eqn | l = \cos x \sin \phi       | r = \frac q M \cos x       | c =  and $\cos x$ }} {{eqn | n = 1       | ll= \implies       | l = \sin \phi       | r = \frac q M       | c =  }} {{eqn | n = 2       | l = \cos \phi       | r = \frac p M       | c =  }} {{eqn | ll= \implies       | l = \sin^2 \phi + \cos^2 \phi       | r = \left({\frac q M}\right)^2 + \left({\frac p M}\right)^2       | c = squaring and adding $(1)$ and $(2)$ }} {{eqn | ll= \implies       | l = 1       | r = \frac {p^2 + q^2} {M^2}       | c =  }} {{eqn | ll= \implies       | r = M       | l = \sqrt {p^2 + q^2}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \tan \phi       | r = \frac {\sin \phi} {\cos \phi}       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {q / M} {p / M}       | c = from $(1)$ and $(2)$ }} {{eqn | r = \frac q p       | c = simplifying }} {{end-eqn}} Hence the result. {{qed}} Category:Sine Function Category:Cosine Function oc56d623463q7jxezd0o7ngan8b9zd8
First note that: {{begin-eqn}} {{eqn | n = 1       | l = \map {\frac \d {\d x} } {\map \sin {a x + \phi} }       | r = a \, \map \cos {a x + \phi}       | c = Derivative of $\sin a x$ etc. }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\cos a x \rd x} {\map \sin {a x + \phi} }       | r = \int \frac {\cos a x \cos \phi \rd x} {\cos \phi \, \map \sin {a x + \phi} }       | c = multiplying top and bottom by $\cos \phi$ }} {{eqn | r = \int \frac {\paren {\cos a x \cos \phi - \sin a x \sin \phi + \sin a x \sin \phi} \rd x} {\cos \phi \, \map \sin {a x + \phi} }       | c =  }} {{eqn | r = \int \frac {\paren {\map \cos {a x + \phi} + \sin a x \sin \phi} \rd x} {\cos \phi \, \map \sin {a x + \phi} }       | c = Cosine of Sum }} {{eqn | r = \frac 1 {\cos \phi} \int \frac {\map \cos {a x + \phi} \rd x} {\map \sin {a x + \phi} } + \frac {\sin \phi} {\cos \phi} \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} }       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {\cos \phi} \ln \size {\map \sin {a x + \phi} } + \frac {\sin \phi} {\cos \phi} \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} }       | c = Primitive of Function under its Derivative and $(1)$ }} {{eqn | r = \frac {\ln \size {\map \sin {a x + \phi} } } {a \cos \phi} + \tan \phi \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} } + C       | c = Tangent is Sine divided by Cosine }} {{end-eqn}} {{qed}} Category:Primitives involving Sine Function Category:Primitives involving Cosine Function h6zl6lhm66diinvu2vcr0bv7k58ai5z