hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \int \coth^2 x \rd x       | r = \int \paren {1 + \csch^2 x} \rd x       | c = Difference of Squares of Hyperbolic Cotangent and Cosecant }} {{eqn | r = \int 1 \rd x + \int \csch^2 x \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \int 1 \rd x - \coth x + C       | c = Primitive of Square of Hyperbolic Cosecant Function }} {{eqn | r = x - \coth x + C       | c = Primitive of Constant }} {{end-eqn}} {{Qed}}
Let the finite subdivision of $\mu_c$ be: :$\left\{{c - M, c, c + M}\right\}$ Let $\epsilon > 0$. For $t \to \left({c - M}\right)^+$, choose $\delta$ to be any number at all, because: :$c - M < t < c - M + \delta \implies \left\vert{ \mu_c \left({t}\right) - 0 }\right\vert = \left\vert{ 0 - 0 }\right\vert < \epsilon$ holds, from True Statement is implied by Every Statement. The case for $t \to c^-$ is proved similarly. For $t \to c^+$, again, choose $\delta$ to be any number at all, because: :$c - \delta < t < c \implies \left\vert{ \mu_c \left({t}\right) - 1 }\right\vert = \left\vert{ 1 - 1 }\right\vert < \epsilon$ holds, from True Statement is implied by Every Statement. The case for $t \to \left({c + M}\right)^-$ is proved similarly. For all other $t$, we invoke Continuity of Heaviside Step Function. {{qed}} Category:Heaviside Step Function Category:Continuous Functions f1rnjh4xr4nyczh29qeqxrivz84a20x
Let $\map f x = \dfrac {\sinh x} x$. By Quotient Rule for Derivatives and Derivative of Hyperbolic Sine: :$\map {f'} x = \dfrac {x \cosh x - \sinh x} {x^2}$ From Hyperbolic Tangent Less than X, we have $\tanh x \le x$ for $x \ge 0$. Since $\cosh x \ge 0$, we can rearrange to get $x \cosh x - \sinh x \ge 0$. Since $x^2 \ge 0$, we have $\map {f'} x \ge 0$ for $x \ge 0$. So by Derivative of Monotone Function it follows that $\map f x$ is increasing for $x \ge 0$. {{qed}} Category:Hyperbolic Sine Function Category:Euler Formula for Sine Function pj2z8lxqa3jxmucl8rypl8bqmyxs3ss
Let $\map f x = x - \tanh x$. By Derivative of Hyperbolic Tangent: :$\map {f'} x = 1 - \sech^2 x$ Since $\cosh x \ge 1$ for all $x \in \R$, we can deduce that: :$\map {f'} x \ge 0$ From Derivative of Monotone Function, $\map f x$ is increasing. By definition of hyperbolic tangent: :$\map f x = 0$ It follows that $\map f x \ge 0$ for $x \ge 0$. Hence the result. {{qed}} Category:Hyperbolic Tangent Function ryxjy4bcq7w21sw5ohci51xqz8sts03
{{begin-eqn}} {{eqn | l = \int \sinh^2 x \rd x       | r = \int \paren {\frac {\cosh 2 x - 1} 2} \rd x       | c = Square of Hyperbolic Sine }} {{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x - \int \frac 1 2 \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x - \frac x 2 + C       | c = Primitive of Constant }} {{eqn | r = \frac 1 2 \paren {\frac {\sinh 2 x} 2} - \frac x 2 + C       | c = Primitive of Function of Constant Multiple and Primitive of Hyperbolic Cosine Function }} {{eqn | r = \frac {\sinh 2 x} 4 - \frac x 2 + C       | c = rearranging }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = \int \cosh^2 x \rd x       | r = \int \paren {\frac {\cosh 2 x + 1} 2} \rd x       | c = Square of Hyperbolic Cosine }} {{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x + \int \frac 1 2 \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x + \frac x 2 + C       | c = Primitive of Constant }} {{eqn | r = \frac 1 2 \paren {\frac {\map \sinh {2 x} } 2} + \frac x 2 + C       | c = Primitive of Function of Constant Multiple and Primitive of Hyperbolic Cosine Function }} {{eqn | r = \frac {\sinh 2 x} 4 + \frac x 2 + C       | c = rearranging }} {{end-eqn}} {{Qed}}
From Derivative of Hyperbolic Secant: :$\dfrac \d {\d x} \sech x = -\sech x \tanh x$ The result follows from the definition of primitive. {{Qed}}
From Derivative of Hyperbolic Cosecant: :$\dfrac \d {\d x} \csch x = -\csch x \coth x$ The result follows from the definition of primitive. {{Qed}}
Let $x > a$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 - a^2}       | r = -\frac 1 a \coth^{-1} {\frac x a} + C       | c = Primitive of Reciprocal of $x^2 - a^2$ in $\coth^{-1}$ form }} {{eqn | r = -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} } } + C       | c = $\coth^{-1} {\dfrac x a}$ in Logarithm Form }} {{eqn | r = -\frac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C       | c = simplifying }} {{eqn | r = \frac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C       | c = Logarithm of Reciprocal }} {{eqn | r = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C       | c = as $\dfrac {x - a} {x + a} > 0$ for $x > a$ }} {{end-eqn}} Let $x < -a$. Let $z = -x$. Then: :$\d x = -\d z$ and we then have: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 - a^2}       | r = \int \frac {-\d z} {\paren {-z}^2 - a^2}       | c = Integration by Substitution }} {{eqn | r = -\int \frac {\d z} {z^2 - a^2}       | c = simplifying }} {{eqn | r = -\frac 1 {2 a} \map \ln {\frac {z - a} {z + a} } + C       | c = from above }} {{eqn | r = \frac 1 {2 a} \map \ln {\frac {z + a} {z - a} } + C       | c = Logarithm of Reciprocal }} {{eqn | r = \frac 1 {2 a} \map \ln {\frac {-x + a} {-x - a} } + C       | c = substituting $-x$ back for $z$ }} {{eqn | r = \frac 1 {2 a} \map \ln {\frac {\size {x - a} } {\size {x + a} } } + C       | c = as $x < -a$ }} {{eqn | r = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C       | c =  }} {{end-eqn}} The result follows. {{qed}}
From the $1$st logarithm form: :$\ds \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a\\ & \\  \dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\ & \\ \text {undefined} & : \size x = a \end {cases}$ From Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma: :$\map \ln {\dfrac {x - a} {x + a} }$ is defined {{iff}} $\size x > a$ :$\map \ln {\dfrac {a - x} {a + x} }$ is defined {{iff}} $\size x < a$ Let $\size x > a$. Then $\map \ln {\dfrac {x - a} {x + a} }$ is defined. We have that: {{begin-eqn}} {{eqn | l = \dfrac {x - a} {x + a}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {x - a} {x + a}       | r = \size {\dfrac {x - a} {x + a} }       | c =  }} {{end-eqn}} So the result holds for $\size x > a$. Let $\size x < a$. Then $\map \ln {\dfrac {a - x} {a + x} }$ is defined. We have: We have that: {{begin-eqn}} {{eqn | l = \dfrac {a - x} {a + x}       | r = -\dfrac {x - a} {x + a}       | c =  }} {{eqn | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {a - x} {a + x}       | r = \size {\dfrac {x - a} {x + a} }       | c =  }} {{end-eqn}} The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 - a^2}       | r = \int \frac {\d x} {\paren {x - a} \paren {x + a} }       | c = Difference of Two Squares }} {{eqn | r = \int \frac {\d x} {2 a \paren {x - a} } - \int \frac {\d x} {2 a \paren {x + a} }       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {2 a} \int \frac {\d x} {x - a} - \frac 1 {2 a} \int \frac {\d x} {x + a}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 {2 a} \ln \size {x - a} - \frac 1 {2 a} \ln \size {x + a} + C       | c = Primitive of Reciprocal }} {{eqn | r = \dfrac 1 {2 a} \ln \size {\dfrac {x - a} {x + a} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
=== Case where $\size x < a$ === {{:Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x less than a}} === Case where $\size x > a$ === {{:Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x greater than a}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {x^2 + a^2} }       | r = \sinh^{-1} {\frac x a} + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ in $\sinh^{-1}$ form }} {{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } - \ln a + C       | c = $\sinh^{-1} {\dfrac x a}$ in Logarithm Form }} {{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } + C       | c = subsuming $-\ln a$ into arbitrary constant }} {{end-eqn}} {{qed}}
Let $\epsilon > 0$. By the definition of limits at infinity, we need to show that there is some $M \in \R$ such that: :$\displaystyle x > M \implies \size {\frac {\map {P_n} x} {e^{a z} } - 0} < \epsilon$ But: {{begin-eqn}} {{eqn | l = \size {\frac {\map {P_n} x}{e^{a z} } - 0}       | r = \frac {\size {\map {P_n} x} } {\size {e^{a z} } }       | c = Modulus of Product }} {{eqn | r = \frac {\size {\map {P_n} x} } {\size {e^{a x \mathop + i a y} } } }} {{eqn | r = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }       | c = Modulus of Exponential is Modulus of Real Part }} {{end-eqn}} This means it is sufficient to find an $M \in \R$ such that: {{begin-eqn}} {{eqn | l = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }       | o = <       | r = \epsilon }} {{end-eqn}} {{questionable|Exponential Dominates Polynomial doesn't say anything about the quotient. (Proof corrected Nov. 2016.)}} Let $\displaystyle \map {P_n} x = \sum_{j \mathop = 0}^n \paren {a_j x^j}$ where $a_j \in \R$ for every $j$ in $\set {0, \ldots, n}$. Let $\displaystyle M' = \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}$. We observe that $M' \ge 0$. Also, for every $x > M'$: :$x > 0$ as $M' \ge 0$ We have for every $x > M'$: {{begin-eqn}} {{eqn | l = M'       | o = <       | r = x }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}       | o = <       | r = x }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x \sum_{j \mathop = 0}^n \size {a_j}       | o = <       | r = \epsilon       | c = as $x \not = 0$ since $x > 0$ }} {{end-eqn}} We have that for every $k \in \N$ there exists $N_k \in \N$ such that: :$x^k < e^{a x}$ for all $x > N_k$ by Exponential Dominates Polynomial Let $N = \map \max {N_1, \ldots, N_{n + 1} }$. We have for every $k \in \set {1, \ldots, n + 1}$: {{begin-eqn}} {{eqn | l = x^k       | o = <       | r = e^{a x}       | c = for all $x > N$ }} {{eqn | ll= \leadstoandfrom       | l = \frac {x^k} {e^{a x} }       | o = <       | r = 1       | c = for all $x > N$ }} {{end-eqn}} Let $M = \map \max {M', N}$. We get for every $x > M$: {{begin-eqn}} {{eqn | l = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }       | r = \frac {\size {\sum_{j \mathop = 0}^n \paren {a_j x^j} } } {e^{a x} } }} {{eqn | o = \le       | r = \frac {\sum_{j \mathop = 0}^n \size {a_j x^j} } {e^{a x} } }} {{eqn | r = \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j}  x^j} } {e^{a x} }       | c = as $x^j > 0$ since $x > 0$ }} {{eqn | r = \frac 1 x \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j} x^\paren {j + 1} } } {e^{a x} }       | c = as $x \not = 0$ since $x > 0$ }} {{eqn | r = \frac 1 x \sum_{j \mathop = 0}^n \paren {\size {a_j} \frac {x^\paren {j + 1} } {e^{a x} } } }} {{eqn | o = \le       | r = \frac 1 x \sum_{j \mathop = 0}^n \size {a_j} 1       | c = as $\dfrac {x^\paren {j + 1} } {e^{a x} } < 1$ for every $j$ in $\set {0, \ldots, n}$ }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} {{qed}} Category:Limits of Functions Category:Polynomial Theory Category:Exponential Function 3r6mkfmu8w1d1jzcycs88kojf9c27bo
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either: :$x > a$ or: :$x < -a$ where it is assumed that $a > 0$. Consider the arcsecant substitution: :$u = \arcsec {\dfrac x a}$ which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$. That is: :$\size x \ge a$ Hence from Shape of Secant Function, this substitution is valid for all for all $x$ such that $\size {\dfrac x a} > 1$. {{begin-eqn}} {{eqn | l = u       | r = \arcsec {\frac x a}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = a \sec u       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d u}       | r = a \sec u \tan u       | c = Derivative of Secant Function }} {{end-eqn}} Let $x > a$. Then: {{begin-eqn}} {{eqn | l = \int \frac 1 {\sqrt {x^2 - a^2} } \rd x       | r = \int \frac {a \sec u \tan u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u       | c = Integration by Substitution }} {{eqn | r = \frac a a \int \frac {\sec u \tan u} {\sqrt {\sec^2 u - 1} } \rd u       | c = simplifying }} {{eqn | r = \int \sec u \frac {\tan u} {\sqrt {\tan^2 u} } \rd u       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \int \sec u \rd u       | c = as $x > 1$ both $\sec u$ and $\tan u$ are greater than $0$ }} {{eqn | r = \ln \size {\sec u + \tan u} + C       | c = Primitive of $\sec x$: Secant plus Tangent Form }} {{eqn | r = \map \ln {\sec u + \tan u} + C       | c = as both $\sec u$ and $\tan u$ are greater than $0$ }} {{eqn | r = \map \ln {\frac x a + \sqrt {\frac {x^2} {a^2} - 1} } + C       | c = substituting back: $\sec u = \dfrac x a$ and Difference of Squares of Secant and Tangent }} {{eqn | r = \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } + C       | c = rearrangement }} {{eqn | r = \map \ln {\frac {x + \sqrt {x^2 - a^2} } a} + C       | c = rearrangement }} {{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} - \ln a} + C       | c = Difference of Logarithms }} {{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } + C       | c = subsuming $-\ln a$ into constant }} {{eqn | r = \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = as $x + \sqrt {x^2 - a^2} > 0$ for $x > a$ }} {{end-eqn}} Now suppose $x < -a$. Let $z = -x$. Then: :$\d x = -\d z$ and we then have: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }       | r = \int \frac {-\d z} {\sqrt {\paren {-z}^2 - a^2} }       | c = Integration by Substitution }} {{eqn | r = -\int \frac {\d z} {\sqrt {z^2 - a^2} }       | c = simplifying }} {{eqn | r = -\map \ln {z + \sqrt {z^2 - a^2} } + C       | c = from above }} {{eqn | r = \map \ln {z - \sqrt {z^2 - a^2} } - \map \ln {a^2} + C       | c = Negative of $\map \ln {z + \sqrt {z^2 - a^2} }$ }} {{eqn | r = \map \ln {z - \sqrt {z^2 - a^2} } + C       | c = subsuming $-\map \ln {a^2}$ into constant }} {{eqn | r = \map \ln {-x - \sqrt {\paren {-x}^2 - a^2} } + C       | c = substituting back for $x$ }} {{eqn | r = \map \ln {-\paren {x + \sqrt {x^2 - a^2} } } + C       | c = simplifying }} {{eqn | r = \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = as $x + \sqrt {x^2 - a^2} < 0$ for $x < -a$ }} {{end-eqn}} The result follows. {{qed}}
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either: :$x > a$ or: :$x < -a$ where it is assumed that $a > 0$. Consider the arcsecant substitution: :$u = \arcsec {\dfrac x a}$ which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$. That is: :$\size x \ge a$ and it is seen that $u = \arcsec {\dfrac x a}$ is defined over the whole domain of the integrand. Hence: {{begin-eqn}} {{eqn | l = u       | r = \arcsec {\frac x a}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = a \sec u       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d u}       | r = a \sec u \tan u       | c = Derivative of Secant Function }} {{end-eqn}} Let $x > a$. {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \sqrt {x^2 - a^2} }       | r = \int \frac {a \sec u \tan u} {a \sec u \sqrt {a^2 \sec^2 u - a^2} } \rd u       | c = Integration by Substitution }} {{eqn | r = \frac a {a^2} \int \frac {\sec u \tan u} {\sec u \sqrt {\sec^2 u - 1} } \rd u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 a \int \frac {\sec u \tan u} {\sec u \tan u} \rd u       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac 1 a \int 1 \rd u }} {{eqn | r = \frac 1 a u + C       | c = Integral of Constant }} {{eqn | r = \frac 1 a \arcsec {\frac x a} + C       | c = Definition of $u$ }} {{eqn | r = \frac 1 a \arcsec {\frac {\size x} a} + C       | c = {{Defof|Absolute Value}}: $\size x = x$ for $x > 0$ }} {{end-eqn}} Now suppose $x < -a$. Let $z = -x$. Then: :$\d x = -\d z$ and we then have: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \sqrt {x^2 - a^2} }       | r = \int \frac {-\d z} {\paren {-z} \sqrt {\paren {-z}^2 - a^2} }       | c = Integration by Substitution }} {{eqn | r = \int \frac {\d z} {z \sqrt {z^2 - a^2} }       | c = simplifying }} {{eqn | r = \frac 1 a \arcsec {\frac z a} + C       | c = from above }} {{eqn | r = \frac 1 a \arcsec {\frac {\paren {-x} } a} + C       | c = substituting back for $x$ }} {{eqn | r = \frac 1 a \arcsec {\frac {\size x} a} + C       | c = {{Defof|Absolute Value}}: $\size x = -x$ for $x < 0$ }} {{end-eqn}} The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \sqrt {x^2 + a^2} }       | r = -\frac 1 a \csch^{-1} {\frac x a} + C       | c = Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$: $\csch^{-1}$ form }} {{eqn | r = -\frac 1 a \map \ln {\frac {a + \sqrt{a^2 + x^2} } x} + C       | c = $\csch^{-1} {\dfrac x a}$ in Logarithm Form }} {{end-eqn}} {{qed}} {{questionable|Negative $x$ not taken into account. The result should end up being absolute valued.}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \sqrt {a^2 - x^2} }       | r = -\frac 1 a \sech^{-1} {\frac x a} + C       | c = Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$: $\sech^{-1}$ form }} {{eqn | r = -\frac 1 a \map \ln {\frac {1 + \sqrt {1 - \paren {\frac x a}^2} } {\frac x a} } + C       | c = {{Defof|Inverse Hyperbolic Secant|subdef = Real|index = 2}} }} {{eqn | r = -\frac 1 a \map \ln {\frac {a + a \sqrt {1 - \paren {\frac x a}^2} } x} + C       | c = multiplying top and bottom by $a$ }} {{eqn | r = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - a^2 \paren {\frac x a}^2} } x} + C       | c = moving $a$ within the square root }} {{eqn | r = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C       | c = simplifying }} {{end-eqn}} {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \int f^{\paren n} g \rd x = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^\paren j + \paren {-1}^n \int f g^{\paren n} \rd x$ === Basis for the Induction === $\map P 1$ is the case: :$\displaystyle \int f' g \rd x = f g - \int f g' \rd x$ which is proved in Integration by Parts. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \int f^{\paren k} g \rd x = \sum_{j \mathop = 0}^{k - 1} \paren {-1}^j f^{\paren {k - j - 1} } g^{\paren j} + \paren {-1}^k \int f g^{\paren k} \rd x$ Then we need to show: :$\displaystyle \int f^{\paren {k + 1} } g \rd x = \sum_{j \mathop = 0}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \int f^{\paren {k + 1} } g \rd x       | r = \int \paren {f^{\paren k} }' g \rd x       | c =  }} {{eqn | r = f^{\paren k } g - \int f^{\paren k} g' \rd x       | c = Basis for the Induction }} {{eqn | r = f^{\paren k} g - \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^j f^{\paren {k - j - 1} } g^\paren {j + 1} + \paren {-1}^k \int f g^{\paren {k + 1} } \rd x}       | c = Induction Hypothesis }} {{eqn | r = f^{\paren k} g + \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^{j + 1} f^{\paren {k - \paren {j + 1} } } g^{\paren {j + 1} } + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x}       | c = moving $-1$ into the parenthesis }} {{eqn | r = f^{\paren k} g + \paren {\sum_{j \mathop = 1}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x}       | c = substituting $j$ for $j + 1$ }} {{eqn | r = \sum_{j \mathop = 0}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x       | c = as $f^{\paren k} g$ is the $0$th element of the summation }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{>0}: \int f^{\paren n} g \rd x = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^{\paren j} + \paren {-1}^n \int f g^{\paren n} \rd x$ assuming that $f$ and $g$ are sufficiently differentiable. {{qed}}
{{begin-eqn}} {{eqn | l = u        | r = a x + b       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = a       | c = Derivative of Function of Constant Multiple: Corollary }} {{eqn | ll= \leadsto       | l = \int \map F {a x + b} \rd x       | r = \int \frac {\map F u} a \d u       | c = Primitive of Composite Function }} {{eqn | r = \frac 1 a \int \map F u \rd u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = u        | r = \sqrt {a x + b}       | c =  }} {{eqn | l = u        | r = \left({a x + b}\right)^{1/2}       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 2 \left({a x + b}\right)^{-1/2} \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)       | c = Chain Rule for Derivatives, Power Rule for Derivatives }} {{eqn | r = \frac 1 {2 u} \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)       | c = substituting for $u$ }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac a {2 u}       | c = Derivative of Function of Constant Multiple: Corollary }} {{eqn | ll= \implies       | l = \int F \left({\sqrt {a x + b} }\right) \ \mathrm d x       | r = \int \frac {2 u} a F \left({u}\right) \ \mathrm d u       | c = Primitive of Composite Function }} {{eqn | r = \frac 1 a \int u \ F \left({u}\right) \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = f \left({x}\right)       | r = \sqrt [n] x       | c =  }} {{eqn | r = x^{1 / n}       | c = Definition of $n$th Root }} {{eqn | ll= \implies       | l = f' \left({x}\right)       | r = \frac 1 n x^{\left({1 / n}\right) - 1}       | c = Power Rule for Derivatives }} {{eqn | r = \frac 1 n x^{\left({1 / n}\right)\left({1 - n}\right)}       | c = rearrangement }} {{eqn | r = \frac 1 {n x^{\left({1 / n}\right) \left({n - 1}\right)} }       | c = Exponent Combination Laws: Negative Power }} {{eqn | r = \frac 1 {n \left({\sqrt [n] x}\right)^{n - 1} }       | c = Definition of $n$th Root }} {{end-eqn}} {{qed}} Category:Derivatives igt3uezk8pi0aw2x6m1ojrw2wb2p5ll
{{begin-eqn}} {{eqn | l = u        | r = \sqrt [n] {a x + b}       | c =  }} {{eqn | l = u        | r = \left({a x + b}\right)^{1/n}       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 {n \left({\sqrt [n] {a x + b} }\right)^{n - 1} } \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)       | c = Chain Rule for Derivatives, Derivative of Nth Root }} {{eqn | r = \frac 1 {n u^{n - 1} } \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)       | c = substituting for $u$ }} {{eqn | r = \frac a {n u^{n - 1} }       | c = Derivative of Function of Constant Multiple: Corollary }} {{eqn | ll= \implies       | l = \int F \left({\sqrt [n] {a x + b} }\right) \ \mathrm d x       | r = \int \frac {n u^{n - 1} } a F \left({u}\right) \ \mathrm d u       | c = Primitive of Composite Function }} {{eqn | r = \frac n a \int u^{n - 1} F \left({u}\right) \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
First note that: {{begin-eqn}} {{eqn | l = x        | r = a \sin u       | c =  }} {{eqn | ll= \implies       | l = \sqrt {a^2 - x^2}       | r = \sqrt {a^2 - \left({a \sin u}\right)^2}       | c = }} {{eqn | r = a \sqrt {1 - \sin^2 u}       | c = taking $a$ outside the square root }} {{eqn | n = 1       | r = a \cos u       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x        | r = a \sin u       | c =  }} {{eqn | ll= \implies       | l = \frac{\mathrm d x} {\mathrm d u}       | r = a \cos u       | c = Derivative of Sine Function }} {{eqn | ll= \implies       | l = \int F \left({\sqrt {a^2 - x^2} }\right) \ \mathrm d x       | r = \int a \cos u \ F \left({\sqrt {a^2 - x^2} }\right) \ \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \int a \cos u \ F \left({a \cos u}\right) \ \mathrm d u       | c = from $(1)$ }} {{eqn | r = a \int \cos u \ F \left({a \cos u}\right) \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}