hcju/proofwikiqa huggingface dataset:
By definition of ideal: : $\operatorname{Rad} \left({\left({1}\right)}\right) \subseteq A$ By Ideal of Ring is Contained in Radical: : $\left({1}\right) = A \subseteq \operatorname{Rad} \left({\left({1}\right)}\right)$. By definition of set equality: : $\operatorname{Rad} \left({\left({1}\right)}\right) = \left({1}\right)$ {{qed}} Category:Radical of Ideals 0jut8ufj93cgdrbc3a51mjle0bq51o9
Because $a$ is nilpotent, there exists a natural number $n > 0$ with $a^n = 0$. By Sum of Geometric Sequence in Ring: :$\paren {1 + a} \cdot \displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k = 1 + \paren {-a}^n$ :$\paren {\displaystyle \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k} \cdot \paren {1 + a} = 1 + \paren {-a}^n$ where $\sum$ denotes summation. By Negative of Nilpotent Ring Element: :$\paren {-a}^n = 0$ Thus $1 + a$ is a unit. {{qed}}
By definition of extension and generated ideal: :$f(\mathfrak a) \subseteq \mathfrak a^e$ By Subset of Domain is Subset of Preimage of Image: :$\mathfrak a \subseteq f^{-1}(f(\mathfrak a))$ By Preimage Preserves Inclusion of Subsets: :$\mathfrak a \subseteq f^{-1}(f(\mathfrak a)) \subseteq f^{-1}(\mathfrak a^e)$ {{qed}} Category:Ideal Theory q992nqmk7bewv1ds54dodlffs6bhxxi
Note that by Field is Integral Domain, $K$ is indeed an integral domain. {{proof wanted}} Category:Irreducible Polynomials e32op3s3m1i7mw9bk3haxpgjj5vsf1o
Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$. By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$. Hence: {{begin-eqn}} {{eqn | l = \int \paren {\lambda \map f t + \mu \map g t} \rd t       | r = \lambda \map F t + \mu \map G t       | c = }} {{eqn | r = \lambda \int \map f t \rd t + \mu \int \map g t \rd t       | c = }} {{end-eqn}} {{qed}}
It is clear that for step functions $s$ and $t$: {{handwaving|"It is clear that ..."}} {{MissingLinks}} :$\displaystyle \int_a^b \lambda \map s x + \mu \map t x \rd x = \lambda \int_a^b \map s x \rd x + \mu \int_a^b \map t x \rd x$ Under any partition, the lower sums and upper sums of $f$ and $g$ are step functions, so the above formula relates the lower and upper sums of $f$ and $g$ to the lower and upper sums of the linear combinations of $f$ and $g$. Because this identity is preserved for all possible partitions of $\closedint a b$, it is preserved for the supremum and infimum of all possible lower and upper sums, so the linear combinations of $f$ and $g$ are integrable. {{begin-eqn}} {{eqn | l = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t       | r = \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren {\lambda f + \mu g} } } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }       | c =  }} {{eqn | r = \lambda \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren f} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} } + \mu \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren g} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }       | c =  }} {{eqn | r = \lambda \int_a^b \map f x \rd x + \mu \int_a^b \map g x \rd x       | c =  }} {{end-eqn}} {{qed}}
Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$. By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$. Hence by the Fundamental Theorem of Calculus: {{begin-eqn}} {{eqn | l = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t       | r = \bigintlimits {\lambda \map F t + \mu \map G t} a b       | c =  }} {{eqn | r = \lambda \bigintlimits {\map F t} a b + \mu \bigintlimits {\map G t} a b       | c =  }} {{eqn | r = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t       | c =  }} {{end-eqn}} {{qed}}
Let: :$\displaystyle \int_a^b \map g x \rd x = 0$ We are given that: :$\forall x \in \closedint a b: \map g x \ge 0$ Hence by Continuous Non-Negative Real Function with Zero Integral is Zero Function: :$\forall x \in \closedint a b: \map g x = 0$ Hence: :$\displaystyle \int_a^b \map f x \cdot 0 \rd x = \map f k \cdot 0$ and so the result holds for any choice of $k$. {{qed|lemma}} Let: :$\displaystyle \int_a^b \map g x \rd x \ne 0$ From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$. By the Extreme Value Theorem, there exist $m, M \in \closedint a b$ such that: :$\displaystyle \map f m = \min_{x \mathop \in \closedint a b} \map f x$ :$\displaystyle \map f M = \max_{x \mathop \in \closedint a b} \map f x$ Then the following inequality holds for all $x$ in $\closedint a b$: :$\map f m \le \map f x \le \map f M$ Multiplying by $\map g x$, and using that: :$\forall x \in \closedint a b: \map g x \ge 0$ we get: :$\map f m \map g x \le \map f x \map g x \le \map f M \map g x$ Integrating from $a$ to $b$ gives: :$\displaystyle \int_a^b \map f m \map g x \rd x \le \int_a^b \map f x \map g x \rd x \le \int_a^b \map f M \map g x \rd x$ By Linear Combination of Definite Integrals: :$\displaystyle \map f m \int_a^b \map g x \rd x \le \int_a^b \map f x \map g x \rd x \le \map f M \int_a^b \map g x \rd x$ Dividing by $\displaystyle \int_a^b \map g x \rd x$ gives: :$\displaystyle \map f m \le \dfrac {\int_a^b \map f x \map g x \rd x} {\int_a^b \map g x \rd x} \le \map f M $ By the Intermediate Value Theorem, there exists some $k \in \openint a b$ such that: {{begin-eqn}} {{eqn | l = \dfrac {\int_a^b \map f x \map g x \rd x} {\int_a^b  \map g x \rd x}       | r = \map f k       | c =  }} {{eqn | ll= \leadsto       | l = \int_a^b \map f x \map g x \rd x       | r = \map f k \int_a^b \map g x \rd x       | c =  }} {{end-eqn}} {{qed}}
{{ProofWanted|This will probably boil down to a graphical approach based on the structure of a Darboux integral. We need to explain rigorously what an "approximation" means, and we may also want to quantify the error.}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \dfrac {\d x} {x^2 + a^2} 	| r = \frac \pi { 2 a^{2 - 1} } \csc \left({\frac \pi 2}\right) 	| c = Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$: Corollary }} {{eqn	| r = \frac \pi {2 a}  	| c = Sine of Right Angle }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^\infty \dfrac {\d x} {x^2 + a^2}       | r = \int_0^{\mathop \to +\infty} \dfrac {\d x} {x^2 + a^2}       | c =  }} {{eqn | r = \lim_{\gamma \mathop \to +\infty} \int_0^\gamma \dfrac {\d x} {x^2 + a^2}       | c = {{Defof|Improper Integral on Closed Interval Unbounded Above}} }} {{eqn | r = \lim_{\gamma \mathop \to +\infty} \intlimits {\frac 1 a \arctan \frac x a} 0 \gamma       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac 1 a \lim_{\gamma \mathop \to +\infty} \paren {\arctan \frac \gamma a - \arctan 0}       | c =  }} {{eqn | r = \frac 1 a \lim_{\gamma \mathop \to +\infty} \arctan \frac \gamma a       | c = Arctangent of Zero is Zero }} {{eqn | r = \frac 1 a \frac \pi 2       | c = Arctangent Tends to Half Pi as Argument Tends to Infinity }} {{end-eqn}} Hence the result. {{qed}}
Let $C_R$ be the semicircular contour of radius $R$ situated on the upper half plane, centred at the origin, traversed anti-clockwise. Let $\Gamma_R = C_R \cup \left[{-R \,.\,.\, R}\right]$. Then, by Contour Integral of Concatenation of Contours: :$\displaystyle \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2} = \int_{C_R} \frac {\d z} {z^2 + a^2} + \int_{-R}^R \frac {\d x} {x^2 + a^2}$ we have:  {{begin-eqn}} {{eqn	| l = \left\vert \int_{C_R} \frac {\d z} {z^2 + a^2} \right\vert 	| o = \le 	| r = \pi R \cdot \max_{0 \le \theta \le \pi} \left\vert \frac 1 {R^2 e^{2 i \theta} + a^2} \right\vert 	| c = Estimation Lemma }} {{eqn	| r = \frac {\pi R} {R^2 + a^2} }} {{eqn	| o = \le 	| r = \frac {\pi R} {R^2} }} {{eqn	| r = \frac \pi R }} {{eqn	| o = \to 	| r = 0 	| c = as $R \to \infty$ }} {{end-eqn}} from which: :$\displaystyle \lim_{R \to \infty} \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2} = \lim_{R \to \infty} \int_{-R}^R \frac {\d x} {x^2 + a^2} = \int_{-\infty}^\infty \frac {\d x} {x^2 + a^2}$ The integrand is meromorphic. So by the Residue Theorem: :$\displaystyle \lim_{R \to \infty} \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2} = 2 \pi i \sum \operatorname{Res} \left({\frac 1 {z^2 + a^2}, z_0}\right)$ where the sum runs over the poles of $\dfrac 1 {z^2 + a^2}$. The integrand can be written as: : $\dfrac 1 {\left({z - ai}\right) \left({z + ai}\right)}$ From this it can be observed that the integrand has only two simple poles: $z_0 = a i$ and $z_0 = -a i$. Only the former of these lies in the contour, so: {{begin-eqn}} {{eqn	| l = \lim_{R \to \infty} \oint_{\Gamma_R} \frac {\d z} {z^2 + a^2} 	| r = 2 \pi i \operatorname{Res} \left({\frac 1 {z^2 + a^2}, ai}\right) }} {{eqn	| r = 2 \pi i \cdot \frac 1 { \frac \d {\d z}\left({z^2 + a^2}\right)_{z \mathop = a i} } 	| c = Residue at Simple Pole }} {{eqn	| r = 2 \pi i \cdot \frac 1 {2 a i} 	| c = Derivative of Power }} {{eqn	| r = \frac \pi a }} {{end-eqn}} So:  :$\displaystyle \int_{-\infty}^\infty \frac {\d x} {x^2 + a^2} = \frac \pi a$ From Definite Integral of Even Function: :$\displaystyle \int_{-\infty}^\infty \frac {\d x} {x^2 + a^2} = 2 \int_0^\infty \frac {\d x} {x^2 + a^2}$ Hence:  :$\displaystyle \int_0^\infty \frac {\d x} {x^2 + a^2} = \frac \pi {2 a}$ {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac 1 {a^n + x^n} \rd x 	| r = \frac 1 {a^n} \int_0^\infty \frac 1 {1 + \paren {\frac x a}^n} \rd x }} {{eqn	| r = \frac 1 {a^n} \cdot \frac 1 {\frac 1 a} \int_0^\infty \frac 1 {1 + \paren {\frac x a}^n} \rd \paren {\frac x a} 	| c = Primitive of Function of Constant Multiple }} {{eqn	| r = \frac a {a^n} \cdot \frac \pi n \csc \paren {\frac \pi n} 	| c = Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$ }} {{eqn	| r = \frac \pi {n a^{n - 1} } \csc \paren {\frac \pi n} }} {{end-eqn}} {{qed}} Category:Definite Integral to Infinity of Reciprocal of 1 plus Power of x 69z3s4t8x3fe0za9rvsqjy8y5frjyeg
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {x^{p - 1} \rd x} {1 + x}  	| r = \int_0^\infty \frac 1 { px^{p - 1} } \cdot \frac{x^{p - 1} \rd t} { 1 + t^{\frac 1 p} } 	| c = substituting $t = x^p$ }} {{eqn	| r = \frac 1 p \int_0^\infty \frac {\rd t} { 1 + t^{\frac 1 p} } }} {{eqn	| r = \frac 1 p \cdot \frac \pi {\frac 1 p} \csc \left({ \frac \pi {\frac 1 p} }\right) 	| c = as $0 < p < 1$, $\frac 1 p > 1$, so we can apply Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$ }} {{eqn	| r = \pi \csc \left({\pi p}\right) }} {{eqn	| r = \frac \pi {\sin \pi p} 	| c = {{Defof|Cosecant}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \dfrac {x^m \rd x} {x^n + a^n} 	| r = \int_0^\infty \dfrac {x^m \rd x} {\paren {x^{m + 1} }^{\frac n {m + 1} } + \paren {a^{m + 1} }^{\frac n {m + 1} } } }} {{eqn	| r = \frac 1 {m + 1} \int_0^\infty \dfrac 1 {u^{\frac n {m + 1} } + \paren {a^{m + 1} }^{\frac n {m + 1} } } \rd u 	| c = substituting $u = x^{m + 1}$ }} {{eqn	| r = \frac 1 {\paren {m + 1} \paren {\frac n {m + 1} \paren {a^{m + 1} }^{\frac n {m + 1} - 1} } } \map \csc {\frac {\paren {m + 1} \pi} n} 	| c = Definite Integral to Infinity of $\dfrac 1 {1 + x^n}$: Corollary }} {{eqn	| r = \frac {a^{m + 1} } {n a^{\paren {m + 1} \frac n {m + 1} } } \map \csc {\frac {\paren {m + 1} \pi} n} }} {{eqn	| r = \frac {\pi a^{m + 1 - n} } {n \map \sin {\paren {m + 1} \frac \pi n} } }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^a \dfrac {\d x} {\sqrt {a^2 - x^2} }       | r = \int_0^{\mathop \to a} \dfrac {\d x} {\sqrt {a^2 - x^2} }       | c = as $\dfrac 1 {\sqrt {a^2 - x^2} }$ does not exist for $x = a$ }} {{eqn | r = \lim_{\gamma \mathop \to a} \int_0^\gamma \dfrac {\d x} {\sqrt {a^2 - x^2} }       | c = {{Defof|Improper Integral on Open Above Interval}} }} {{eqn | r = \lim_{\gamma \mathop \to a} \intlimits {\arcsin \frac x a} 0 \gamma       | c = Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ }} {{eqn | r = \lim_{\gamma \mathop \to a} \paren {\arcsin \frac \gamma a - \arcsin 0}       | c =  }} {{eqn | r = \lim_{\gamma \mathop \to a} \arcsin \frac \gamma a       | c = Arcsine of Zero is Zero }} {{eqn | r = \arcsin \frac a a       | c =  }} {{eqn | r = \arcsin 1       | c =  }} {{eqn | r = \frac \pi 2       | c = Arcsine of One is Half Pi }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^a \sqrt {a^2 - x^2} \rd x       | r = \intlimits {\frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a} 0 a       | c = Primitive of $\sqrt {a^2 - x^2}$ }} {{eqn | r = \paren {\frac {a \sqrt {a^2 - a^2} } 2 + \frac {a^2} 2 \arcsin \frac a a} - \paren {\frac {0 \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac 0 a}       | c =  }} {{eqn | r = \frac {a^2} 2 \arcsin 1 - \frac {a^2} 2 \arcsin 0       | c = removing vanishing terms }} {{eqn | r = \frac {a^2} 2 \arcsin 1       | c = Sine of Zero is Zero }} {{eqn | r = \frac \pi 2 \frac {a^2} 2       | c = Sine of Right Angle }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^a \sqrt {a^2 - x^2} \rd x 	| r = \frac {a^{1 + \frac 2 2} } 2 \frac {\map \Gamma {\frac 1 2} \map \Gamma {1 + \frac 1 2} } {\map \Gamma {\frac 1 2 + \frac 1 2 + 1} }  	| c = Definite Integral from 0 to a of $x^m \paren {a^n - x^n}^p$ }} {{eqn	| r = \frac {a^2} 2 \frac {\map \Gamma {\frac 3 2} \map \Gamma {\frac 1 2} } {\map \Gamma 2} }} {{eqn	| r = \frac {a^2} {2 \times 1!} \paren {\frac 1 2} \paren {\map \Gamma {\frac 1 2} }^2  	| c = Gamma Difference Equation, Gamma Function Extends Factorial }} {{eqn	| r = \frac {\pi a^2} 4 	| c = Gamma Function of One Half }} {{end-eqn}} {{qed}}
By Universal Property of Quotient Group, it suffices to show that the kernel $\ker f$ contains the commutator subgroup $\sqbrk {G, G}$. By definition of generated subgroup, this is equivalent to showing that $\ker f$ contains all commutators of $G$. Let $g, h \in G$. Then by Group Homomorphism Preserves Inverses: :$\map f {g^{-1} h^{-1} g h} = \map f g^{-1} \, \map f h^{-1} \, \map f g \, \map f g = 1 \in H$ Thus the commutator $\sqbrk {g, h} = g^{-1} h^{-1} g h$ is in $\ker f$. {{qed}} Category:Group Theory Category:Universal Properties 8764e4u82n2uz4bnszehkmekdsfugd9
{{begin-eqn}} {{eqn	| l = \int_0^a x^m \paren {a^n - x^n}^p \rd x 	| r = a \int_0^1 \paren {a u}^m \paren {a^n - \paren {a u}^n}^p \rd u 	| c = substituting $x = a u$ }} {{eqn	| r = a \times a^m \times a^{n p} \int_0^1 u^m \paren {1 - u^n}^p \rd u }} {{eqn	| r = \frac {a^{m + 1 + n p} } n \int_0^1 \frac {u^m} {u^{n - 1} } \paren {1 - t}^p \rd u 	| c = substituting $t = u^n$ }} {{eqn	| r = \frac {a^{m + 1 + n p} } n \int_0^1 t^{\frac {m - n + 1} n} \paren {1 - t}^p \rd u }} {{eqn	| r = \frac {a^{m + 1 + n p} } n \map \Beta {\frac {m + 1} n, p + 1} 	| c = {{Defof|Beta Function|index = 1}} }} {{eqn	| r = \frac {a^{m + 1 + n p} \, \map \Gamma {\frac {m + 1} n} \map \Gamma {p + 1} } {n \map \Gamma {\frac {m + 1} n + p + 1} } 	| c = {{Defof|Beta Function|index = 3}} }} {{end-eqn}} {{qed}}
=== Definition $1$ implies Definition $2$ === Let $G = H K$. Then $H K$ is a group. By Subset Product of Subgroups: :$H K = K H$ Thus $K H = G$. {{qed|lemma}} === Definition $2$ implies Definition $1$ === Let $G = K H$. Then $K H$ is a group. By Subset Product of Subgroups: : $H K = K H$ Thus $H K = G$. {{qed}} Category:Subgroup Complements c900owbpqdvf0uaeqhjvqx7562d7pk7
Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the identity mapping on $S$. Then: {{begin-eqn}} {{eqn | l = \tau_1       | o = \supseteq       | r = \tau_2       | c = {{Defof|Finer Topology|index = 1}} }} {{eqn | ll= \leadstoandfrom       | lo= \forall U \subseteq S:       | l = U \in \tau_2       | o = \implies       | r = U \in \tau_1       | c = {{Defof|Superset}} }} {{eqn | ll= \leadstoandfrom       | lo= \forall U \subseteq S:       | l = U \in \tau_2       | o = \implies       | r = \map {I_S^{-1} } U \in \tau_1       | c = as $\map {I_S^{-1} } U = U$: {{Defof|Identity Mapping}} }} {{end-eqn}} $\: \: \: \leadstoandfrom \: \: I_S$ is continuous, by definition of a continuous mapping, which is {{Defof|Finer Topology|index = 2}}. {{qed}} Category:Topology ki3rw68okdoiz0t1c03ywp96b7a6k1o
Let $m \ne n$. {{begin-eqn}} {{eqn | l = \int \sin m x \sin n x \rd x       | r = \frac {\sin \paren {m - n} x} {2 \paren {m - n} } - \frac {\sin \paren {m + n} x} {2 \paren {m + n} } + C       | c = Primitive of $\sin m x \sin n x$ }} {{eqn | ll= \leadsto       | l = \int_0^\pi \sin m x \sin n x \rd x       | r = \intlimits {\frac {\sin \paren {m - n} x} {2 \paren {m - n} } - \frac {\sin \paren {m + n} x} {2 \paren {m + n} } } 0 \pi       | c =  }} {{eqn | r = \paren {\frac {\sin \paren {m - n} \pi} {2 \paren {m - n} } - \frac {\sin \paren {m + n} \pi} {2 \paren {m + n} } }       | c =  }} {{eqn | o =        | ro= -       | r = \paren {\frac {\sin 0} {2 \paren {m - n} } - \frac {\sin 0} {2 \paren {m + n} } }       | c =  }} {{eqn | r = \frac {\sin \paren {m - n} \pi} {2 \paren {m - n} } - \frac {\sin \paren {m + n} \pi} {2 \paren {m + n} }       | c = Sine of Zero is Zero }} {{eqn | r = 0       | c = Sine of Multiple of Pi }} {{end-eqn}} {{qed|lemma}} When $m = n$ we have: {{begin-eqn}} {{eqn | l = \int \sin m x \sin m x \rd x       | r = \int \sin^2 m x \rd x       | c =  }} {{eqn | r = \frac x 2 - \frac {\sin 2 m x} {4 m} + C       | c = Primitive of $\sin^2 m x$ }} {{eqn | ll= \leadsto       | l = \int_0^\pi \sin m x \sin m x \rd x       | r = \intlimits {\frac x 2 - \frac {\sin 2 m x} {4 m} } 0 \pi       | c =  }} {{eqn | r = \paren {\frac \pi 2 - \frac {\sin \paren {2 m \pi} } {4 m} } - \paren {\frac 0 2 - \frac {\sin 0} {4 m} }       | c =  }} {{eqn | r = \frac \pi 2 - \frac {\sin \paren {2 m \pi} } {4 m}       | c = Sine of Zero is Zero }} {{eqn | r = \frac \pi 2       | c = Sine of Multiple of Pi }} {{end-eqn}} {{qed}}
Let $m \ne n$. {{begin-eqn}} {{eqn | l = \int \cos m x \cos n x \rd x       | r = \frac {\sin \left({m - n}\right) x} {2 \left({m - n}\right)} + \frac {\sin \left({m + n}\right) x} {2 \left({m + n}\right)} + C       | c = Primitive of $\cos m x \cos n x$ }} {{eqn | ll= \leadsto       | l = \int_0^\pi \cos m x \cos n x \rd x       | r = \left[{\frac {\sin \left({m - n}\right) x} {2 \left({m - n}\right)} + \frac {\sin \left({m + n}\right) x} {2 \left({m + n}\right)} }\right]_0^\pi       | c =  }} {{eqn | r = \left({\frac {\sin \left({\left({m - n}\right) \pi}\right)} {2 \left({m - n}\right)} + \frac {\sin \left({\left({m + n}\right) \pi}\right)} {2 \left({m + n}\right)} }\right)       | c =  }} {{eqn | o =        | ro= -       | r = \left({\frac {\sin 0} {2 \left({m - n}\right)} + \frac {\sin 0} {2 \left({m + n}\right)} }\right)       | c =  }} {{eqn | r = \frac {\sin \left({\left({m - n}\right) \pi}\right)} {2 \left({m - n}\right)} + \frac {\sin \left({\left({m + n}\right) \pi}\right)} {2 \left({m + n}\right)}       | c = Sine of Zero is Zero }} {{eqn | r = 0       | c = Sine of Multiple of Pi }} {{end-eqn}} {{qed|lemma}} When $m = n$ we have: {{begin-eqn}} {{eqn | l = \int \cos m x \cos m x \rd x       | r = \int \cos^2 m x \rd x       | c =  }} {{eqn | r = \frac x 2 + \frac {\sin 2 m x} {4 m} + C       | c = Primitive of $\cos^2 m x$ }} {{eqn | ll= \leadsto       | l = \int_0^\pi \cos m x \cos m x \rd x       | r = \left[{\frac x 2 + \frac {\sin 2 m x} {4 m} }\right]_0^\pi       | c =  }} {{eqn | r = \left({\frac \pi 2 + \frac {\sin \left({2 m \pi}\right)} {4 m} }\right) - \left({\frac 0 2 + \frac {\sin 0} {4 m} }\right)       | c =  }} {{eqn | r = \frac \pi 2 + \frac {\sin \left({2 m \pi}\right)} {4 m}       | c = Sine of Zero is Zero }} {{eqn | r = \frac \pi 2       | c = Sine of Multiple of Pi }} {{end-eqn}} {{qed}}
We have:  {{begin-eqn}} {{eqn	| l = \int_0^{\frac \pi 2} \sin^2 x \rd x 	| r = \int_0^{\frac \pi 2} \sin^2 \paren {\frac \pi 2 - x} \rd x 	| c = Integral between Limits is Independent of Direction }} {{eqn	| r = \int_0^{\frac \pi 2} \cos^2 x \rd x 	| c = Sine of Complement equals Cosine }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = 2 \int_0^{\frac \pi 2} \sin^2 x \rd x 	| r = \int_0^{\frac \pi 2} \paren {\sin^2 x + \cos^2 x} \rd x }} {{eqn	| r = \int_0^{\frac \pi 2} \rd x 	| c = Sum of Squares of Sine and Cosine }} {{eqn	| r = \frac \pi 2 	| c = Primitive of Constant }} {{end-eqn}} giving: :$\displaystyle \int_0^{\frac \pi 2} \sin^2 x \rd x = \frac \pi 4$ {{qed}}
{{begin-eqn}} {{eqn | l = \int \sin^2 x \rd x       | r = \frac x 2 - \frac {\sin 2 x} 4 + C       | c = Primitive of $\sin^2 x$ }} {{eqn | ll= \leadsto       | l = \int_0^{\frac \pi 2} \sin^2 x \rd x       | r = \intlimits {\frac x 2 - \frac {\sin 2 x} 4} 0 {\frac \pi 2}       | c =  }} {{eqn | r = \paren {\frac \pi 4 - \frac {\sin \pi} 4} - \paren {\frac 0 2 - \frac {\sin 0} 4}       | c =  }} {{eqn | r = \frac \pi 4       | c = Sine of Multiple of Pi and simplifying }} {{end-eqn}} {{qed}}
We have:  {{begin-eqn}} {{eqn	| l = \int_0^{\frac \pi 2} \cos^2 x \rd x 	| r = \int_0^{\frac \pi 2} \map {\cos^2} {\frac \pi 2 - x} \rd x 	| c = Integral between Limits is Independent of Direction }} {{eqn	| r = \int_0^{\frac \pi 2} \sin^2 x \rd x 	| c = Cosine of Complement equals Sine }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = 2 \int_0^{\frac \pi 2} \cos^2 x \rd x 	| r = \int_0^{\frac \pi 2} \paren {\sin^2 x + \cos^2 x} \rd x }} {{eqn	| r = \int_0^{\frac \pi 2} \rd x 	| c = Sum of Squares of Sine and Cosine }} {{eqn	| r = \frac \pi 2 	| c = Primitive of Constant }} {{end-eqn}} giving: :$\displaystyle \int_0^{\frac \pi 2} \cos^2 x \rd x = \frac \pi 4$ {{qed}}
{{begin-eqn}} {{eqn | l = \int \cos^2 x \rd x       | r = \frac x 2 + \frac {\sin 2 x} 4 + C       | c = Primitive of $\cos^2 x$ }} {{eqn | ll= \leadsto       | l = \int_0^{\frac \pi 2} \cos^2 x \rd x       | r = \intlimits {\frac x 2 + \frac {\sin 2 x} 4} 0 {\frac \pi 2}       | c =  }} {{eqn | r = \paren {\frac \pi 4 + \frac {\sin \pi} 4} - \paren {\frac 0 2 + \frac {\sin 0} 4}       | c =  }} {{eqn | r = \frac \pi 4       | c = Sine of Multiple of Pi and simplifying }} {{end-eqn}} {{qed}}
From Shape of Cosine Function: :$\forall x \in \left[{0 \,.\,.\, \dfrac \pi 2}\right]: 0 \le \cos x \le 1$ and so on the same interval: :$0 \le \cos^{n + 1} x \le \cos^n x$ therefore: :$\forall n \in \N: 0 < I_{n + 1} < I_n$ From Reduction Formula for Integral of Power of Cosine: :$\displaystyle \int \cos^n x \rd x = \dfrac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n \int \cos^{n - 2} x \rd x$ Thus: {{begin-eqn}} {{eqn | l = I_n       | r = \int_0^{\frac \pi 2} \cos^n x \rd x       | c =  }} {{eqn | r = \dfrac {n - 1} n \int_0^{\frac \pi 2} \cos^{n - 2} x \rd x - \left[{\dfrac {\cos^{n - 1} x \sin x} n}\right]_0^{\pi / 2}       | c =  }} {{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({\cos^{n - 1} \frac \pi 2 \sin \frac \pi 2 - \cos^{n - 1} 0 \sin 0}\right)} n       | c =  }} {{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - \cos^{n - 1} 0 \sin 0}\right)} n       | c = Cosine of Right Angle }} {{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - 0}\right)} n       | c = Sine of Zero is Zero }} {{eqn | ll= \implies       | l = n I_n       | r = \left({n - 1}\right) I_{n - 2}       | c = multiplying both sides by $n$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^{\frac \pi 2} \cos^{2 n} x \rd x 	| r = \int_0^{\frac \pi 2} \paren {\sin x}^{\frac 2 2 - 1} \paren {\cos x}^{2 \paren {n + \frac 1 2} - 1} \rd x }} {{eqn	| r = \frac 1 2 \Beta \paren {\frac 1 2, n + \frac 1 2} 	| c = {{Defof|Beta Function|index = 2}} }} {{eqn	| r = \frac 1 2 \cdot \frac {\map \Gamma {n + \frac 1 2} \, \map \Gamma {\frac 1 2} } {\map \Gamma {n + 1} } 	| c = {{Defof|Beta Function|index = 3}} }} {{eqn	| r = \frac {\map \Gamma {n + \frac 1 2} \sqrt \pi} {2 \paren {n!} } 	| c = Gamma Function of One Half }} {{eqn	| r = \frac {\paren {2 n}! \paren {\sqrt \pi}^2} {2 \cdot 2^{2 n} \paren {n!}^2} 	| c = Gamma Function of Positive Half-Integer }} {{eqn	| r = \frac {\paren {2 n}!} {\paren {2^n n!}^2} \frac \pi 2 }} {{end-eqn}} {{qed}}