hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({1 - \sin a x}\right)^2}       | r = \int \left({\frac 1 2 \sec^2 \left({\frac \pi 4 + \frac {a x} 2}\right)}\right)^2 \ \mathrm d x       | c = Reciprocal of One Minus Sine }} {{eqn | r = \frac 1 4 \int \sec^4 \left({\frac \pi 4 + \frac {a x} 2}\right) \ \mathrm d x       | c = simplifying }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = \frac \pi 4 + \frac {a x} 2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = \frac a 2       | c = simplifying }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 4 \int \sec^4 \left({\frac \pi 4 + \frac {a x} 2}\right) \ \mathrm d x       | r = \frac 1 4 \int \frac 2 a \sec^4 z \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac 1 {2 a} \int \sec^4 z \ \mathrm d z       | c = simplifying }} {{eqn | r = \frac 1 {2 a} \left({\frac{\sec^2 z \tan z} 3 + \frac 2 3 \int \sec^2 z \ \mathrm d z}\right) + C       | c = Primitive of $\sec^n a x$ }} {{eqn | r = \frac 1 {2 a} \left({\frac{\sec^2 z \tan z} 3 + \frac 2 3 \tan z}\right) + C       | c = Primitive of $\sec^2 a x$ }} {{eqn | r = \frac 1 {2 a} \left({\frac{\left({1 + \tan^2 z}\right) \tan z} 3 + \frac 2 3 \tan z}\right) + C       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac 1 {2 a} \tan z + \frac 1 {6 a} \tan^3 z + C       | c = simplifying }} {{eqn | r = \frac 1 {2 a} \tan \left({\frac \pi 4 + \frac {a x} 2}\right) + \frac 1 {6 a} \tan^3 \left({\frac \pi 4 + \frac {a x} 2}\right) + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 1 - \sin x       | r = \sin \frac \pi 2 - \sin x       | c = Sine of Right Angle }} {{eqn | r = 2 \map \cos {\frac 1 2 \paren {\frac \pi 2 + x} } \map \sin {\frac 1 2 \paren {\frac \pi 2 - x} }       | c = Prosthaphaeresis Formula for Sine minus Sine }} {{eqn | r = 2 \map \cos {\frac \pi 4 + \frac x 2} \map \sin {\frac \pi 4 - \frac x 2}       | c = simplifying }} {{eqn | r = 2 \map \cos {\frac \pi 4 + \frac x 2} \map \cos {\frac \pi 2 - \paren {\frac \pi 4 - \frac x 2} }       | c = Cosine of Complement equals Sine }} {{eqn | r = 2 \map \cos {\frac \pi 4 + \frac x 2} \map \cos {\frac \pi 4 + \frac x 2}       | c = simplifying }} {{eqn | r = 2 \map {\cos^2} {\frac \pi 4 + \frac x 2}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \frac 1 {1 - \sin x}       | r = \frac 1 2 \map {\sec^2} {\frac \pi 4 + \frac x 2}       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 1 + \sin x       | r = \sin \frac \pi 2 + \sin x       | c = Sine of Right Angle }} {{eqn | r = 2 \map \sin {\frac 1 2 \paren {\frac \pi 2 + x} } \map \cos {\frac 1 2 \paren {\frac \pi 2 - x} }       | c = Prosthaphaeresis Formula for Sine minus Sine }} {{eqn | r = 2 \map \sin {\frac \pi 4 + \frac x 2} \map \cos {\frac \pi 4 - \frac x 2}       | c = simplifying }} {{eqn | r = 2 \map \cos {\frac \pi 2 - \paren {\frac \pi 4 - \frac x 2} } \map \cos {\frac \pi 4 - \frac x 2}       | c = Cosine of Complement equals Sine }} {{eqn | r = 2 \map \cos {\frac \pi 4 - \frac x 2} \map \cos {\frac \pi 4 - \frac x 2}       | c = simplifying }} {{eqn | r = 2 \map {\cos^2} {\frac \pi 4 - \frac x 2}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \frac 1 {1 + \sin x}       | r = \frac 1 2 \map {\sec^2} {\frac \pi 4 - \frac x 2}       | c = {{Defof|Secant Function}} }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Let $f$ be periodic with period $L$. Then: {{begin-eqn}} {{eqn | l = f \left({x}\right)       | r = f \left({x + L}\right)       | c = {{Defof|Periodic Function}} }} {{eqn | ll = \implies       | l = f \left({x}\right) + k       | r = f \left({x + L}\right) + k }} {{end-eqn}} Thus $f + k$ has been shown to be periodic with period $L$. {{qed|lemma}} === Necessary Condition === Let $f + k$ be periodic with period $L$. Then: {{begin-eqn}} {{eqn | l = f \left({x}\right) + k       | r = f \left({x + L}\right) + k       | c = {{Defof|Periodic Function}} }} {{eqn | ll = \implies       | l = f \left({x}\right)       | r = f \left({x + L}\right) }} {{end-eqn}} Thus $f$ has been shown to be periodic with period $L$. {{qed}} {{ProofWanted|Proof of period?}} Category:Periodic Functions l3mvqqw23gy77gd9ksqd3c0feje65pp
Let $f$ be differentiable on all of $\R$. Let $f$ be periodic with period $L$. Then taking the derivative of both sides using the Chain Rule for Derivatives yields: :$\map f x = \map f {x + L} \implies \map {f'} x = \map {f'} {x + L}$ Let $L'$ be the period of $f'$. Suppose that $\size {L'} < \size L$. $f$ is differentiable and therefore continuous, by Differentiable Function is Continuous. From Image of Closed Real Interval is Bounded, it follows that $f$ is bounded on $\closedint 0 {\size L}$. But from the General Periodicity Property, it follows that $f$ is bounded on all of $\R$. Then from Primitive of Periodic Function, it follows that $L'$ is the period of $f$. But we had previously established that $L$ was the period of $f$. This is a contradiction, therefore $L' = L$. Hence the result. {{qed}}
Let $f$ be periodic with period $L$. Let $f$ have a primitive $F$ that is bounded on all of $\R$. By definition of a periodic function, it is seen that: :$\map f x = \map f {x + L}$. Then: :$\displaystyle \int \map f x \rd x$ and: :$\displaystyle \int \map f {x + L} \rd x$ are both primitives of the same function. So by Primitives which Differ by Constant: {{begin-eqn}} {{eqn | l = \int \map f x \rd x + C       | r = \int \map f {x + L} \rd x }} {{eqn | r = \int \map f {x + L} \frac {\map \d {x + L} } {\d x} \rd x       | c = as $\dfrac {\map \d {x + L} } {\d x} = 1$ }} {{eqn | r = \int \map f {x + L} \map \d {x + L}       | c = Integration by Substitution }} {{eqn | ll= \leadsto       | l = \map F x + C       | r = \map F {x + L} }} {{end-eqn}} {{AimForCont}} that $C \ne 0$. Then it is to be shown that for all $k \in \N_{> 0}$: :$\map F x + k C = \map F {x + k L}$ The case for $k = 1$ has already been proven, so this will be proven by induction. Suppose that for some $n \in \N_{> 0}$ we have: :$\map F x + n C = \map F {x + n L}$ Then {{begin-eqn}} {{eqn | l = \map F {x + \paren {n + 1} L}       | r = \map F {x + L + n L} }} {{eqn | r = \map F {x + L} + n C       | c = by the induction hypothesis }} {{eqn | r = \map F x + \paren {n + 1} C       | c = by the base case }} {{end-eqn}} {{qed|lemma}} Holding $x$ fixed yields: :$\displaystyle \lim_{k \mathop \to \infty} \size {\map F {x + k L} } = \lim_{k \mathop \to \infty} \size {\map F x + k C} = \infty$ and so $\map F x$ is unbounded. But we had previously established that $\map F x$ was bounded. This is a contradiction Therefore our assumption that $C \ne 0$ was false. Hence $C = 0$ and so: :$\map F x = \map F {x + L}$ It has been shown that $F$ is periodic. {{qed|lemma}} Let $L'$ be the period of $F$. Suppose that $\size {L'} < \size L$. Then: {{begin-eqn}} {{eqn | l = \map F x       | r = \map F {x + L'} }} {{eqn | ll= \leadsto       | l = \map {F'} x       | r = \map {F'} {x + L'}       | c = Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \map f x       | r = \map f {x + L'}       | c = {{Defof|Primitive of Real Function}} }} {{end-eqn}} But it was previously established that $L$ was the period of $f$. This is a contradiction, therefore $L' = L$. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 \left({x^3 + a^3}\right)}       | r = \int \frac {a^3 \ \mathrm d x} {a^3 x^2 \left({x^3 + a^3}\right)}       | c = multiplying top and bottom by $a^3$ }} {{eqn | r = \int \frac {\left({x^3 + a^3 - x^3}\right) \ \mathrm d x} {a^3 x^2 \left({x^3 + a^3}\right)}       | c =  }} {{eqn | r = \frac 1 {a^3} \int \frac {\left({x^3 + a^3}\right) \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)} - \frac 1 {a^3} \int \frac {x^3 \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^3} \int \frac {\mathrm d x} {x^2} - \frac 1 {a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)}       | c = simplifying }} {{eqn | r = \frac 1 {a^3} \left({\frac {-1} x}\right) - \frac 1 {a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)}       | c = Primitive of Power }} {{eqn | r = \frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)}       | c = simplifying }} {{end-eqn}} {{qed}} Category:Primitives involving x cubed plus a cubed/Lemmata t31pctz64q23aw881rz6hwlffbntil0
{{begin-eqn}} {{eqn | l = \int \csc x \rd x        | r = \int \frac 1 {\sin x} \rd x       | c = Cosecant is Reciprocal of Sine }} {{end-eqn}} We make the Weierstrass Substitution: {{begin-eqn}} {{eqn | l = u       | r = \tan \frac x 2 }} {{eqn | ll= \leadsto       | l = \sin x       | r = \frac {2 u} {u^2 + 1} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d u}       | r = \frac 2 {u^2 + 1} }} {{eqn | ll= \leadsto       | l = \int \frac 1 {\sin x} \rd x       | r = \int \frac {u^2 + 1} {2 u} \frac 2 {u^2 + 1} \rd u }} {{eqn | r = \int \frac 1 u \rd u }} {{eqn | r = \ln \size u + C       | c = Primitive of Reciprocal }} {{eqn | r = \ln \size {\tan \frac x 2} + C       | c = substituting back for $u$ }} {{end-eqn}} {{qed}}
=== For finite languages === For any ''finite'' regular language $L_{fin}$, the proof is simple. Let $s_{maxlen} \in L_{fin}$. Thus: :$\forall s \in L_{fin}: \left|{s}\right| \le \left|{s_{maxlen}}\right|$ Now choose $n_0 > \left|{s_{maxlen}}\right|$. The implication now trivially holds because the premise: :$\left({\left|{z}\right| > n_0}\right)$ is false. {{qed|lemma}} === For infinite languages === For any one ''infinite'' regular language $L_{inf}$, the following holds: :$(1): \quad$ There exists a finite automaton: ::::$F = \left({\Sigma, Q, q_0, A, \delta}\right)$ :::such that: :::: $\mathcal L \left({F}\right) = L_{inf}$. See Equivalence of Finite Automata and Regular Languages for a demonstration of this. :$(2): \quad$ There exists an infinite number of words $s \in L_{inf}$ with: ::::$\left|{s}\right| \ge \left|{Q}\right|$ Let $s \in L_{inf}$ be one of these words. Let $n_0 = \left|{Q}\right| + 1$. Consider all prefixes of this $s$. We have: :$\left|{s}\right| > \left|{Q}\right|$ Thus from the Pigeonhole Principle, for at least two of them $F$ needs to end in the same state. Let the first two indices in $s$ for which this is the case be $i$ and $j$. The difference between $prefix_i$ and $prefix_j$ ($v$ in this formulation of the theorem) is a "loop" in $F$. This loop can be traversed any number of times (including 0) and $F$ will still be in the same state immediately afterwards. Hence: :$\forall i \N_0: u v^i w \in L_{inf}$ Furthermore: :$\left|{u v}\right| = j - 1 < n_0$ And since $i \ne j$ it follows that: :$\left|{v}\right| > 0$ {{qed|lemma}}   === General case === Since it could be shown that the theorem holds for both all finite and all infinite regular languages, it can be stated that it holds for all regular languages. {{qed}}
{{begin-eqn}} {{eqn | l = \frac {1 + \tan \frac x 2}{1 - \tan \frac x 2}       | r = \frac {1 + \frac {\sin \frac x 2}{\cos \frac x 2} }{1 - \frac {\sin \frac x 2}{\cos \frac x 2} }       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {\cos \frac x 2 + \sin \frac x 2}{\cos \frac x 2 - \sin \frac x 2}       | c = multiply top and bottom by $\cos \dfrac x 2$ }} {{eqn | r = \frac {\cos \frac x 2 + \sin \frac x 2}{\cos \frac x 2 - \sin \frac x 2} \times \frac {\cos \frac x 2 + \sin \frac x 2}{\cos \frac x 2 + \sin \frac x 2} }} {{eqn | r = \frac{\cos^2 \frac x 2 + \sin^2 \frac x 2 + 2 \cos \frac x 2 \sin \frac x 2}{\cos^2 \frac x 2 - \sin^2 \frac x 2} }} {{eqn | r = \frac{1 + 2 \cos \frac x 2 \sin \frac x 2}{\cos^2 \frac x 2 - \sin^2 \frac x 2}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \frac{1 + \sin x}{\cos x}       | c = Double Angle Formula for Sine, Double Angle Formula for Cosine }} {{eqn | r = \sec x + \tan x       | c = Sum of Secant and Tangent }} {{end-eqn}} {{qed}} Category:Trigonometric Identities 2ylde7dt30nahj4y1u0pwz17afz24fx
Let: {{begin-eqn}} {{eqn | l = u       | r = \tan x + \sec x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac \d {\d x} \tan x + \frac \d {\d x} \sec x        | c = Linear Combination of Derivatives }} {{eqn | r = \sec^2 x + \frac \d {\d x} \sec x        | c = Derivative of Tangent Function }} {{eqn | r = \sec^2 x + \sec x \tan x        | c = Derivative of Secant Function }} {{eqn | r = \sec x \paren {\sec x + \tan x}       | c = factorising }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \sec x \rd x       | r = \int \frac {\sec x \paren {\sec x + \tan x} } {\sec x + \tan x} \rd x       | c = multiplying top and bottom by $\sec x + \tan x$ }} {{eqn | r = \ln \size {\sec x + \tan x} + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \sec x \rd x        | r = \int \frac 1 {\cos x} \rd x       | c = Secant is Reciprocal of Cosine }} {{end-eqn}} We make the Weierstrass Substitution: {{begin-eqn}} {{eqn | l = u       | r = \tan \frac x 2 }} {{eqn | ll= \leadsto       | l = \cos x       | r = \frac {1 - u^2} {1 + u^2} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d u}       | r = \frac 2 {u^2 + 1} }} {{eqn | ll= \leadsto       | l = \int \frac 1 {\cos x} \rd x       | r = \int \frac {1 + u^2} {1 - u^2} \frac 2 {u^2 + 1} \rd u }} {{eqn | r = 2 \int \frac 1 {1 - u^2} \rd u }} {{eqn | r =  \ln \size {\frac {1 + u} {1 - u} } + C       | c = Primitive of $\dfrac 1 {a^2 - u^2}$: Logarithm Form }} {{eqn | r = \ln \size {\frac {1 + \tan \frac x 2} {1 - \tan \frac x 2} } + C }} {{eqn | r = \ln \size {\sec x + \tan x} + C       | c = One Plus Tangent Half Angle over One Minus Tangent Half Angle }} {{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 = \csc^{n - 2} a x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = -a \left({n - 2}\right) \csc^{n - 3} a x \csc a x \cot a x       | c = Derivative of Power, Derivative of $\csc$, Chain Rule for Derivatives }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = -a \left({n - 2}\right) \csc^{n - 2} a x \cot a x       | c =  }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = \csc^2 a x       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {-\cot a x} a       | c = Primitive of $\csc^2 a x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \csc^n a x \ \mathrm d x       | r = \int \csc^{n - 2} a x \csc^2 a x \ \mathrm d x       | c =  }} {{eqn | r = \csc^{n - 2} a x \left({\frac {-\cot a x} a}\right) - \int \left({\frac {-\cot a x} a}\right) \left({-a \left({n - 2}\right) \csc^{n - 2} a x \cot a x }\right) \ \mathrm d x       | c = Integration by Parts }} {{eqn | r = \frac {-\csc^{n - 2} a x \cot a x} a - \left({n - 2}\right) \int \cot^2 a x \csc^{n - 2} a x \ \mathrm d x       | c = simplifying }} {{eqn | r = \frac {-\csc^{n - 2} a x \cot a x} a - \left({n - 2}\right) \int \left({\csc^2 a x - 1}\right) \csc^{n - 2} a x \ \mathrm d x       | c = Difference of $\csc^2$ and $\cot^2$ }} {{eqn | r = \frac {-\csc^{n - 2} a x \cot a x} a - \left({n - 2}\right) \int \csc^n a x \ \mathrm d x       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= +       | r = \left({n - 2}\right) \int \csc^{n - 2} a x \ \mathrm d x       | c =  }} {{eqn | l = \left({n - 1}\right) \int \csc^n a x \ \mathrm d x       | r = \frac {-\csc^{n - 2} a x \cot a x} a + \left({n - 2}\right) \int \csc^{n - 2} a x \ \mathrm d x       | c = gathering terms }} {{eqn | l = \int \csc^n a x \ \mathrm d x       | r = \frac{-\csc^{n - 2} a x \cot a x} {a \left({n - 1}\right)} + \frac {n - 2} {n - 1} \int \csc^{n - 2} a x \ \mathrm d x       | c = dividing by $n - 1$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \csc x \rd x       | r = \ln \size {\tan \frac x 2}       | c = Primitive of $\csc x$: Tangent Form }} {{eqn | ll= \leadsto       | l = \int \csc a x \rd x       | r = \frac 1 a \ln \size {\tan \frac {a x} 2} + C       | c = Primitive of Function of Constant Multiple }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \csc x \rd x       | r = \ln \size {\csc x - \cot x}       | c = Primitive of $\csc x$: Cosecant minus Cotangent Form }} {{eqn | ll= \leadsto       | l = \int \csc a x \rd x       | r = \frac 1 a \ln \size {\csc a x - \cot a x} + C       | c = Primitive of Function of Constant Multiple }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({1 + \sin a x}\right)^2}       | r = \int \left({\frac 1 2 \sec^2 \left({\dfrac \pi 4 - \frac {a x} 2}\right)}\right)^2 \ \mathrm d x       | c = Reciprocal of One Plus Sine }} {{eqn | r = \frac 1 4 \int \sec^4 \left({\frac \pi 4 - \frac {a x} 2}\right) \ \mathrm d x       | c = simplifying }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = \frac \pi 4 - \frac {a x} 2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = \frac {-a} 2       | c = simplifying }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac 1 4 \int \sec^4 \left({\frac \pi 4 - \frac {a x} 2}\right) \ \mathrm d x       | r = \frac 1 4 \int \frac {-2} a \sec^4 z \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac {-1} {2 a} \int \sec^4 z \ \mathrm d z       | c = simplifying }} {{eqn | r = \frac {-1} {2 a} \left({\frac{\sec^2 z \tan z} 3 + \frac 2 3 \int \sec^2 z \ \mathrm d z}\right) + C       | c = Primitive of $\sec^n a x$ }} {{eqn | r = \frac {-1} {2 a} \left({\frac{\sec^2 z \tan z} 3 + \frac 2 3 \tan z}\right) + C       | c = Primitive of $\sec^2 a x$ }} {{eqn | r = \frac {-1} {2 a} \left({\frac{\left({1 + \tan^2 z}\right) \tan z} 3 + \frac 2 3 \tan z}\right) + C       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac {-1} {2 a} \tan z - \frac 1 {6 a} \tan^3 z + C       | c = simplifying }} {{eqn | r = \frac {-1} {2 a} \tan \left({\frac \pi 4 - \frac {a x} 2}\right) - \frac 1 {6 a} \tan^3 \left({\frac \pi 4 - \frac {a x} 2}\right) + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {p + q \sin a x}       | r =  \frac 2 a \int \frac {\d u} {p u^2 + 2 q u + p}       | c = Weierstrass Substitution: $u = \tan \dfrac {a x} 2$ }} {{end-eqn}} The discriminant of $p u^2 + 2 q u + p$ is $4 q^2 - 4 p^2$. Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {p + q \sin a x}       | r = \begin{cases}  \displaystyle \frac 2 a \paren {\dfrac 2 {\sqrt {4 p^2 - 4 q^2} } \map \arctan {\dfrac {2 p u + 2 q} {\sqrt {4 p^2 - 4 q^2} } } } + C & : 4 q^2 - 4 p^2 < 0 \\ \displaystyle \frac 2 a \paren {\dfrac 1 {\sqrt {4 q^2 - 4 p^2} } \ln \size {\dfrac {2 p x + 2 q - \sqrt {4 q^2 - 4 p^2} } {2 p x + 2 q + \sqrt {4 q^2 - 4 p^2} } } } + C & : 4 q^2 - 4 p^2 > 0 \\ \end{cases}       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ }} {{eqn | r = \begin{cases}  \displaystyle \dfrac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\dfrac {p u + q} {\sqrt {p^2 - q^2} } } + C & : q^2 - p^2 < 0 \\ \displaystyle \dfrac 1 {a \sqrt {q^2 - p^2} } \ln \size {\dfrac {p x + q - \sqrt {q^2 - p^2} } {p x + q + \sqrt {q^2 - p^2} } } + C & : q^2 - p^2 > 0 \\ \end{cases}       | c = simplifying }} {{eqn | r = \begin{cases}  \displaystyle \frac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\frac {p \tan \dfrac {a x} 2 + q} {\sqrt {p^2 - q^2} } } + C & : q^2 - p^2 < 0 \\ \displaystyle \frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\frac {p \tan \dfrac {a x} 2 + q - \sqrt {p^2 - q^2} } {p \tan \dfrac {a x} 2 + q + \sqrt {p^2 - q^2} } } + C & : q^2 - p^2 > 0 \\ \end{cases}       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {\paren {p + q \sin a x}^2}       | r = \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}       | c = Weierstrass Substitution: $u = \tan \dfrac {a x} 2$ }} {{eqn | r = \frac 2 a \int \frac {u^2 \rd u} {\paren {p u^2 + 2 q u + p}^2} + \frac 2 a \int \frac {\d u} {\paren {p u^2 + 2 q u + p}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 2 a \paren {\frac {\paren {4 q^2 - 2 p^2} u + 2 p q} {p \paren {4 p^2 - 4 q^2} \paren {p u^2 + 2 q u + p} } + \frac {2 p} {4 p^2 - 4 q^2} \int \frac {\d u} {p u^2 + 2 q u + p} }       | c = Primitive of $\dfrac {x^2} {\paren {a x^2 + b x + c}^2}$ }} {{eqn | o =        | ro= +       | r = \frac 2 a \paren {\frac {2 p u + 2 q} {\paren {4 p^2 - 4 q^2} \paren {p u^2 + 2 q u + p} } + \frac {2 p} {4 p^2 - 4 q^2} \int \frac {\d u} {p u^2 + 2 q u + p} }       | c = Primitive of $\dfrac 1 {\paren {a x^2 + b x + c}^2}$ }} {{eqn | r = \frac {\paren {2 q^2 - p^2} u + p q} {a p \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac p {a \paren {p^2 - q^2} } \int \frac {\d u} {p u^2 + 2 q u + p}       | c = simplifying ... }} {{eqn | o =        | ro= +       | r = \frac {p^2 u + p q} {a p \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac p {a \paren {p^2 - q^2} } \int \frac {\d u} {p u^2 + 2 q u + p}       | c = ... and creating common denominators }} {{eqn | r = \frac {\paren {2 q^2 - p^2} u + p q + p^2 u + p q} {a p \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac {2 p} {a \paren {p^2 - q^2} } \int \frac {\d u} {p u^2 + 2 q u + p}       | c = summing }} {{eqn | r = \frac {2 q \paren {q u + p} } {a p \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac {2 p} {a \paren {p^2 - q^2} } \int \frac {\d u} {p u^2 + 2 q u + p}       | c = simplifying }} {{end-eqn}} {{proof wanted|Anyone care to finish it off?}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {p^2 + q^2 \sin^2 a x}       | r = \int \frac {\sec^2 a x \rd x} {p^2 \sec^2 a x + q^2 \tan^2 a x}       | c = multiplying numerator and denominator by $\sec^2 a x$ }} {{eqn | r = \int \frac {\sec^2 a x \rd x} {p^2 + \paren {p^2 + q^2} \tan^2 a x}       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \int \frac {\paren {\tan a x}' \rd x} {a p^2 + a \paren {p^2 + q^2} \tan^2 a x}       | c = Derivative of Tangent Function }} {{eqn | r = \int \frac {\d t} {a p^2 + a \paren {p^2 + q^2} t^2}       | c = substituting $t = \tan a x$ }} {{eqn | r = \frac 1 {a \paren {p^2 + q^2} } \int \frac {\rd t} {\paren {\tfrac p {\sqrt {p^2 + q^2} } }^2 + t^2}       | c =  }} {{eqn | r = \frac 1 {a \paren {p^2 + q^2} } \frac {\sqrt {p^2 + q^2} } p \map \arctan {\frac {\sqrt {p^2 + q^2} } p t} + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac 1 {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} } p t} + C       | c =  }} {{eqn | r = \frac 1 {a p \sqrt {p^2 + q^2} } \map \arctan {\frac {\sqrt {p^2 + q^2} \tan a x} p} + C       | c = substituting $t = \tan a x$ }} {{end-eqn}} {{qed}}