hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn 	| l = \int x \sec a x \rd x  	| r = \frac 1 {a^2} \int \theta \sec \theta \rd \theta  	| c = Substitution of $a x \to \theta$ }} {{eqn 	| r = \frac 1 {a^2} \int \theta \sum_{n \mathop = 0}^\infty \frac{ \paren {-1}^n E_{2 n} \theta^{2 n} } {\paren {2 n}!} \rd \theta 	| c = Power Series Expansion for Secant Function }} {{eqn 	| r = \frac 1 {a^2} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n E_{2 n} } {\paren {2 n}!} \int \theta^{2 n + 1} \rd \theta 	| c = Fubini's Theorem }} {{eqn 	| r = \frac 1 {a^2} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n E_{2 n} \paren {a x}^{2 n + 2} } {\paren {2 n + 2} \paren {2 n}!}  + C 	| c = Substituting back $\theta \to ax$ }}  {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\csc a x} x \ \mathrm d x       | r = \int \frac 1 x \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n E_{2 n} \paren {a x}^{2 n} } {\paren {2 n}!} \ \rd x       | c = Power Series Expansion for Cosecant Function }} {{eqn | r = \int \frac {E_0} x \ \rd x + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n E_{2 n} a^{2 n} } {\paren {2 n}!} \int  x^{2 n - 1} \ \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \int \frac 1 x \ \rd x + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n E_{2 n} a^{2 n} } {\paren {2 n}!} \paren {\frac {x^{2 n} } {2 n} } + C       | c = Primitive of Power }} {{eqn | r = \ln \size x + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n E_{2 n} \paren {a x}^{2 n} } {\paren {2 n} \paren {2 n}!} + C       | c = Primitive 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 = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = 1       | c = Derivative of Identity Function }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = \sec^2 a x       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {\tan a x} a       | c = Primitive of $\sec^2 a x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \sec^2 a x \ \mathrm d x       | r = \frac {x \tan a x} a - \int \frac {\tan a x} a \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x \tan a x} a - \frac 1 a \left({\frac {-1} a \ln \left\vert{\cos a x}\right\vert}\right) + C       | c = Primitive of $\tan a x$ }} {{eqn | r = \frac {x \tan a x} a + \frac  1 {a^2} \ln \left\vert{\cos a x}\right\vert + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {q + p \sec a x}       | r = \frac 1 q \int \frac {q \rd x} {q + p \sec a x}       | c = multiplying top and bottom by $q$ }} {{eqn | r = \frac 1 q \int \frac {\paren {q + p \sec a x - p \sec a x} \rd x} {q + p \sec a x}       | c =  }} {{eqn | r = \frac 1 q \int \frac {\paren {q + p \sec a x} \rd x} {q + p \sec a x} - \frac p q \int \frac {\sec a x \rd x} {q + p \sec a x}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 q \int \d x - \frac p q \int \frac {\sec a x \rd x} {q + p \sec a x}       | c = simplifying }} {{eqn | r = \frac x q - \frac p q \int \frac {\sec a x \rd x} {q + p \sec a x} + C       | c = Primitive of Constant }} {{eqn | r = \frac x q - \frac p q \int \frac {\cos a x \sec a x \rd x} {q \cos a x + p \cos a x \sec a x} + C       | c = multiplying top and bottom by $\cos a x$ }} {{eqn | r = \frac x q - \frac p q \int \frac {\d x} {p + q \cos a x} + C       | c = Secant is Reciprocal of Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \csc^3 x \rd x       | r = \frac {-\csc a x \cot a x} {2 a} + \frac 1 2 \int \csc a x \rd x       | c = Primitive of $\csc^n a x$ where $n = 3$ }} {{eqn | r = \frac {-\csc a x \cot a x} {2 a} + \frac 1 2 \paren {\frac 1 a \ln \size {\tan \dfrac {a x} 2} }       | c = Primitive of $\csc a x$: Tangent Form }} {{eqn | r = \frac {-\csc a x \cot a x} {2 a} + \frac 1 {2 a} \ln \size {\tan \dfrac {a x} 2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {\csc a x}       | r = \int \sin a x \rd x       | c = Cosecant is Reciprocal of Sine }} {{eqn | r = \frac {-\cos a x} a + C       | c = Primitive of $\sin a x$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn 	| l = \int x \csch a x \rd x  	| r = \frac 1 {a^2} \int \theta \csch \theta \rd \theta  	| c = Substitution of $a x \to \theta$ }} {{eqn 	| r = \frac 1 {a^2} \int \theta \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} \, \theta^{2 n - 1} } {\paren {2 n}!} \rd \theta 	| c = Power Series Expansion for Cosecant Function }} {{eqn 	| r = \frac 1 {a^2} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} } {\paren {2 n}!} \int \theta^{2 n} \rd \theta 	| c = Fubini's Theorem }} {{eqn 	| r = \frac 1 {a^2} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} \paren {a x}^{2 n + 1} } {\paren {2 n + 1}!}  + C 	| c = Substituting back $\theta \to ax$ }}  {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\csc a x} x \ \mathrm d x       | r = \int \frac 1 x \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} \paren {a x}^{2 n - 1} } {\paren {2 n}!} \ \mathrm d x       | c = Power Series Expansion for Cosecant Function }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} a^{2 n - 1} } {\paren {2 n}!} \int  x^{2 n - 2} \ \mathrm d x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} a^{2 n - 1} } {\paren {2 n}!} \paren {\frac {x^{2 n - 1} } {2 n - 1} } + C       | c = Primitive of Power }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} \paren {a x}^{2 n - 1} } {\paren {2 n - 1} \paren {2 n}!} + C       | c =  }} {{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 = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = 1       | c = Derivative of Identity Function }} {{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 x \csc^2 a x \ \mathrm d x       | r = \frac {-x \cot a x} a - \int \frac {-\cot a x} a \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {-x \cot a x} a + \frac 1 a \int \cot a x \ \mathrm d x + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {-x \cot a x} a + \frac 1 a \left({\frac 1 a \ln \left\vert{\sin a x}\right\vert}\right) + C       | c = Primitive of $\cot a x$ }} {{eqn | r = \frac {-x \cot a x} a + \frac  1 {a^2} \ln \left\vert{\sin a x}\right\vert + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {q + p \csc a x}       | r = \frac 1 q \int \frac {q \ \mathrm d x} {q + p \csc a x}       | c = multiplying top and bottom by $q$ }} {{eqn | r = \frac 1 q \int \frac {\left({q + p \csc a x - p \csc a x}\right) \ \mathrm d x} {q + p \csc a x}       | c =  }} {{eqn | r = \frac 1 q \int \frac {\left({q + p \csc a x}\right) \ \mathrm d x} {q + p \csc a x} - \frac p q \int \frac {\csc a x \ \mathrm d x} {q + p \csc a x}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 q \int \mathrm d x - \frac p q \int \frac {\csc a x \ \mathrm d x} {q + p \csc a x}       | c = simplifying }} {{eqn | r = \frac x q - \frac p q \int \frac {\csc a x \ \mathrm d x} {q + p \csc a x} + C       | c = Primitive of Constant }} {{eqn | r = \frac x q - \frac p q \int \frac {\sin a x \, \csc a x \ \mathrm d x} {q \sin a x + p \sin a x \, \csc a x} + C       | c = multiplying top and bottom by $\sin a x$ }} {{eqn | r = \frac x q - \frac p q \int \frac {\mathrm d x} {p + q \sin a x} + C       | c = Cosecant is Reciprocal of Sine }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \arcsin \frac x a       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = \sin u        | r = \frac x a       | c = {{Defof|Arcsine}} }} {{eqn | n = 2       | ll= \leadsto       | l = \cos u        | r = \sqrt {1 - \frac {x^2} {a^2} }       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \arcsin \frac x a \rd x       | r = a \int u \cos u \rd u       | c = Primitive of Function of Arcsine }} {{eqn | r = a \paren {\cos u + u \sin u} + C       | c = Primitive of $x \cos a x$ }} {{eqn | r = a \paren {\cos u + u \frac x a} + C       | c = Substitution for $\sin u$ from $\paren 1$ }} {{eqn | r = a \paren {\sqrt {1 - \frac {x^2} {a^2} } + u \frac x a} + C       | c = Substitution for $\cos u$ from $\paren 2$ }} {{eqn | r = a \paren {\sqrt {1 - \frac {x^2} {a^2} } + \frac x a \arcsin \frac x a} + C       | c = Substitution for $u$ }} {{eqn | r = x \arcsin \frac x a + \sqrt {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 = \arcsin \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac 1 {\sqrt {a^2 - x^2} }       | c = Derivative of $\arcsin \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = x       | c = Primitive of Constant }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \arcsin \frac x a \rd x       | r = x \arcsin \frac x a - \int x \paren {\frac 1 {\sqrt {a^2 - x^2} } } \rd x + C       | c = Integration by Parts }} {{eqn | r = x \arcsin \frac x a - \int \frac {x \rd x} {\sqrt {a^2 - x^2} } + C       | c = simplifying }} {{eqn | r = x \arcsin \frac x a - \paren {-\sqrt {a^2 - x^2} } + C       | c = Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$ }} {{eqn | r = x \arcsin \frac x a + \sqrt {a^2 - x^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \arcsin \frac x a       | c =  }} {{eqn | n = 1       | ll= \implies       | l = \sin u        | r = \frac x a       | c = Definition of Arcsine }} {{eqn | n = 2       | ll= \implies       | l = \cos u        | r = \sqrt {1 - \frac {x^2} {a^2} }       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \arcsin \frac x a \ \mathrm d x       | r = a \int u \left({a \sin u}\right) \cos u \ \mathrm d u       | c = Primitive of Function of Arcsine }} {{eqn | r = a \int a u \frac {\sin 2 u} 2 \ \mathrm d u       | c = Double Angle Formula for Sine }} {{eqn | r = \frac {a^2} 2 \int u \sin 2 u \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {a^2} 2 \left({\frac {\sin 2 u} 4 - \frac {u \cos 2 u} 2}\right) + C       | c = Primitive of $x \sin a x$ where $a = 2$ }} {{eqn | r = \frac {a^2 \sin 2 u} 8 - \frac {a^2 u \cos 2 u} 4 + C       | c = simplifying }} {{eqn | r = \frac {a^2 \sin u \cos u} 4 - \frac {a^2 u \cos 2 u} 4 + C       | c = Double Angle Formula for Sine }} {{eqn | r = \frac {a^2 \sin u \cos u} 4 - \frac {a^2 u \left({\cos^2 u - \sin^2 u}\right)} 4 + C       | c = Double Angle Formula for Cosine }} {{eqn | r = \frac {a^2 \sin u \cos u} 4 - \frac {a^2 \arcsin \frac x a \left({\cos^2 u - \sin^2 u}\right)} 4 + C       | c = substituting for $u$ }} {{eqn | r = \frac {a^2 \frac x a \cos u} 4 - \frac {a^2 \arcsin \frac x a \left({\cos^2 u - \frac {x^2} {a^2} }\right)} 4 + C       | c = substituting for $\sin u$ from $(1)$ }} {{eqn | r = \frac {a^2 \frac x a \sqrt {1 - \frac {x^2} {a^2} } } 4 - \frac {a^2 \arcsin \frac x a \left({\left({1 - \frac {x^2} {a^2} }\right) - \frac {x^2} {a^2} }\right)} 4 + C       | c = substituting for $\cos u$ from $(2)$ }} {{eqn | r = \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} } 4 + 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 = \arcsin \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 {\sqrt {a^2 - x^2} }       | c = Derivative of $\arcsin \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = x       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^2} 2       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \arcsin \frac x a \ \mathrm d x       | r = \frac {x^2} 2 \arcsin \frac x a - \int \frac {x^2} 2 \left({\frac 1 {\sqrt {a^2 - x^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \arcsin \frac x a - \frac 1 2 \int \frac {x^2 \ \mathrm d x} {\sqrt {a^2 - x^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \arcsin \frac x a - \frac 1 2 \left({\frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a}\right) + C       | c = Primitive of $\dfrac {x^2} {\sqrt {a^2 - x^2} }$ }} {{eqn | r = \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arcsin \frac x a + \frac {x \sqrt {a^2 - x^2} } 4 + 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 = \arcsin \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac 1 {\sqrt {a^2 - x^2} }       | c = Derivative of $\arcsin \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = x^2       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {x^3} 3       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^2 \arcsin \frac x a \rd x       | r = \frac {x^3} 3 \arcsin \frac x a - \int \frac {x^3} 3 \paren {\frac 1 {\sqrt {a^2 - x^2} } } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^3} 3 \arcsin \frac x a - \frac 1 3 \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^3} 3 \arcsin \frac x a - \frac 1 3 \paren {\frac {\paren {\sqrt {a^2 - x^2} }^3} 3 - a^2 \sqrt {a^2 - x^2} } + C       | c = Primitive of $\dfrac {x^3} {\sqrt {a^2 - x^2} }$ }} {{eqn | r = \frac {x^3} 3 \arcsin \frac x a + \frac {\paren {x^2 + 2 a^2} \sqrt {a^2 - x^2} } 9 + 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 = \arcsin \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 {\sqrt {a^2 - x^2} }       | c = Derivative of $\arcsin \dfrac x a$ }} {{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 {\arcsin \frac x a \ \mathrm d x} {x^2}       | r = \arcsin \frac x a \left({\frac {-1} x}\right) - \int \left({\frac {-1} x}\right) \left({\frac 1 {\sqrt {a^2 - x^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {-\arcsin \frac x a} x + \int \frac {\mathrm d x} {x \sqrt {a^2 - x^2} }  \ \mathrm d x + C       | c = simplifying }} {{eqn | r = \frac {-\arcsin \frac x a} x - \frac 1 a \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C       | c = Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \arccos \frac x a       | c =  }} {{eqn | n = 1       | ll= \implies       | l = \cos u        | r = \frac x a       | c = {{Defof|Arccosine}} }} {{eqn | n = 2       | ll= \implies       | l = \cos u        | r = \sqrt {1 - \frac {x^2} {a^2} }       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \arccos \frac x a \rd x       | r = -a \int u \sin u \rd u       | c = Primitive of Function of Arccosine }} {{eqn | r = -a \left({\sin u - u \cos u}\right) + C       | c = Primitive of $x \sin a x$ }} {{eqn | r = -a \left({\sin u - u \frac x a}\right) + C       | c = Substitution for $\cos u$ from $(1)$ }} {{eqn | r = -a \left({\sqrt {1 - \frac {x^2} {a^2} } - u \frac x a}\right) + C       | c = Substitution for $\sin u$ from $(2)$ }} {{eqn | r = -a \left({\sqrt {1 - \frac {x^2} {a^2} } - \frac x a \arccos \frac x a}\right) + C       | c = Substitution for $u$ }} {{eqn | r = x \arccos \frac x a - \sqrt {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 = \arccos \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac {-1} {\sqrt {a^2 - x^2} }       | c = Derivative of $\arccos \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = 1       | c =  }} {{eqn | ll= \implies       | l = v       | r = x       | c = Primitive of Constant }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \arccos \frac x a \rd x       | r = x \arccos \frac x a - \int x \left({\frac {-1} {\sqrt {a^2 - x^2} } }\right) \rd x + C       | c = Integration by Parts }} {{eqn | r = x \arccos \frac x a + \int \frac {x \rd x} {\sqrt {a^2 - x^2} } + C       | c = simplifying }} {{eqn | r = x \arccos \frac x a + \left({-\sqrt {a^2 - x^2} }\right) + C       | c = Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$ }} {{eqn | r = x \arccos \frac x a - \sqrt {a^2 - x^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \arccos \frac x a       | c =  }} {{eqn | n = 1       | ll= \implies       | l = \cos u        | r = \frac x a       | c = Definition of Arccosine }} {{eqn | n = 2       | ll= \implies       | l = \sin u        | r = \sqrt {1 - \frac {x^2} {a^2} }       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \arccos \frac x a \ \mathrm d x       | r = -a \int u \left({a \cos u}\right) \sin u \ \mathrm d u       | c = Primitive of Function of Arcsine }} {{eqn | r = -a \int a u \frac {\sin 2 u} 2 \ \mathrm d u       | c = Double Angle Formula for Sine }} {{eqn | r = -\frac {a^2} 2 \int u \sin 2 u \ \mathrm d u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -\frac {a^2} 2 \left({\frac {\sin 2 u} 4 - \frac {u \cos 2 u} 2}\right) + C       | c = Primitive of $x \sin a x$ where $a = 2$ }} {{eqn | r = -\frac {a^2 \sin 2 u} 8 + \frac {a^2 u \cos 2 u} 4 + C       | c = simplifying }} {{eqn | r = -\frac {a^2 \sin u \cos u} 4 + \frac {a^2 u \cos 2 u} 4 + C       | c = Double Angle Formula for Sine }} {{eqn | r = -\frac {a^2 \sin u \cos u} 4 + \frac {a^2 u \left({\cos^2 u - \sin^2 u}\right)} 4 + C       | c = Double Angle Formula for Cosine }} {{eqn | r = -\frac {a^2 \sin u \cos u} 4 + \frac {a^2 \arccos \frac x a \left({\cos^2 u - \sin^2 u}\right)} 4 + C       | c = substituting for $u$ }} {{eqn | r = -\frac {a^2 \sin u \frac x a} 4 + \frac {a^2 \arccos \frac x a \left({\frac {x^2} {a^2} - \sin^2 u}\right)} 4 + C       | c = substituting for $\cos u$ from $(1)$ }} {{eqn | r = -\frac {a^2 \sqrt {1 - \frac {x^2} {a^2} } \frac x a} 4 + \frac {a^2 \arccos \frac x a \left({\frac {x^2} {a^2} - \left({1 - \frac {x^2} {a^2} }\right)}\right)} 4 + C       | c = substituting for $\sin u$ from $(2)$ }} {{eqn | r = \left({\frac {x^2} 2 - \frac {a^2} 4}\right) \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + 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 = \arccos \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac {-1} {\sqrt {a^2 - x^2} }       | c = Derivative of $\arccos \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {x^2} 2       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \arccos \frac x a \rd x       | r = \frac {x^2} 2 \arccos \frac x a - \int \frac {x^2} 2 \paren {\frac {-1} {\sqrt {a^2 - x^2} } } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \arccos \frac x a + \frac 1 2 \int \frac {x^2 \rd x} {\sqrt {a^2 - x^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \arccos \frac x a + \frac 1 2 \paren {\frac {-x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a} + C       | c = Primitive of $\dfrac {x^2} {\sqrt {a^2 - x^2} }$ }} {{eqn | r = \frac {x^2} 2 \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + \frac {a^2} 4 \paren {\frac \pi 2 - \arccos \frac x a} + C       | c = Sum of Arcsine and Arccosine }} {{eqn | r = \paren {\frac {x^2} 2 - \frac {a^2} 4} \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + \frac {\pi a^2} 8 + C       | c = simplifying }} {{eqn | r = \paren {\frac {x^2} 2 - \frac {a^2} 4} \arccos \frac x a - \frac {x \sqrt {a^2 - x^2} } 4 + C       | c = subsuming $\dfrac {\pi a^2} 8$ into the arbitrary constant }} {{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 = \arccos \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac {-1} {\sqrt {a^2 - x^2} }       | c = Derivative of $\arccos \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = x^2       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {x^3} 3       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^2 \arccos \frac x a \rd x       | r = \frac {x^3} 3 \arccos \frac x a - \int \frac {x^3} 3 \paren {\frac {-1} {\sqrt {a^2 - x^2} } } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^3} 3 \arccos \frac x a + \frac 1 3 \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^3} 3 \arccos \frac x a + \frac 1 3 \paren {\frac {\paren {\sqrt {a^2 - x^2} }^3} 3 - a^2 \sqrt {a^2 - x^2} } + C       | c = Primitive of $\dfrac {x^3} {\sqrt {a^2 - x^2} }$ }} {{eqn | r = \frac {x^3} 3 \arccos \frac x a - \frac {\paren {x^2 + 2 a^2} \sqrt {a^2 - x^2} } 9 + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\arccos \frac x a \rd x} x       | r = \int \frac {\paren {\frac \pi 2 - \arcsin \frac x a} \rd x} x       | c = Sum of Arcsine and Arccosine }} {{eqn | r = \frac \pi 2 \int \frac {\d x} x - \int \frac {\arcsin \frac x a \rd x} x       | c = Linear Combination of Integrals }} {{eqn | r = \frac \pi 2 \ln \size x - \int \frac {\arcsin \frac x a \rd x} x + C       | c = Primitive of Reciprocal }} {{end-eqn}} {{qed}}