hcju/proofwikiqa huggingface dataset:
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = \arccos \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 $\arccos \dfrac x a$ }} {{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 {\arccos \frac x a \ \mathrm d x} {x^2}       | r = \arccos \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 {-\arccos \frac x a} x - \int \frac {\mathrm d x} {x \sqrt {a^2 - x^2} }  \ \mathrm d x + C       | c = simplifying }} {{eqn | r = \frac {-\arccos \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= \leadsto       | l = \cos u        | r = \frac x a       | c = {{Defof|Arccosine}} }} {{eqn | n = 2       | ll= \leadsto       | 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 \paren {\arccos \frac x a}^2 \rd x       | r = -a \int u^2 \sin u \rd x       | c = Primitive of Function of Arccosine }} {{eqn | r = -a \paren {2 u \sin u + \paren {2 - u^2} \cos u} + C       | c = Primitive of $x^2 \sin a x$ where $a = 1$ }} {{eqn | r = -a \paren {2 \arccos \frac x a \sqrt {1 - \frac {x^2} {a^2} } + \paren {2 - \paren {\arccos \frac x a}^2} \frac x a} + C       | c = substituting for $u$, $\sin u$ and $\cos u$ }} {{eqn | r = x \paren {\arccos \frac x a}^2 - 2 x - 2 \sqrt {a^2 - x^2} \arccos \frac x a + C       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \arctan \frac x a       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = \tan u        | r = \frac x a       | c = {{Defof|Arctangent}} }} {{eqn | n = 2       | ll= \leadsto       | l = \sec u        | r = \sqrt {1 + \frac {x^2} {a^2} }       | c = Difference of Squares of Secant and Tangent }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \arctan \frac x a \rd x       | r = a \int u \sec^2 u \rd u       | c = Primitive of Function of Arctangent }} {{eqn | r = a \paren {u \tan u + \ln \size {\cos u} } + C       | c = Primitive of $x \sec^2 a x$ with $a = 1$ }} {{eqn | r = a u \tan u - a \ln \size {\sec u} + C       | c = Logarithm of Reciprocal and Secant is Reciprocal of Cosine }} {{eqn | r = a u \frac x a - a \ln \size {\sec u} + C       | c = Substitution for $\tan u$ from $\paren 1$ }} {{eqn | r = a u \frac x a - a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C       | c = Substitution for $\sec u$ from $\paren 2$ }} {{eqn | r = x \arctan \frac x a - a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C       | c = Substitution for $u$ }} {{eqn | r = x \arctan \frac x a - \frac a 2 \ln \size{\frac {x^2 + a^2 } {a^2} } + C       | c = Logarithm of Power and simplifying }} {{eqn | r = x \arctan \frac x a - \frac a 2 \ln \size{x^2 + a^2} - \ln \size {a^2} + C       | c = Difference of Logarithms }} {{eqn | r = x \arctan \frac x a - \frac a 2 \ln \size {x^2 + a^2} + C       | c = subsuming $\ln \left\vert{a^2}\right\vert$ into arbitrary constant }} {{eqn | r = x \arctan \frac x a - \frac a 2 \ln \paren {x^2 + a^2} + C       | c = $x^2 + a^2$ always positive }} {{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 = \arctan \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac a {x^2 + a^2}       | c = Derivative of $\arctan \dfrac x a$ }} {{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 \arctan \frac x a \rd x       | r = x \arctan \frac x a - \int x \paren {\frac a {x^2 + a^2} } \rd x + C       | c = Integration by Parts }} {{eqn | r = x \arctan \frac x a - a \int \frac {x \rd x} {x^2 + a^2} + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = x \arctan \frac x a - a \paren {\frac 1 2 \ln \paren {x^2 + a^2} } + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | r = x \arctan \frac x a - \frac a 2 \ln \paren {x^2 + a^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
<onlyinclude> 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 = \arctan \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac a {x^2 + a^2}       | c = Derivative of $\arctan \dfrac x a$ }} {{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 \arctan \frac x a \ \mathrm d x       | r = \frac {x^2} 2 \arctan \frac x a - \int \frac {x^2} 2 \left({\frac a {x^2 + a^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \arctan \frac x a - \frac a 2 \int \frac {x^2 \ \mathrm d x} {x^2 + a^2} + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \arctan \frac x a - \frac a 2 \left({x - a \arctan{\frac x a} }\right) + C       | c = Primitive of $\dfrac {x^2} {x^2 + a^2}$ }} {{eqn | r = \frac {x^2 + a^2} 2 \arctan \frac x a - \frac {a x} 2 + C       | c = simplifying }} {{end-eqn}} {{qed}}
<onlyinclude> 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 = \arccot \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac {-a} {x^2 + a^2}       | c = Derivative of $\arccot \dfrac x a$ }} {{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 \arccot \frac x a \rd x       | r = \frac {x^2} 2 \arccot \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x^2 + a^2} } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \arccot \frac x a + \frac a 2 \int \frac {x^2 \rd x} {x^2 + a^2} + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \arccot \frac x a + \frac a 2 \paren {x - a \arctan {\frac x a} } + C       | c = Primitive of $\dfrac {x^2} {x^2 + a^2}$ }} {{eqn | r = \frac {x^2} 2 \arccot \frac x a + \frac a 2 \paren {x - a \paren {\frac \pi 2 - \arccot {\frac x a} } } + C       | c = Sum of Arctangent and Arccotangent }} {{eqn | r = \frac {x^2 + a^2} 2 \arccot \frac x a + \frac {a x} 2 - \frac {a^2 \pi} 2 + C       | c = simplifying }} {{eqn | r = \frac {x^2 + a^2} 2 \arccot \frac x a + \frac {a x} 2 + C       | c = subsuming $\dfrac {a^2 \pi} 2$ into the arbitrary constant }} {{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 = \arctan \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac a {x^2 + a^2}       | c = Derivative of $\arctan \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = x^2       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^3} 3       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^2 \arctan \frac x a \ \mathrm d x       | r = \frac {x^3} 3 \arctan \frac x a - \int \frac {x^3} 3 \left({\frac a {x^2 + a^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^3} 3 \arctan \frac x a - \frac a 3 \int \frac {x^3 \ \mathrm d x} {x^2 + a^2} + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^3} 3 \arctan \frac x a - \frac a 3 \left({\frac {x^2} 2 - \frac {a^2} 2 \ln \left ({x^2 + a^2}\right) }\right) + C       | c = Primitive of $\dfrac {x^3} {x^2 + a^2}$ }} {{eqn | r = \frac {x^3} 3 \arctan \frac x a - \frac {a x^2} 6 + \frac {a^3} 6 \ln \left({x^2 + a^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = \operatorname{arccot} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac {-a} {x^2 + a^2}       | c = Derivative of $\operatorname{arccot} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = x^2       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^3} 3       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^2 \operatorname{arccot} \frac x a \ \mathrm d x       | r = \frac {x^3} 3 \operatorname{arccot} \frac x a - \int \frac {x^3} 3 \left({\frac {-a} {x^2 + a^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^3} 3 \operatorname{arccot} \frac x a + \frac a 3 \int \frac {x^3 \ \mathrm d x} {x^2 + a^2} + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^3} 3 \operatorname{arccot} \frac x a + \frac a 3 \left({\frac {x^2} 2 - \frac {a^2} 2 \ln \left ({x^2 + a^2}\right) }\right) + C       | c = Primitive of $\dfrac {x^3} {x^2 + a^2}$ }} {{eqn | r = \frac {x^3} 3 \operatorname{arccot} \frac x a + \frac {a x^2} 6 - \frac {a^3} 6 \ln \left({x^2 + a^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn 	| l = \int \frac 1 x \arctan \paren {\frac x a} \rd x  	| r = \int \frac 1 x \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac 1 {2 k + 1} \paren {\frac x a}^{2 k + 1} \rd x 	| c = Power Series Expansion for Real Arctangent Function }} {{eqn 	| r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac 1 {\paren {2 k + 1} a^{2 k + 1} } \int x^{2 k} \rd x  	| c = Fubini's Theorem }} {{eqn 	| r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac{x^{2 k + 1} }{\paren {2 k + 1}^2 a^{2 k + 1} } + C }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\operatorname{arccot} \frac x a \ \mathrm d x} x       | r = \int \frac {\left({\frac \pi 2 - \arctan \frac x a}\right) \ \mathrm d x} x       | c = Sum of Arctangent and Arccotangent }} {{eqn | r = \frac \pi 2 \int \frac {\mathrm d x} x - \int \frac {\arctan \frac x a \ \mathrm d x} x       | c = Linear Combination of Integrals }} {{eqn | r = \frac \pi 2 \ln \left\vert{x}\right\vert  - \int \frac {\arctan \frac x a \ \mathrm d x} x + 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 = \arctan \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac a {x^2 + a^2}       | c = Derivative of $\arctan \dfrac x a$ }} {{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 {\arctan \frac x a \ \mathrm d x} {x^2}       | r = \arctan \frac x a \left({\frac {-1} x}\right) - \int \left({\frac {-1} x}\right) \left({\frac a {x^2 + a^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {-\arctan \frac x a} x + a \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)}  \ \mathrm d x + C       | c = simplifying }} {{eqn | r = \frac {-\arctan \frac x a} x + \frac 1 a \left({\frac 1 {2 a^2} \ln \left({\frac {x^2} {x^2 + a^2} }\right)}\right) + C       | c = Primitive of $\dfrac 1 {x \left({x^2 + a^2}\right)}$ }} {{eqn | r = \frac {-\arctan \frac x a} x - \frac 1 {2 a} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C       | c = Logarithm of Reciprocal }} {{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 = \arccot \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac {-a} {x^2 + a^2}       | c = Derivative of $\arccot \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = \frac 1 {x^2}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {-1} x       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\arccot \frac x a \rd x} {x^2}       | r = \arccot \frac x a \paren {\frac {-1} x} - \int \paren {\frac {-1} x} \paren {\frac {-a} {x^2 + a^2} } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {-\arccot \frac x a} x - a \int \frac {\d x} {x \paren {x^2 + a^2} } \rd x + C       | c = simplifying }} {{eqn | r = \frac {-\arccot \frac x a} x - \frac 1 a \paren {\frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } } + C       | c = Primitive of $\dfrac 1 {x \paren {x^2 + a^2} }$ }} {{eqn | r = \frac {-\arccot \frac x a} x + \frac 1 {2 a} \map \ln {\frac {x^2 + a^2} {x^2} } + C       | c = Logarithm of Reciprocal }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \arccot \frac x a       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = \cot u        | r = \frac x a       | c = {{Defof|Arccotangent}} }} {{eqn | n = 2       | ll= \leadsto       | l = \csc u        | r = \sqrt {1 + \frac {x^2} {a^2} }       | c = Difference of Squares of Cosecant and Cotangent }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \arccot \frac x a \rd x       | r = -a \int u \csc^2 u \rd u       | c = Primitive of Function of Arccotangent }} {{eqn | r = -a \paren {-u \cot u + \ln \size {\sin u} } + C       | c = Primitive of $x \csc^2 a x$ with $a = 1$ }} {{eqn | r = a u \cot u - a \ln \size {\sin u} + C       | c = simplifying }} {{eqn | r = a u \cot u + a \ln \size {\csc u} + C       | c = Logarithm of Reciprocal and Cosecant is Reciprocal of Sine }} {{eqn | r = a u \frac x a + a \ln \size {\csc u} + C       | c = Substitution for $\cot u$ from $(1)$ }} {{eqn | r = x u + a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C       | c = Substitution for $\csc u$ from $(2)$ }} {{eqn | r = x \arccot \frac x a + a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C       | c = Substitution for $u$ }} {{eqn | r = x \arccot \frac x a + \frac a 2 \ln \size {\frac {x^2 + a^2 } {a^2} } + C       | c = Logarithm of Power and simplifying }} {{eqn | r = x \arccot \frac x a + \frac a 2 \ln \size {x^2 + a^2} - \ln \size {a^2} + C       | c = Difference of Logarithms }} {{eqn | r = x \arccot \frac x a + \frac a 2 \ln \size {x^2 + a^2} + C       | c = subsuming $\ln \size {a^2}$ into arbitrary constant }} {{eqn | r = x \arccot \frac x a + \frac a 2 \, \map \ln {x^2 + a^2} + C       | c = $x^2 + a^2$ always positive }} {{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 = \operatorname{arccot} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac {-a} {x^2 + a^2}       | c = Derivative of $\operatorname{arccot} \dfrac x a$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = 1       | c =  }} {{eqn | ll= \implies       | l = v       | r = x       | c = Primitive of Constant }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \operatorname{arccot} \frac x a \ \mathrm d x       | r = x \operatorname{arccot} \frac x a - \int x \left({\frac {-a} {x^2 + a^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = x \operatorname{arccot} \frac x a + a \int \frac {x \ \mathrm d x} {x^2 + a^2} + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = x \operatorname{arccot} \frac x a + a \left({\frac 1 2 \ln \left ({x^2 + a^2}\right)}\right) + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | r = x \operatorname{arccot} \frac x a + \frac a 2 \ln \left({x^2 + a^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \operatorname{arccsc} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \begin{cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arccsc} \dfrac x a < \dfrac \pi 2 \\ \dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \operatorname{arccsc} \dfrac x a < 0 \\ \end{cases}       | c = Derivative of $\operatorname{arccsc} \dfrac x a$ }} {{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}} First let $\operatorname{arccsc} \dfrac x a$ be in the interval $\left({0 \,.\,.\,\dfrac \pi 2}\right)$. Then: {{begin-eqn}} {{eqn | l = \int \operatorname{arccsc} \frac x a \rd x       | r = x \operatorname{arccsc} \frac x a - \int x \left({\frac {-a} {x \sqrt {x^2 - a^2} } }\right) \rd x + C       | c = Integration by Parts }} {{eqn | r = x \operatorname{arccsc} \frac x a + a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = x \operatorname{arccsc} \frac x a + a \ln \left({x + \sqrt {x^2 - a^2} }\right) + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ }} {{end-eqn}} Similarly, let $\operatorname{arccsc} \dfrac x a$ be in the interval $\left({-\dfrac \pi 2 \,.\,.\, 0}\right)$. Then: {{begin-eqn}} {{eqn | l = \int \operatorname{arccsc} \frac x a \rd x       | r = x \operatorname{arccsc} \frac x a - \int x \left({\frac a {x \sqrt {x^2 - a^2} } }\right) \rd x + C       | c = Integration by Parts }} {{eqn | r = x \operatorname{arccsc} \frac x a - a \int \frac {\d x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = x \operatorname{arccsc} \frac x a - a \ln \left({x + \sqrt {x^2 - a^2} }\right) + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = \operatorname{arcsec} \frac x a       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \begin{cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}       | c = Derivative of $\operatorname{arcsec} \dfrac x a$ }} {{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}} First let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({0 \,.\,.\,\dfrac \pi 2}\right)$. Then: {{begin-eqn}} {{eqn | l = \int x \operatorname{arcsec} \frac x a \ \mathrm d x       | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a - \int \frac {x^2} 2 \left({\frac a {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a - \frac a 2 \int \frac {x \ \mathrm d x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a - \frac a 2 \sqrt {x^2 - a^2} + C       | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$ }} {{eqn | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a - \frac {a \sqrt{x^2 - a^2} } 2 + C       | c = simplifying }} {{end-eqn}} Similarly, let $\operatorname{arcsec} \dfrac x a$ be in the interval $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$. Then: {{begin-eqn}} {{eqn | l = \int x \operatorname{arcsec} \frac x a \ \mathrm d x       | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a - \int \frac {x^2} 2 \left({\frac {-a} {x \sqrt {x^2 - a^2} } }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a + \frac a 2 \int \frac {x \ \mathrm d x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a + \frac a 2 \sqrt {x^2 - a^2} + C       | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$ }} {{eqn | r = \frac {x^2} 2 \operatorname{arcsec} \frac x a + \frac {a \sqrt{x^2 - a^2} } 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 = \arccsc \frac x a       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \begin{cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\ \end{cases}       | c = Derivative of $\arccsc \dfrac x a$ }} {{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}} First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$. Then: {{begin-eqn}} {{eqn | l = \int x \arccsc \frac x a \rd x       | r = \frac {x^2} 2 \arccsc \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \arccsc \frac x a + \frac a 2 \int \frac {x \ \mathrm d x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \arccsc \frac x a + \frac a 2 \sqrt {x^2 - a^2} + C       | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$ }} {{eqn | r = \frac {x^2} 2 \arccsc \frac x a + \frac {a \sqrt{x^2 - a^2} } 2 + C       | c = simplifying }} {{end-eqn}} Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$. Then: {{begin-eqn}} {{eqn | l = \int x \arccsc \frac x a \rd x       | r = \frac {x^2} 2 \arccsc \frac x a - \int \frac {x^2} 2 \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \arccsc \frac x a - \frac a 2 \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \arccsc \frac x a - \frac a 2 \sqrt {x^2 - a^2} + C       | c = Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$ }} {{eqn | r = \frac {x^2} 2 \arccsc \frac x a - \frac {a \sqrt{x^2 - a^2} } 2 + C       | c = simplifying }} {{end-eqn}} {{qed}}