Caso I: Integrando conteniendo a 2 − x 2 {\displaystyle a^{2}-x^{2}}
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Se hace el cambio de variable x = a sen θ {\displaystyle x=a\operatorname {sen} \theta } identidad trigonométrica sen 2 ( θ ) + cos 2 ( θ ) = 1 {\displaystyle \operatorname {sen} ^{2}(\theta )+\cos ^{2}(\theta )=1}
Construcción geométrica para Caso I {\displaystyle {\text{I}}} Integral Indefinida
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Para calcular la integral
∫ d x a 2 − x 2 {\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}} se puede realizar el cambio de variable
x = a sen θ d x = a cos θ d θ θ = arcsen ( x a ) {\displaystyle {\begin{aligned}x&=a\operatorname {sen} \theta \\dx&=a\cos \theta \;d\theta \\\theta &={\text{arcsen}}\left({\frac {x}{a}}\right)\end{aligned}}} entonces
∫ d x a 2 − x 2 = ∫ a cos θ a 2 − a 2 sen 2 θ d θ = ∫ a cos θ a 2 ( 1 − sen 2 θ ) d θ = ∫ a cos θ a 2 cos 2 θ d θ = ∫ d θ = θ + C = arcsen ( x a ) + C {\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos \theta }{\sqrt {a^{2}-a^{2}\operatorname {sen} ^{2}\theta }}}\;d\theta \\[6pt]&=\int {\frac {a\cos \theta }{\sqrt {a^{2}(1-\operatorname {sen} ^{2}\theta )}}}\;d\theta \\[6pt]&=\int {\frac {a\cos \theta }{\sqrt {a^{2}\cos ^{2}\theta }}}\;d\theta \\[6pt]&=\int d\theta \\[6pt]&=\theta +C\\[6pt]&={\text{arcsen}}\left({\frac {x}{a}}\right)+C\end{aligned}}} Los pasos anteriores requirieron que a > 0 {\displaystyle a>0} cos θ > 0 {\displaystyle \cos \theta >0}
Es posible escoger a {\displaystyle a} a 2 {\displaystyle a^{2}} − π / 2 < θ < π / 2 {\displaystyle -\pi /2<\theta <\pi /2} arco seno .
Para una integral definida, se debe averiguar cómo cambian los límites de la integración. Por ejemplo, cuando x {\displaystyle x} 0 {\displaystyle 0} a / 2 {\displaystyle a/2} sen θ {\displaystyle \operatorname {sen} \theta } 0 {\displaystyle 0} 1 / 2 {\displaystyle 1/2} θ {\displaystyle \theta } 0 {\displaystyle 0} π / 6 {\displaystyle \pi /6}
∫ 0 a / 2 d x a 2 − x 2 = ∫ 0 π / 6 d θ = π 6 . {\displaystyle \int _{0}^{a/2}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\pi /6}d\theta ={\frac {\pi }{6}}.} Se necesita elegir los límites con cuidado. Debido a que la integración anterior requiere que − π / 2 < θ < π / 2 {\displaystyle -\pi /2<\theta <\pi /2} θ {\displaystyle \theta } 0 {\displaystyle 0} π / 6 {\displaystyle \pi /6} θ {\displaystyle \theta } π {\displaystyle \pi } 5 π / 6 {\displaystyle 5\pi /6}
Alternativamente, se deben evaluar completamente las integrales indefinidas antes de aplicar las condiciones de contorno. En ese caso, la antiderivada da
∫ 0 a / 2 d x a 2 − x 2 = arcsen ( x a ) | 0 a / 2 = arcsen ( 1 2 ) − arcsen ( 0 ) = π 6 {\displaystyle {\begin{aligned}\int _{0}^{a/2}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}&=\operatorname {arcsen} \left({\frac {x}{a}}\right){\Biggl |}_{0}^{a/2}\\&=\operatorname {arcsen} \left({\frac {1}{2}}\right)-\operatorname {arcsen} (0)\\&={\frac {\pi }{6}}\end{aligned}}} como antes.
La integral
∫ a 2 − x 2 d x {\displaystyle \int {\sqrt {a^{2}-x^{2}}}\,dx} puede ser evaluada haciendo el cambio de variable
x = a sen θ d x = a cos θ d θ θ = arcsen ( x a ) {\displaystyle {\begin{aligned}x&=a\operatorname {sen} \theta \\dx&=a\cos \theta \;d\theta \\\theta &={\text{arcsen}}\left({\frac {x}{a}}\right)\end{aligned}}} donde a > 0 {\displaystyle a>0} a 2 = a {\displaystyle {\sqrt {a^{2}}}=a}
− π 2 ≤ θ ≤ π 2 {\displaystyle -{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}} porque cos θ ≥ 0 {\displaystyle \cos \theta \geq 0} cos 2 θ = cos θ {\displaystyle {\sqrt {\cos ^{2}\theta }}=\cos \theta }
Luego
∫ a 2 − x 2 d x = ∫ a 2 − a 2 sen 2 θ ( a cos θ ) d θ = ∫ a 2 ( 1 − sen 2 θ ) ( a cos θ ) d θ = ∫ a 2 ( cos 2 θ ) ( a cos θ ) d θ = ∫ ( a cos θ ) ( a cos θ ) d θ = a 2 ∫ cos 2 θ d θ = a 2 ∫ ( 1 + cos 2 θ 2 ) d θ = a 2 2 ( θ + 1 2 sen 2 θ ) + C = a 2 2 ( θ + sen θ cos θ ) + C = a 2 2 ( arcsen ( x a ) + x a 1 − x 2 a 2 ) + C = a 2 2 arcsen ( x a ) + x 2 a 2 − x 2 + C {\displaystyle {\begin{aligned}\int {\sqrt {a^{2}-x^{2}}}\,dx&=\int {\sqrt {a^{2}-a^{2}\operatorname {sen} ^{2}\theta }}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1-\operatorname {sen} ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(\cos ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int (a\cos \theta )(a\cos \theta )\,d\theta \\[6pt]&=a^{2}\int \cos ^{2}\theta \,d\theta \\[6pt]&=a^{2}\int \left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&={\frac {a^{2}}{2}}\left(\theta +{\frac {1}{2}}\operatorname {sen} 2\theta \right)+C\\[6pt]&={\frac {a^{2}}{2}}(\theta +\operatorname {sen} \theta \cos \theta )+C\\[6pt]&={\frac {a^{2}}{2}}\left({\text{arcsen}}\left({\frac {x}{a}}\right)+{\frac {x}{a}}{\sqrt {1-{\frac {x^{2}}{a^{2}}}}}\right)+C\\[6pt]&={\frac {a^{2}}{2}}\;{\text{arcsen}}\left({\frac {x}{a}}\right)+{\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+C\end{aligned}}} Integral Definida
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Para una integral definida, los límites de integración cambian una vez que se realiza la sustitución y estos están determinados por
θ = arcsen ( x a ) {\displaystyle \theta ={\text{arcsen}}\left({\frac {x}{a}}\right)} con valores para θ {\displaystyle \theta }
− π 2 ≤ θ ≤ π 2 {\displaystyle -{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}} Considérese la integral definida
∫ − 1 1 4 − x 2 d x {\displaystyle \int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx} que puede ser evaluada haciendo el cambio de variable
x = 2 sen θ d x = 2 cos θ d θ {\displaystyle {\begin{aligned}x&=2\operatorname {sen} \theta \\dx&=2\cos \theta \,d\theta \end{aligned}}} y en este caso, los límites de integración estarán determinados por
θ = arcsen ( x 2 ) {\displaystyle \theta ={\text{arcsen}}\left({\frac {x}{2}}\right)} Tenemos que
si x = − 1 {\displaystyle x=-1} θ = arcsen ( − 1 2 ) = − π 6 {\displaystyle \theta ={\text{arcsen}}\left(-{\frac {1}{2}}\right)=-{\frac {\pi }{6}}}
y si x = 1 {\displaystyle x=1} θ = arcsen ( 1 2 ) = π 6 {\displaystyle \theta ={\text{arcsen}}\left({\frac {1}{2}}\right)={\frac {\pi }{6}}}
entonces
∫ − 1 1 4 − x 2 d x = ∫ − π / 6 π / 6 4 − 4 sen 2 θ ( 2 cos θ ) d θ = ∫ − π / 6 π / 6 4 ( 1 − sen 2 θ ) ( 2 cos θ ) d θ = ∫ − π / 6 π / 6 4 ( cos 2 θ ) ( 2 cos θ ) d θ = ∫ − π / 6 π / 6 ( 2 cos θ ) ( 2 cos θ ) d θ = 4 ∫ − π / 6 π / 6 cos 2 θ d θ = 4 ∫ − π / 6 π / 6 ( 1 + cos 2 θ 2 ) d θ = 2 [ θ + 1 2 sen 2 θ ] − π / 6 π / 6 = [ 2 θ + sen 2 θ ] | − π / 6 π / 6 = ( π 3 + sen π 3 ) − ( − π 3 + sen ( − π 3 ) ) = 2 π 3 + 3 {\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\int _{-\pi /6}^{\pi /6}{\sqrt {4-4\operatorname {sen} ^{2}\theta }}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(1-\operatorname {sen} ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(\cos ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}(2\cos \theta )(2\cos \theta )\,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\cos ^{2}\theta \,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&=2\left[\theta +{\frac {1}{2}}\operatorname {sen} 2\theta \right]_{-\pi /6}^{\pi /6}\\[6pt]&=[2\theta +\operatorname {sen} 2\theta ]{\Biggl |}_{-\pi /6}^{\pi /6}\\[6pt]&=\left({\frac {\pi }{3}}+\operatorname {sen} {\frac {\pi }{3}}\right)-\left(-{\frac {\pi }{3}}+\operatorname {sen} \left(-{\frac {\pi }{3}}\right)\right)\\[6pt]&={\frac {2\pi }{3}}+{\sqrt {3}}\end{aligned}}} Por otro lado, si aplicamos directamente los límites de integración para la fórmula de la antiderivada obtenemos
∫ − 1 1 4 − x 2 d x = [ 2 arcsen ( x 2 ) + x 2 4 − x 2 ] − 1 1 = ( 2 arcsen ( 1 2 ) + 1 2 3 ) − ( 2 arcsen ( − 1 2 ) − 1 2 3 ) = ( 2 ⋅ π 6 + 3 2 ) − ( 2 ⋅ ( − π 6 ) − 3 2 ) = 2 π 3 + 3 {\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\left[2\;{\text{arcsen}}\left({\frac {x}{2}}\right)+{\frac {x}{2}}{\sqrt {4-x^{2}}}\right]_{-1}^{1}\\[6pt]&=\left(2\;{\text{arcsen}}\left({\frac {1}{2}}\right)+{\frac {1}{2}}{\sqrt {3}}\right)-\left(2\;{\text{arcsen}}\left(-{\frac {1}{2}}\right)-{\frac {1}{2}}{\sqrt {3}}\right)\\[6pt]&=\left(2\cdot {\frac {\pi }{6}}+{\frac {\sqrt {3}}{2}}\right)-\left(2\cdot \left(-{\frac {\pi }{6}}\right)-{\frac {\sqrt {3}}{2}}\right)\\[6pt]&={\frac {2\pi }{3}}+{\sqrt {3}}\end{aligned}}} Caso II: Integrando conteniendo a 2 + x 2 {\displaystyle a^{2}+x^{2}}
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Se hace el cambio de variable x = a tan θ {\displaystyle x=a\tan \theta } identidad trigonométrica sec 2 ( θ ) − tan 2 ( θ ) = 1 {\displaystyle \sec ^{2}(\theta )-\tan ^{2}(\theta )=1}
Integral Indefinida
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Construcción geométrica para Caso II {\displaystyle {\text{II}}} En la integral
∫ d x a 2 + x 2 {\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}} hacemos el cambio de variable
x = a tan θ d x = a sec 2 θ d θ θ = arctan ( x a ) {\displaystyle {\begin{aligned}x&=a\tan \theta \\dx&=a\sec ^{2}\theta \;d\theta \\\theta &=\arctan \left({\frac {x}{a}}\right)\end{aligned}}} de modo que la integral se convierte en
∫ d x a 2 + x 2 = ∫ a sec 2 θ a 2 + a 2 tan 2 θ d θ = ∫ a sec 2 θ a 2 ( 1 + tan 2 θ ) d θ = ∫ a sec 2 θ a 2 sec 2 θ d θ = 1 a ∫ d θ = θ a + C = 1 a arctan ( x a ) + C {\displaystyle {\begin{aligned}\int {\frac {dx}{a^{2}+x^{2}}}&=\int {\frac {a\sec ^{2}\theta }{a^{2}+a^{2}\tan ^{2}\theta }}\,d\theta \\[6pt]&=\int {\frac {a\sec ^{2}\theta }{a^{2}(1+\tan ^{2}\theta )}}\;d\theta \\[6pt]&=\int {\frac {a\sec ^{2}\theta }{a^{2}\sec ^{2}\theta }}\,d\theta \\[6pt]&={\frac {1}{a}}\int d\theta \\[6pt]&={\frac {\theta }{a}}+C\\[6pt]&={\frac {1}{a}}\arctan \left({\frac {x}{a}}\right)+C\end{aligned}}} para a ≠ 0 {\displaystyle a\neq 0}
La integral
∫ a 2 + x 2 d x {\displaystyle \int {\sqrt {a^{2}+x^{2}}}\,{dx}} puede ser evaluada haciendo el cambio de variable
x = a tan θ d x = a sec 2 θ d θ θ = arctan ( x a ) {\displaystyle {\begin{aligned}x&=a\tan \theta \\dx&=a\sec ^{2}\theta \,d\theta \\\theta &=\arctan \left({\frac {x}{a}}\right)\end{aligned}}} donde a > 0 {\displaystyle a>0} a 2 = a {\displaystyle {\sqrt {a^{2}}}=a}
− π 2 < θ < π 2 {\displaystyle -{\frac {\pi }{2}}<\theta <{\frac {\pi }{2}}} por lo que sec θ > 0 {\displaystyle \sec \theta >0} sec 2 θ = sec θ {\displaystyle {\sqrt {\sec ^{2}\theta }}=\sec \theta }
Entonces
∫ a 2 + x 2 d x = ∫ a 2 + a 2 tan 2 θ ( a sec 2 θ ) d θ = ∫ a 2 ( 1 + tan 2 θ ) ( a sec 2 θ ) d θ = ∫ a 2 sec 2 θ ( a sec 2 θ ) d θ = ∫ ( a sec θ ) ( a sec 2 θ ) d θ = a 2 ∫ sec 3 θ d θ . {\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&=\int {\sqrt {a^{2}+a^{2}\tan ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1+\tan ^{2}\theta )}}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}\sec ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int (a\sec \theta )(a\sec ^{2}\theta )\,d\theta \\[6pt]&=a^{2}\int \sec ^{3}\theta \,d\theta .\\[6pt]\end{aligned}}} La integral de la secante cúbica puede ser evaluada utilizando integración por partes , dando como resultado
∫ a 2 + x 2 d x = a 2 2 ( sec θ tan θ + ln | sec θ + tan θ | ) + C = a 2 2 ( 1 + x 2 a 2 ⋅ x a + ln | 1 + x 2 a 2 + x a | ) + C = 1 2 ( x a 2 + x 2 + a 2 ln | x + a 2 + x 2 | ) + C . {\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\sqrt {1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {x}{a}}+\ln \left|{\sqrt {1+{\frac {x^{2}}{a^{2}}}}}+{\frac {x}{a}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {a^{2}+x^{2}}}+a^{2}\ln \left|x+{\sqrt {a^{2}+x^{2}}}\right|\right)+C.\end{aligned}}} Integral Definida
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Para una integral definida, los límites de integración cambian una vez que se hace la sustitución y estos están determinados por
θ = arctan ( x a ) {\displaystyle \theta =\arctan \left({\frac {x}{a}}\right)} con valores para θ {\displaystyle \theta }
− π 2 < θ < π 2 {\displaystyle -{\frac {\pi }{2}}<\theta <{\frac {\pi }{2}}} Considérese la integral definida
∫ 0 1 4 1 + x 2 d x {\displaystyle \int _{0}^{1}{\frac {4}{1+x^{2}}}\,dx} esta puede ser evaluada haciendo el cambio de variable
x = tan θ d x = sec 2 θ d θ {\displaystyle {\begin{aligned}x&=\tan \theta \\dx&=\sec ^{2}\theta \,d\theta \end{aligned}}}
con los límites de integración determinados por θ = arctan x {\displaystyle \theta =\arctan x}
Tenemos que
si x = 0 {\displaystyle x=0} θ = arctan ( 0 ) = 0 {\displaystyle \theta =\arctan(0)=0}
y si x = 1 {\displaystyle x=1} θ = arctan ( 1 ) = π 4 {\displaystyle \theta =\arctan(1)={\frac {\pi }{4}}}
de modo que
∫ 0 1 4 1 + x 2 d x = 4 ∫ 0 π / 4 sec 2 θ 1 + tan 2 θ d θ = 4 ∫ 0 π / 4 sec 2 θ sec 2 θ d θ = 4 ∫ 0 π / 4 d θ = 4 θ | 0 π / 4 = 4 ( π 4 ) = π {\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {4}{1+x^{2}}}\;dx&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta }{1+\tan ^{2}\theta }}\;d\theta \\[6pt]&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta }{\sec ^{2}\theta }}\;d\theta \\[6pt]&=4\int _{0}^{\pi /4}d\theta \\[6pt]&=4\theta {\Bigg |}_{0}^{\pi /4}\\&=4\left({\frac {\pi }{4}}\right)\\&=\pi \end{aligned}}} Caso III: Integrando conteniendo x 2 − a 2 {\displaystyle x^{2}-a^{2}}
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Se hace el cambio de variable x = a sec θ {\displaystyle x=a\sec \theta } identidad trigonométrica sec 2 ( θ ) − tan 2 ( θ ) = 1 {\displaystyle \sec ^{2}(\theta )-\tan ^{2}(\theta )=1}
Integral Indefinida
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Construcción geométrica para Caso III {\displaystyle {\text{III}}} La integral
∫ d x x 2 − a 2 {\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}} también puede ser evaluada utilizando fracciones parciales en lugar de utilizar sustitución trigonométrica. Sin embargo, la integral
∫ x 2 − a 2 d x {\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx} no. En este caso, una sustitución apropiada es
x = a sec θ d x = a sec θ tan θ d θ θ = arcsec ( x a ) {\displaystyle {\begin{aligned}x&=a\sec \theta \\dx&=a\sec \theta \tan \theta \,d\theta \\\theta &=\operatorname {arcsec} \left({\frac {x}{a}}\right)\end{aligned}}} donde a > 0 {\displaystyle a>0} a 2 = a {\displaystyle {\sqrt {a^{2}}}=a}
0 ≤ θ < π 2 {\displaystyle 0\leq \theta <{\frac {\pi }{2}}} suponiendo que x > 0 {\displaystyle x>0} tan θ ≥ 0 {\displaystyle \tan \theta \geq 0} tan 2 θ = tan θ {\displaystyle {\sqrt {\tan ^{2}\theta }}=\tan \theta }
Entonces,
∫ x 2 − a 2 d x = ∫ a 2 sec 2 θ − a 2 ⋅ a sec θ tan θ d θ = ∫ a 2 ( sec 2 θ − 1 ) ⋅ a sec θ tan θ d θ = ∫ a 2 tan 2 θ ⋅ a sec θ tan θ d θ = ∫ a 2 sec θ tan 2 θ d θ = a 2 ∫ ( sec θ ) ( sec 2 θ − 1 ) d θ = a 2 ∫ ( sec 3 θ − sec θ ) d θ {\displaystyle {\begin{aligned}\int {\sqrt {x^{2}-a^{2}}}\,dx&=\int {\sqrt {a^{2}\sec ^{2}\theta -a^{2}}}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int {\sqrt {a^{2}(\sec ^{2}\theta -1)}}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int {\sqrt {a^{2}\tan ^{2}\theta }}\cdot a\sec \theta \tan \theta \,d\theta \\&=\int a^{2}\sec \theta \tan ^{2}\theta \,d\theta \\&=a^{2}\int (\sec \theta )(\sec ^{2}\theta -1)\,d\theta \\&=a^{2}\int (\sec ^{3}\theta -\sec \theta )\,d\theta \end{aligned}}} Uno puede evaluar la integral de la función secante multiplicando tanto el numerador como el denominador por ( sec θ + tan θ ) {\displaystyle (\sec \theta +\tan \theta )} [ 3]
∫ x 2 − a 2 d x = a 2 2 ( sec θ tan θ + ln | sec θ + tan θ | ) − a 2 ln | sec θ + tan θ | + C = a 2 2 ( sec θ tan θ − ln | sec θ + tan θ | ) + C = a 2 2 ( x a ⋅ x 2 a 2 − 1 − ln | x a + x 2 a 2 − 1 | ) + C = 1 2 ( x x 2 − a 2 − a 2 ln | x + x 2 − a 2 a | ) + C . {\displaystyle {\begin{aligned}\int {\sqrt {x^{2}-a^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)-a^{2}\ln |\sec \theta +\tan \theta |+C\\[6pt]&={\frac {a^{2}}{2}}(\sec \theta \tan \theta -\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\frac {x}{a}}\cdot {\sqrt {{\frac {x^{2}}{a^{2}}}-1}}-\ln \left|{\frac {x}{a}}+{\sqrt {{\frac {x^{2}}{a^{2}}}-1}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {x^{2}-a^{2}}}-a^{2}\ln \left|{\frac {x+{\sqrt {x^{2}-a^{2}}}}{a}}\right|\right)+C.\end{aligned}}} Sustituciones que eliminan funciones trigonométricas
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La sustitución de una nueva variable por una función trigonométrica en ocasiones puede ser usada para facilitar el cálculo de la integral, dejando el integrando sin funciones trigonométricas.
∫ f ( sen ( x ) , cos ( x ) ) d x = ∫ 1 ± 1 − u 2 f ( u , ± 1 − u 2 ) d u u = sen ( x ) ∫ f ( sen ( x ) , cos ( x ) ) d x = ∫ 1 ∓ 1 − u 2 f ( ± 1 − u 2 , u ) d u u = cos ( x ) ∫ f ( sen ( x ) , cos ( x ) ) d x = ∫ 2 1 + u 2 f ( 2 u 1 + u 2 , 1 − u 2 1 + u 2 ) d u u = tan ( x 2 ) {\displaystyle {\begin{aligned}\int f(\operatorname {sen}(x),\cos(x))\,dx&=\int {\frac {1}{\pm {\sqrt {1-u^{2}}}}}\;f\left(u,\pm {\sqrt {1-u^{2}}}\right)\,du&&u=\operatorname {sen}(x)\\[6pt]\int f(\operatorname {sen}(x),\cos(x))\,dx&=\int {\frac {1}{\mp {\sqrt {1-u^{2}}}}}\;f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du&&u=\cos(x)\\[6pt]\int f(\operatorname {sen}(x),\cos(x))\,dx&=\int {\frac {2}{1+u^{2}}}\;f\left({\frac {2u}{1+u^{2}}},{\frac {1-u^{2}}{1+u^{2}}}\right)\,du&&u=\tan \left({\tfrac {x}{2}}\right)\\[6pt]\end{aligned}}} La última sustitución es conocida como la Sustitución de Weierstrass , que hace uso de las fórmulas de la tangente del ángulo mitad .
Considérese la integral
∫ 4 cos x ( 1 + cos x ) 3 d x {\displaystyle \int {\frac {4\cos x}{(1+\cos x)^{3}}}\;dx} Si utilizamos la sustitución de Weierstrass entonces
∫ 4 cos x ( 1 + cos x ) 3 d x = ∫ 2 1 + u 2 4 ( 1 − u 2 1 + u 2 ) ( 1 + 1 − u 2 1 + u 2 ) 3 d u = ∫ ( 1 − u 2 ) ( 1 + u 2 ) d u = ∫ ( 1 − u 4 ) d u = u − u 5 5 + C = tan ( x 2 ) − 1 5 tan 5 ( x 2 ) + C {\displaystyle {\begin{aligned}\int {\frac {4\cos x}{(1+\cos x)^{3}}}\,dx&=\int {\frac {2}{1+u^{2}}}{\frac {4\left({\frac {1-u^{2}}{1+u^{2}}}\right)}{\left(1+{\frac {1-u^{2}}{1+u^{2}}}\right)^{3}}}\,du\\&=\int (1-u^{2})(1+u^{2})\,du\\&=\int (1-u^{4})\,du\\&=u-{\frac {u^{5}}{5}}+C\\&=\tan \left({\frac {x}{2}}\right)-{\frac {1}{5}}\tan ^{5}\left({\frac {x}{2}}\right)+C\end{aligned}}} Sustitución hiperbólica
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También se pueden utilizar sustituciones mediante funciones hiperbólicas para simplificar determinadas integrales.[ 4]
Por ejemplo, en la integral
∫ 1 a 2 + x 2 d x {\displaystyle \int {\frac {1}{\sqrt {a^{2}+x^{2}}}}\,dx}
se realiza la sustitución x = a sinh u {\displaystyle x=a\sinh {u}} d x = a cosh u d u . {\displaystyle dx=a\cosh u\,du.}
Entonces, usando las identidades cosh 2 ( x ) − sinh 2 ( x ) = 1 {\displaystyle \cosh ^{2}(x)-\sinh ^{2}(x)=1} sinh − 1 x = ln ( x + x 2 + 1 ) , {\displaystyle \sinh ^{-1}{x}=\ln(x+{\sqrt {x^{2}+1}}),}
∫ 1 a 2 + x 2 d x = ∫ a cosh u a 2 + a 2 sinh 2 u d u = ∫ a cosh u a 1 + sinh 2 u d u = ∫ a cosh u a cosh u d u = u + C = sinh − 1 x a + C = ln ( x 2 a 2 + 1 + x a ) + C = ln ( x 2 + a 2 + x a ) + C {\displaystyle {\begin{aligned}\int {\frac {1}{\sqrt {a^{2}+x^{2}}}}\,dx&=\int {\frac {a\cosh u}{\sqrt {a^{2}+a^{2}\sinh ^{2}u}}}\,du\\[6pt]&=\int {\frac {a\cosh {u}}{a{\sqrt {1+\sinh ^{2}{u}}}}}\,du\\[6pt]&=\int {\frac {a\cosh {u}}{a\cosh u}}\,du\\[6pt]&=u+C\\[6pt]&=\sinh ^{-1}{\frac {x}{a}}+C\\[6pt]&=\ln \left({\sqrt {{\frac {x^{2}}{a^{2}}}+1}}+{\frac {x}{a}}\right)+C\\[6pt]&=\ln \left({\frac {{\sqrt {x^{2}+a^{2}}}+x}{a}}\right)+C\end{aligned}}}
Véase también
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![{\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos \theta }{\sqrt {a^{2}-a^{2}\operatorname {sen} ^{2}\theta }}}\;d\theta \\[6pt]&=\int {\frac {a\cos \theta }{\sqrt {a^{2}(1-\operatorname {sen} ^{2}\theta )}}}\;d\theta \\[6pt]&=\int {\frac {a\cos \theta }{\sqrt {a^{2}\cos ^{2}\theta }}}\;d\theta \\[6pt]&=\int d\theta \\[6pt]&=\theta +C\\[6pt]&={\text{arcsen}}\left({\frac {x}{a}}\right)+C\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c33cee5a81fe5b1fc55154b51e4ebb497aa176b)
![{\displaystyle {\begin{aligned}\int {\sqrt {a^{2}-x^{2}}}\,dx&=\int {\sqrt {a^{2}-a^{2}\operatorname {sen} ^{2}\theta }}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1-\operatorname {sen} ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(\cos ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int (a\cos \theta )(a\cos \theta )\,d\theta \\[6pt]&=a^{2}\int \cos ^{2}\theta \,d\theta \\[6pt]&=a^{2}\int \left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&={\frac {a^{2}}{2}}\left(\theta +{\frac {1}{2}}\operatorname {sen} 2\theta \right)+C\\[6pt]&={\frac {a^{2}}{2}}(\theta +\operatorname {sen} \theta \cos \theta )+C\\[6pt]&={\frac {a^{2}}{2}}\left({\text{arcsen}}\left({\frac {x}{a}}\right)+{\frac {x}{a}}{\sqrt {1-{\frac {x^{2}}{a^{2}}}}}\right)+C\\[6pt]&={\frac {a^{2}}{2}}\;{\text{arcsen}}\left({\frac {x}{a}}\right)+{\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+C\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1d4616bf2e80282227306399174c44ba7787196)
![{\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\int _{-\pi /6}^{\pi /6}{\sqrt {4-4\operatorname {sen} ^{2}\theta }}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(1-\operatorname {sen} ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(\cos ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}(2\cos \theta )(2\cos \theta )\,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\cos ^{2}\theta \,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&=2\left[\theta +{\frac {1}{2}}\operatorname {sen} 2\theta \right]_{-\pi /6}^{\pi /6}\\[6pt]&=[2\theta +\operatorname {sen} 2\theta ]{\Biggl |}_{-\pi /6}^{\pi /6}\\[6pt]&=\left({\frac {\pi }{3}}+\operatorname {sen} {\frac {\pi }{3}}\right)-\left(-{\frac {\pi }{3}}+\operatorname {sen} \left(-{\frac {\pi }{3}}\right)\right)\\[6pt]&={\frac {2\pi }{3}}+{\sqrt {3}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdb3b420d72073bb0cc67483388d2dd9e7d9b1c4)
![{\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\left[2\;{\text{arcsen}}\left({\frac {x}{2}}\right)+{\frac {x}{2}}{\sqrt {4-x^{2}}}\right]_{-1}^{1}\\[6pt]&=\left(2\;{\text{arcsen}}\left({\frac {1}{2}}\right)+{\frac {1}{2}}{\sqrt {3}}\right)-\left(2\;{\text{arcsen}}\left(-{\frac {1}{2}}\right)-{\frac {1}{2}}{\sqrt {3}}\right)\\[6pt]&=\left(2\cdot {\frac {\pi }{6}}+{\frac {\sqrt {3}}{2}}\right)-\left(2\cdot \left(-{\frac {\pi }{6}}\right)-{\frac {\sqrt {3}}{2}}\right)\\[6pt]&={\frac {2\pi }{3}}+{\sqrt {3}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a30c5d7cd0d8b39715021f705e76be11dcfa5e1)
![{\displaystyle {\begin{aligned}\int {\frac {dx}{a^{2}+x^{2}}}&=\int {\frac {a\sec ^{2}\theta }{a^{2}+a^{2}\tan ^{2}\theta }}\,d\theta \\[6pt]&=\int {\frac {a\sec ^{2}\theta }{a^{2}(1+\tan ^{2}\theta )}}\;d\theta \\[6pt]&=\int {\frac {a\sec ^{2}\theta }{a^{2}\sec ^{2}\theta }}\,d\theta \\[6pt]&={\frac {1}{a}}\int d\theta \\[6pt]&={\frac {\theta }{a}}+C\\[6pt]&={\frac {1}{a}}\arctan \left({\frac {x}{a}}\right)+C\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5b087324f1c492895eac14832e631d2478e8fff)
![{\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&=\int {\sqrt {a^{2}+a^{2}\tan ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1+\tan ^{2}\theta )}}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}\sec ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int (a\sec \theta )(a\sec ^{2}\theta )\,d\theta \\[6pt]&=a^{2}\int \sec ^{3}\theta \,d\theta .\\[6pt]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/108a5f1becea83b5cb41021d81544ff3e1bab889)
![{\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\sqrt {1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {x}{a}}+\ln \left|{\sqrt {1+{\frac {x^{2}}{a^{2}}}}}+{\frac {x}{a}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {a^{2}+x^{2}}}+a^{2}\ln \left|x+{\sqrt {a^{2}+x^{2}}}\right|\right)+C.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b72bbed5d6bbb2d3dff666fda9985029351c25a)
![{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {4}{1+x^{2}}}\;dx&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta }{1+\tan ^{2}\theta }}\;d\theta \\[6pt]&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta }{\sec ^{2}\theta }}\;d\theta \\[6pt]&=4\int _{0}^{\pi /4}d\theta \\[6pt]&=4\theta {\Bigg |}_{0}^{\pi /4}\\&=4\left({\frac {\pi }{4}}\right)\\&=\pi \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09938a7ac146b25472d7a851dd9a7d68b57de4df)
![{\displaystyle {\begin{aligned}\int {\sqrt {x^{2}-a^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)-a^{2}\ln |\sec \theta +\tan \theta |+C\\[6pt]&={\frac {a^{2}}{2}}(\sec \theta \tan \theta -\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\frac {x}{a}}\cdot {\sqrt {{\frac {x^{2}}{a^{2}}}-1}}-\ln \left|{\frac {x}{a}}+{\sqrt {{\frac {x^{2}}{a^{2}}}-1}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {x^{2}-a^{2}}}-a^{2}\ln \left|{\frac {x+{\sqrt {x^{2}-a^{2}}}}{a}}\right|\right)+C.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d551bea9f1a33df981d45ab8cf11a1443d6da85)
![{\displaystyle {\begin{aligned}\int f(\operatorname {sen}(x),\cos(x))\,dx&=\int {\frac {1}{\pm {\sqrt {1-u^{2}}}}}\;f\left(u,\pm {\sqrt {1-u^{2}}}\right)\,du&&u=\operatorname {sen}(x)\\[6pt]\int f(\operatorname {sen}(x),\cos(x))\,dx&=\int {\frac {1}{\mp {\sqrt {1-u^{2}}}}}\;f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du&&u=\cos(x)\\[6pt]\int f(\operatorname {sen}(x),\cos(x))\,dx&=\int {\frac {2}{1+u^{2}}}\;f\left({\frac {2u}{1+u^{2}}},{\frac {1-u^{2}}{1+u^{2}}}\right)\,du&&u=\tan \left({\tfrac {x}{2}}\right)\\[6pt]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4b07688dbf8f93a45c673360d666c378ecf2f09)
![{\displaystyle {\begin{aligned}\int {\frac {1}{\sqrt {a^{2}+x^{2}}}}\,dx&=\int {\frac {a\cosh u}{\sqrt {a^{2}+a^{2}\sinh ^{2}u}}}\,du\\[6pt]&=\int {\frac {a\cosh {u}}{a{\sqrt {1+\sinh ^{2}{u}}}}}\,du\\[6pt]&=\int {\frac {a\cosh {u}}{a\cosh u}}\,du\\[6pt]&=u+C\\[6pt]&=\sinh ^{-1}{\frac {x}{a}}+C\\[6pt]&=\ln \left({\sqrt {{\frac {x^{2}}{a^{2}}}+1}}+{\frac {x}{a}}\right)+C\\[6pt]&=\ln \left({\frac {{\sqrt {x^{2}+a^{2}}}+x}{a}}\right)+C\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4de72234865476739112fe15f4849d934ebb1622)
Datos: Q1191963