Integral na naglalaman ng a 2 − x 2
baguhin
Sa integral na:
∫
d
x
a
2
−
x
2
{\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}}
maaari nating gamitin ang:
x
=
a
sin
(
θ
)
,
d
x
=
a
cos
(
θ
)
d
θ
,
θ
=
arcsin
(
x
a
)
{\displaystyle x=a\sin(\theta ),\quad dx=a\cos(\theta )\,d\theta ,\quad \theta =\arcsin \left({\frac {x}{a}}\right)}
Ang integral ay naging:
∫
d
x
a
2
−
x
2
=
∫
a
cos
(
θ
)
d
θ
a
2
−
a
2
sin
2
(
θ
)
=
∫
a
cos
(
θ
)
d
θ
a
2
(
1
−
sin
2
(
θ
)
)
=
∫
a
cos
(
θ
)
d
θ
a
2
cos
2
(
θ
)
=
∫
d
θ
=
θ
+
C
=
arcsin
(
x
a
)
+
C
{\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}-a^{2}\sin ^{2}(\theta )}}}=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}(1-\sin ^{2}(\theta ))}}}\\[8pt]&=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}\cos ^{2}(\theta )}}}=\int d\theta =\theta +C=\arcsin \left({\frac {x}{a}}\right)+C\end{aligned}}}
Dapat tandaan na ang hakbang ay nangangailangang ang a > 0 at ang cos(θ ) > 0; maaaring gamitin na ang a ay maging positibong ugat ng kwadrado ng a 2 ; at magtakda ng restriksiyon sa θ na maging −π/2 < θ < π/2 gamit ang arcsin na punsiyon.
Para sa depinitong integral, dapat alamin kung paano ang mga hangganan ng integrasyon ay nagbabago. Halimbawa, habang ang x ay patungo mula 0 hanggang a /2, ang sin(θ) ay patungo mula 0 hanggang 1/2, kaya ang θ ay patungo mula 0 hanggang π/6. Ngayon meron tayong:
∫
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}}.}
Kailangan ng pag-iingat sa pagpili ng mga hangganan. Ang integrasyon sa itaas ay nangangailangan na ang −π/2 < θ < π/2, kaya ang θ na patungo mula 0 hanggang π/6 ang tanging mapapagpipilian. Kung makakaligtaan ang restriksiyong ito, baka mapili ang θ na patungo mula π hanggang 5π/6, na magreresulta sa negatibong resulta.
Integral na naglalaman ng a 2 + x 2
baguhin
Sa integral na:
∫
d
x
a
2
+
x
2
{\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}}
Maaari itong isulat na:
x
=
a
tan
(
θ
)
,
d
x
=
a
sec
2
(
θ
)
d
θ
{\displaystyle x=a\tan(\theta ),\quad dx=a\sec ^{2}(\theta )\,d\theta }
θ
=
arctan
(
x
a
)
{\displaystyle \theta =\arctan \left({\frac {x}{a}}\right)}
Ang integral ay naging:
∫
d
x
a
2
+
x
2
=
∫
a
sec
2
(
θ
)
d
θ
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
)
=
∫
d
θ
a
=
θ
a
+
C
=
1
a
arctan
(
x
a
)
+
C
{\displaystyle {\begin{aligned}&{}\qquad \int {\frac {dx}{a^{2}+x^{2}}}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}+a^{2}\tan ^{2}(\theta )}}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}(1+\tan ^{2}(\theta ))}}\\[8pt]&{}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}\sec ^{2}(\theta )}}=\int {\frac {d\theta }{a}}={\frac {\theta }{a}}+C={\frac {1}{a}}\arctan \left({\frac {x}{a}}\right)+C\end{aligned}}}
(provided a ≠ 0).
Integral na naglalaman ng x 2 − a 2
baguhin
Ang integral tulad ng:
∫
d
x
x
2
−
a
2
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}}
ay dapat hanapin gamit ang parsiyal na praksiyon imbis na ang trigonometrikong substitusyon. Ngunit ang integral na:
∫
x
2
−
a
2
d
x
{\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx}
ay maaaring hanapin gamit ang trigonometrikong substitusyon:
x
=
a
sec
(
θ
)
,
d
x
=
a
sec
(
θ
)
tan
(
θ
)
d
θ
{\displaystyle x=a\sec(\theta ),\quad dx=a\sec(\theta )\tan(\theta )\,d\theta }
θ
=
arcsec
(
x
a
)
{\displaystyle \theta =\operatorname {arcsec} \left({\frac {x}{a}}\right)}
∫
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}&{}\qquad \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}}}
Ngayon, pwede na itong lutasin gamit ang pormula para sa integral ng sekant na kinubiko.
Mga substitusyon na nag-aalis ng trigonometrikong punsiyon
baguhin
Ang isang substitusyon ay maaaring gamitin upang alisin ang mga trigonometrikong punsiyon. Sa partikular, ang ginagamit ay Substitusyong Weierstrass .
Halimbawa:
∫
f
(
sin
x
,
cos
x
)
d
x
=
∫
1
±
1
−
u
2
f
(
u
,
±
1
−
u
2
)
d
u
,
u
=
sin
x
{\displaystyle \int f(\sin x,\cos x)\,dx=\int {\frac {1}{\pm {\sqrt {1-u^{2}}}}}f\left(u,\pm {\sqrt {1-u^{2}}}\right)\,du,\qquad \qquad u=\sin x}
∫
f
(
sin
x
,
cos
x
)
d
x
=
∫
−
1
±
1
−
u
2
f
(
±
1
−
u
2
,
u
)
d
u
u
=
cos
x
{\displaystyle \int f(\sin x,\cos x)\,dx=\int {\frac {-1}{\pm {\sqrt {1-u^{2}}}}}f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du\qquad \qquad u=\cos x}
(dapat maging maingat sa mga senyas/sign)
∫
f
(
sin
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 \int f(\sin 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\qquad \qquad u=\tan {\frac {x}{2}}}
∫
cos
x
(
1
+
cos
x
)
3
d
x
=
∫
2
1
+
u
2
1
−
u
2
1
+
u
2
(
1
+
1
−
u
2
1
+
u
2
)
3
d
u
=
∫
1
−
u
2
1
+
u
2
d
u
{\displaystyle \int {\frac {\cos x}{(1+\cos x)^{3}}}\,dx=\int {\frac {2}{1+u^{2}}}{\frac {\frac {1-u^{2}}{1+u^{2}}}{\left(1+{\frac {1-u^{2}}{1+u^{2}}}\right)^{3}}}\,du=\int {\frac {1-u^{2}}{1+u^{2}}}\,du}
![{\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}-a^{2}\sin ^{2}(\theta )}}}=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}(1-\sin ^{2}(\theta ))}}}\\[8pt]&=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}\cos ^{2}(\theta )}}}=\int d\theta =\theta +C=\arcsin \left({\frac {x}{a}}\right)+C\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d8b8f907b9d9358f8ec777fa6cbdb8e02d4d234)
![{\displaystyle {\begin{aligned}&{}\qquad \int {\frac {dx}{a^{2}+x^{2}}}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}+a^{2}\tan ^{2}(\theta )}}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}(1+\tan ^{2}(\theta ))}}\\[8pt]&{}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}\sec ^{2}(\theta )}}=\int {\frac {d\theta }{a}}={\frac {\theta }{a}}+C={\frac {1}{a}}\arctan \left({\frac {x}{a}}\right)+C\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97e2ad790410647fbb324a0105185d5610acc5e8)
