Mula sa Wikipedia, ang malayang ensiklopedya
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.
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).
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.
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)
