Integral dapat dianggap sebagai perhitungan luas daerah di bawah kurva ƒ (x ), antara dua titik a dan b . Integral merupakan suatu objek matematika yang dapat diinterpretasikan sebagai luas wilayah ataupun generalisasi suatu wilayah.
Integral Tak Tentu
Integral tak tentu mempunyai rumus umum:
∫
F
(
x
)
d
x
=
F
(
x
)
+
c
{\displaystyle \int F(x)dx=F(x)+c}
Keterangan:
Pengintegralan standar
Jika
f
(
x
)
=
a
{\displaystyle f(x)=a}
∫
a
d
x
=
a
x
+
c
{\displaystyle \int a\operatorname {d} x=ax+c}
Jika
f
(
x
)
=
a
x
n
{\displaystyle f(x)=ax^{n}}
∫
a
x
n
d
x
=
(
a
x
n
+
1
n
+
1
)
+
c
{\displaystyle \int ax^{n}\operatorname {d} x=({\frac {ax^{n+1}}{n+1}})+c}
Jika
f
(
x
)
=
(
a
x
+
b
)
n
{\displaystyle f(x)=(ax+b)^{n}}
∫
(
a
x
+
b
)
n
d
x
=
(
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
)
+
c
{\displaystyle \int (ax+b)^{n}\operatorname {d} x=({\frac {(ax+b)^{n+1}}{a(n+1)}})+c}
Pengintegralan khusus
∫
1
x
d
x
=
ln
|
x
|
+
k
{\displaystyle \int {\frac {1}{x}}\ dx=\ln |x|+k}
∫
f
′
(
x
)
f
(
x
)
d
x
=
ln
f
(
x
)
+
k
{\displaystyle \int {\frac {f'(x)}{f(x)}}dx=\ln f(x)+k}
∫
1
x
d
x
=
ln
|
x
|
+
k
{\displaystyle \int {\frac {1}{x}}\ dx=\ln |x|+k}
Sifat-sifat
∫
a
f
(
x
)
d
x
=
a
∫
f
(
x
)
d
x
+
k
{\displaystyle \int af(x)\operatorname {d} x=a\int f(x)\operatorname {d} x+k}
∫
(
f
(
x
)
±
g
(
x
)
)
d
x
=
∫
f
(
x
)
d
x
±
∫
g
(
x
)
d
x
{\displaystyle \int (f(x)\pm g(x))\operatorname {d} x=\int f(x)\operatorname {d} x\pm \int g(x)\operatorname {d} x}
Integral Tentu
Integral tentu digunakan untuk mengintegralkan suatu fungsi f(x) tertentu yang memiliki batas atas dan batas bawah. Integral tentu mempunyai rumus umum:
∫
a
b
F
(
x
)
d
x
=
F
(
b
)
−
F
(
a
)
{\displaystyle \int _{a}^{b}F(x)dx=F(b)-F(a)}
Keterangan:
konstanta c tidak lagi dituliskan dalam integral tentu.
Integral trigonometri
∫
sin
(
a
x
)
d
x
=
−
1
a
cos
(
a
x
)
+
k
{\displaystyle \int \sin(ax)\ dx=-{\frac {1}{a}}\ \cos(ax)+k}
∫
cos
(
a
x
)
d
x
=
1
a
sin
(
a
x
)
+
k
{\displaystyle \int \cos(ax)\ dx={\frac {1}{a}}\ \sin(ax)+k}
∫
sec
(
a
x
)
t
a
n
(
a
x
)
d
x
=
1
a
sec
(
a
x
)
+
k
{\displaystyle \int \sec(ax)\ tan(ax)\ dx={\frac {1}{a}}\ \sec(ax)+k}
∫
sec
2
(
a
x
)
d
x
=
1
a
tan
(
a
x
)
+
k
{\displaystyle \int \sec ^{2}(ax)\ dx={\frac {1}{a}}\ \tan(ax)+k}
∫
csc
2
(
a
x
)
d
x
=
−
1
a
cot
(
a
x
)
+
k
{\displaystyle \int \csc ^{2}(ax)dx=-{\frac {1}{a}}\ \cot(ax)+k}
∫
csc
(
a
x
)
cot
(
a
x
)
d
x
=
−
1
a
csc
(
a
x
)
+
k
{\displaystyle \int \csc(ax)\ \cot(ax)dx=-{\frac {1}{a}}\ \csc(ax)+k}
∫
cos
(
a
x
+
b
)
d
x
=
1
a
sin
(
a
x
+
b
)
+
k
{\displaystyle \int \cos(ax+b)dx={\frac {1}{a}}\ \sin(ax+b)+k}
∫
sin
(
a
x
+
b
)
d
x
=
−
1
a
cos
(
a
x
+
b
)
+
k
{\displaystyle \int \sin(ax+b)dx=-{\frac {1}{a}}\ \cos(ax+b)+k}
∫
sec
2
(
a
x
+
b
)
d
x
=
1
a
tan
(
a
x
+
b
)
+
k
{\displaystyle \int \sec ^{2}(ax+b)dx={\frac {1}{a}}\ \tan(ax+b)+k}
∫
sec
(
x
)
d
x
=
ln
|
sec
(
x
)
+
t
a
n
(
x
)
|
+
k
{\displaystyle \int \sec(x)dx=\ln \left\vert \sec(x)+tan(x)\right\vert +k}
∫
csc
(
x
)
d
x
=
−
ln
|
csc
(
x
)
+
c
o
t
(
x
)
|
+
k
{\displaystyle \int \csc(x)dx=-\ln \left\vert \csc(x)+cot(x)\right\vert +k}
∫
tan
(
x
)
d
x
=
−
ln
|
cos
(
x
)
|
+
k
{\displaystyle \int \tan(x)dx=-\ln \left\vert \cos(x)\right\vert +k}
∫
tan
(
x
)
d
x
=
ln
|
sec
(
x
)
|
+
k
{\displaystyle \int \tan(x)dx=\ln \left\vert \sec(x)\right\vert +k}
∫
cot
(
x
)
d
x
=
ln
|
sin
(
x
)
|
+
k
{\displaystyle \int \cot(x)dx=\ln \left\vert \sin(x)\right\vert +k}
∫
cot
(
x
)
d
x
=
−
ln
|
csc
(
x
)
|
+
k
{\displaystyle \int \cot(x)dx=-\ln \left\vert \csc(x)\right\vert +k}
Ingat-ingat juga beberapa sifat-sifat trigonometri, karena mungkin akan digunakan:
sin
2
A
+
cos
2
A
=
1
{\displaystyle \sin ^{2}A+\cos ^{2}A=1\,}
1
+
tan
2
A
=
1
cos
2
A
=
sec
2
A
{\displaystyle 1+\tan ^{2}A={\frac {1}{\cos ^{2}A}}=\sec ^{2}A\,}
1
+
cot
2
A
=
1
sin
2
A
=
csc
2
A
{\displaystyle 1+\cot ^{2}A={\frac {1}{\sin ^{2}A}}=\csc ^{2}A\,}
sin
2
A
=
2
sin
A
cos
A
{\displaystyle \sin 2A=2\sin A\cos A\,}
cos
2
A
=
cos
2
A
−
sin
2
A
=
2
cos
2
A
−
1
=
1
−
2
sin
2
A
{\displaystyle \cos 2A=\cos ^{2}A-\sin ^{2}A=2\cos ^{2}A-1=1-2\sin ^{2}A\,}
2
sin
A
×
cos
B
=
sin
(
A
+
B
)
+
sin
(
A
−
B
)
,
{\displaystyle 2\sin A\times \cos B=\sin(A+B)+\sin(A-B),}
2
cos
A
×
sin
B
=
sin
(
A
+
B
)
−
sin
(
A
−
B
)
,
{\displaystyle 2\cos A\times \sin B=\sin(A+B)-\sin(A-B),}
2
cos
A
×
cos
B
=
cos
(
A
+
B
)
+
cos
(
A
−
B
)
,
{\displaystyle 2\cos A\times \cos B=\cos(A+B)+\cos(A-B),}
2
sin
A
×
sin
B
=
−
sin
(
A
+
B
)
+
cos
(
A
−
B
)
,
{\displaystyle 2\sin A\times \sin B=-\sin(A+B)+\cos(A-B),}
Substitusi trigonometri
a 2 − x 2 Pada integral
∫
d
x
a
2
−
x
2
{\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}}
kita dapat menggunakan
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)}
∫
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}}}
Catatan: semua langkah diatas haruslah memenuhi syarat a > 0 dan cos(θ ) > 0;
a 2 + x 2 Pada integral
∫
d
x
a
2
+
x
2
{\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}}
kita dapat menuliskan
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)}
maka integralnya menjadi
∫
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}}}
(syarat: a ≠ 0).
x 2 − a 2 Pada integral
∫
x
2
−
a
2
d
x
{\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx}
dapat diselesaikan dengan substitusi:
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}}}
Teknik pemecahan sebagian pada pengintegralan
Polinomial tingkat pertama pada penyebut
Misalkan u = ax + b , maka du = a dx akan menjadikan integral
∫
1
a
x
+
b
d
x
{\displaystyle \int {1 \over ax+b}\,dx}
menjadi
∫
1
u
d
u
a
=
1
a
∫
d
u
u
=
1
a
ln
|
u
|
+
C
=
1
a
ln
|
a
x
+
b
|
+
C
.
{\displaystyle \int {1 \over u}\,{du \over a}={1 \over a}\int {du \over u}={1 \over a}\ln \left|u\right|+C={1 \over a}\ln \left|ax+b\right|+C.}
Contoh lain:
Dengan pemisalan yang sama di atas, misalnya dengan integral
∫
1
(
a
x
+
b
)
8
d
x
{\displaystyle \int {1 \over (ax+b)^{8}}\,dx}
akan berubah menjadi
∫
1
u
8
d
u
a
=
1
a
∫
u
−
8
d
u
=
1
a
⋅
u
−
7
(
−
7
)
+
C
=
−
1
7
a
u
7
+
C
=
−
1
7
a
(
a
x
+
b
)
7
+
C
.
{\displaystyle \int {1 \over u^{8}}\,{du \over a}={1 \over a}\int u^{-8}\,du={1 \over a}\cdot {u^{-7} \over (-7)}+C={-1 \over 7au^{7}}+C={-1 \over 7a(ax+b)^{7}}+C.}
Integral Parsial
Jika dimisalkan u = f (x ), v = g (x ), dan diferensialnya du = f '(x ) dx dan dv = g '(x ) dx , maka integral parsial menyatakan bahwa:
∫
f
(
x
)
g
′
(
x
)
d
x
=
f
(
x
)
g
(
x
)
−
∫
f
′
(
x
)
g
(
x
)
d
x
{\displaystyle \int f(x)g'(x)\,dx=f(x)g(x)-\int f'(x)g(x)\,dx\!}
atau dapat ditulis juga:
∫
u
d
v
=
u
v
−
∫
v
d
u
{\displaystyle \int u\,dv=uv-\int v\,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)
