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Integral adalah kebalikan dari proses diferensiasi . Integral ditemukan menyusul ditemukannya masalah dalam diferensiasi di mana matematikawan harus berpikir bagaimana menyelesaikan masalah yang berkebalikan dengan solusi diferensiasi. Lambang integral adalah
∫
{\displaystyle \int \,}
Integral terbagi dua yaitu integral tak tentu integral tertentu
Mencari nilai integral Substitusi Contoh soal:
Cari nilai dari:
∫
l
n
x
x
d
x
{\displaystyle \int {\frac {lnx}{x}}\,dx\,}
t
=
ln
x
,
d
t
=
d
x
x
{\displaystyle t=\ln x,dt={\frac {dx}{x}}}
∫
l
n
x
x
d
x
=
∫
t
d
t
{\displaystyle \int {\frac {lnx}{x}}\,dx\,=\int t\,dt}
=
1
2
t
2
+
C
{\displaystyle ={\frac {1}{2}}t^{2}+C}
=
1
2
l
n
2
x
+
C
{\displaystyle ={\frac {1}{2}}ln^{2}x+C}
Integrasi parsial Integral parsial menggunakan rumus sebagai berikut:
∫
f
(
x
)
g
(
x
)
d
x
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
{\displaystyle \int f(x)g(x)\,dx=f'(x)g(x)-f(x)g'(x)}
Contoh soal:
Cari nilai dari:
∫
ln
x
d
x
{\displaystyle \int \ln x\,dx\,}
f
′
(
x
)
=
1
,
f
(
x
)
=
x
,
g
(
x
)
=
l
n
x
,
g
′
(
x
)
=
1
x
{\displaystyle f'(x)=1,f(x)=x,g(x)=lnx,g'(x)={\frac {1}{x}}\,}
Gunakan rumus di atas
∫
ln
x
d
x
=
x
l
n
x
−
∫
x
1
x
d
x
{\displaystyle \int \ln x\ dx=xlnx-\int x{\frac {1}{x}}\,dx\,}
=
x
l
n
x
−
∫
1
d
x
{\displaystyle =xlnx-\int 1\,dx\,}
=
x
l
n
x
−
x
+
C
{\displaystyle =xlnx-x+C\,}
Substitusi trigonometri
Bentuk Gunakan
a
2
−
b
2
x
2
{\displaystyle {\sqrt {a^{2}-b^{2}x^{2}}}\,}
x
=
a
b
sin
α
{\displaystyle x={\frac {a}{b}}\sin \alpha \,}
a
2
+
b
2
x
2
{\displaystyle {\sqrt {a^{2}+b^{2}x^{2}}}\,}
x
=
a
b
tan
α
{\displaystyle \!\,x={\frac {a}{b}}\tan \alpha \,}
b
2
x
2
−
a
2
{\displaystyle {\sqrt {b^{2}x^{2}-a^{2}}}\,}
x
=
a
b
sec
α
{\displaystyle \,x={\frac {a}{b}}\sec \alpha \,}
Contoh soal:
Cari nilai dari:
∫
d
x
x
2
x
2
+
4
{\displaystyle \int {\frac {dx}{x^{2}{\sqrt {x^{2}+4}}}}\,}
x
=
2
tan
A
,
d
x
=
2
sec
2
A
d
A
{\displaystyle x=2\tan A,dx=2\sec ^{2}A\,dA\,}
∫
d
x
x
2
x
2
+
4
{\displaystyle \int {\frac {dx}{x^{2}{\sqrt {x^{2}+4}}}}\,}
=
∫
2
s
e
c
2
A
d
A
(
2
t
a
n
A
)
2
4
+
(
2
t
a
n
A
)
2
{\displaystyle =\int {\frac {2sec^{2}A\,dA}{(2tanA)^{2}{\sqrt {4+(2tanA)^{2}}}}}\,}
=
∫
2
s
e
c
2
A
d
A
4
t
a
n
2
A
4
+
4
t
a
n
2
A
{\displaystyle =\int {\frac {2sec^{2}A\,dA}{4tan^{2}A{\sqrt {4+4tan^{2}A}}}}\,}
=
∫
2
s
e
c
2
A
d
A
4
t
a
n
2
A
4
(
1
+
t
a
n
2
A
)
{\displaystyle =\int {\frac {2sec^{2}A\,dA}{4tan^{2}A{\sqrt {4(1+tan^{2}A)}}}}\,}
=
∫
2
s
e
c
2
A
d
A
4
t
a
n
2
A
4
s
e
c
2
A
{\displaystyle =\int {\frac {2sec^{2}A\,dA}{4tan^{2}A{\sqrt {4sec^{2}A}}}}\,}
=
∫
2
s
e
c
2
A
d
A
4
t
a
n
2
A
.2
s
e
c
A
{\displaystyle =\int {\frac {2sec^{2}A\,dA}{4tan^{2}A.2secA}}\,}
=
∫
s
e
c
A
d
A
4
t
a
n
2
A
{\displaystyle =\int {\frac {secA\,dA}{4tan^{2}A}}\,}
=
1
4
∫
s
e
c
A
d
A
t
a
n
2
A
{\displaystyle ={\frac {1}{4}}\int {\frac {secA\,dA}{tan^{2}A}}\,}
=
1
4
∫
c
o
s
A
s
i
n
2
A
d
A
{\displaystyle ={\frac {1}{4}}\int {\frac {cosA}{sin^{2}A}}\,dA\,}
Cari nilai dari:
∫
c
o
s
A
s
i
n
2
A
d
A
{\displaystyle \int {\frac {cosA}{sin^{2}A}}\,dA\,}
t
=
s
i
n
A
,
d
t
=
c
o
s
A
d
A
{\displaystyle t=sinA,dt=cosA\,dA\,}
∫
c
o
s
A
s
i
n
2
A
d
A
{\displaystyle \int {\frac {cosA}{sin^{2}A}}\,dA\,}
=
∫
d
t
t
2
{\displaystyle =\int {\frac {dt}{t^{2}}}\,}
=
∫
t
−
2
d
t
{\displaystyle =\int t^{-2}\,dt\,}
=
−
t
−
1
+
C
=
−
1
s
i
n
A
+
C
{\displaystyle =-t^{-1}+C=-{\frac {1}{sinA}}+C\,}
Masukkan nilai tersebut:
=
1
4
∫
c
o
s
A
s
i
n
2
A
d
A
{\displaystyle ={\frac {1}{4}}\int {\frac {cosA}{sin^{2}A}}\,dA\,}
=
1
4
.
−
1
s
i
n
A
+
C
{\displaystyle ={\frac {1}{4}}.-{\frac {1}{sinA}}+C\,}
=
−
1
4
s
i
n
A
+
C
{\displaystyle =-{\frac {1}{4sinA}}+C\,}
Nilai sin A adalah
x
x
2
+
4
{\displaystyle {\frac {x}{\sqrt {x^{2}+4}}}}
=
−
1
4
s
i
n
A
+
C
{\displaystyle =-{\frac {1}{4sinA}}+C\,}
=
−
x
2
+
4
4
x
+
C
{\displaystyle =-{\frac {\sqrt {x^{2}+4}}{4x}}+C\,}
Integrasi pecahan parsial Contoh soal:
Cari nilai dari:
∫
d
x
x
2
−
4
{\displaystyle \int {\frac {dx}{x^{2}-4}}\,}
1
x
2
−
4
=
A
x
+
2
+
B
x
−
2
{\displaystyle {\frac {1}{x^{2}-4}}={\frac {A}{x+2}}+{\frac {B}{x-2}}\,}
=
A
(
x
−
2
)
+
B
(
x
+
2
)
x
2
−
4
{\displaystyle ={\frac {A(x-2)+B(x+2)}{x^{2}-4}}\,}
=
A
x
−
2
A
+
B
x
+
2
B
x
2
−
4
{\displaystyle ={\frac {Ax-2A+Bx+2B}{x^{2}-4}}\,}
=
(
A
+
B
)
x
−
2
(
A
−
B
)
x
2
−
4
{\displaystyle ={\frac {(A+B)x-2(A-B)}{x^{2}-4}}\,}
Akan diperoleh dua persamaan yaitu
A
+
B
=
0
{\displaystyle A+B=0\,}
A
−
B
=
−
1
2
{\displaystyle A-B=-{\frac {1}{2}}}
Dengan menyelesaikan kedua persamaan akan diperoleh hasil
A
=
−
1
4
,
B
=
1
4
{\displaystyle A=-{\frac {1}{4}},B={\frac {1}{4}}\,}
∫
d
x
x
2
−
4
{\displaystyle \int {\frac {dx}{x^{2}-4}}\,}
=
1
4
∫
(
1
x
−
2
−
1
x
+
2
)
d
x
{\displaystyle ={\frac {1}{4}}\int ({\frac {1}{x-2}}-{\frac {1}{x+2}})\,dx\,}
=
1
4
(
l
n
|
x
−
2
|
−
l
n
|
x
+
2
|
)
+
C
{\displaystyle ={\frac {1}{4}}(ln|x-2|-ln|x+2|)+C\,}
=
1
4
l
n
|
x
−
2
x
+
2
|
+
C
{\displaystyle ={\frac {1}{4}}ln|{\frac {x-2}{x+2}}|+C\,}
Rumus integrasi dasar Umum Bilangan natural
∫
e
u
d
u
=
e
u
+
C
{\displaystyle \int e^{u}du=e^{u}+C\,}
Logaritma
∫
log
b
(
x
)
d
x
=
x
log
b
(
x
)
−
x
ln
(
b
)
+
C
=
x
log
b
(
x
e
)
+
C
{\displaystyle \int \log _{b}(x)\,dx=x\log _{b}(x)-{\frac {x}{\ln(b)}}+C=x\log _{b}\left({\frac {x}{e}}\right)+C}
Trigonometri
∫
sin
x
d
x
=
−
cos
x
+
C
{\displaystyle \int \sin x\,dx=-\cos x+C\,}
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos x\,dx=\sin x+C\,}
∫
tan
x
d
x
=
ln
|
sec
x
|
+
C
{\displaystyle \int \tan x\,dx=\ln |\sec x|+C\,}
∫
cot
x
d
x
=
ln
|
sin
x
|
+
C
{\displaystyle \int \cot x\,dx=\ln |\sin x|+C\,}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
C
{\displaystyle \int \sec x\,dx=\ln |\sec x+\tan x|+C\,}
∫
csc
x
d
x
=
ln
|
csc
x
−
cot
x
|
+
C
{\displaystyle \int \csc x\,dx=\ln |\csc x-\cot x|+C\,}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}x\,dx=\tan x+C\,}
∫
csc
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int \csc ^{2}x\,dx=-\cot x+C\,}
∫
sec
x
tan
x
d
x
=
sec
x
+
C
{\displaystyle \int \sec x\tan x\,dx=\sec x+C\,}
∫
csc
x
cot
x
d
x
=
−
csc
x
+
C
{\displaystyle \int \csc x\cot x\,dx=-\csc x+C\,}
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