Integral merupakan suatu objek matematika yang dapat diinterpretasikan sebagai luas wilayah ataupun generalisasi suatu wilayah.
Integral dapat dianggap sebagai perhitungan luas daerah di bawah kurva ƒ (x ), antara dua titik a dan b .
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} maka:
∫ 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}} maka:
∫ 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}} maka:
∫ ( 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
Integral yang mengandung 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;
Integral yang mengandung 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).
Integral yang mengandung 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\!}