Ringkasan Materi Integral Cara Substitusi Matematika SMA
Ringkasan Materi Integral Cara Substitusi Matematika SMA. ada pertanyaan sebagai berikut : "Bagaimana bentuk rumus Integral Cara Substitusi ? ". untuk menjawab pertanyaan tersebut kalian dapat membaca Artikel Materi Matematika SMA tentang Rumus Integral Cara Substitusi, Contoh dan pembahasan soal Integral Cara Substitusi Latihan soal Integral Cara Substitusi.di Langsung Klik, langsung klik wadah tempat belajar untuk SMA, SMK dan MA. Langsung klik menyediakan materi, contoh, soal, rangkuman, ringkasan, buku saku, motivasi, saran untuk mempermudah belajar para siswa-siswi yang sedang menempuh pembelajaran di tingkat SMA,SMK dan MA.
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| Integral Cara Substitusi ( Matematika SMA ) |
Integral Cara Substitusi
A. Pendahuluan
Integral dengan teknik/metode substitusi digunakan ketika proses pengintegralan tidak bisa diselesaikan dengan rumus-rumus dasar integral, atau seandainya bisa diselesaikan namun akan memerlukan proses yang cukup panjang.
Dalam pengintegralan dengan metode substitusi, tentunya kita harus sudah menguasai konsep-konsep turunan, dimana \(\mathrm{\frac{du}{dx}}\) adalah turunan u terhadap x..
Misalkan u = 2x + 1, turunan u terhadap x ditulis :
\(\mathrm{\frac{du}{dx}}\) = 2 ⇔ du = 2 dx
Dalam pengintegralan dengan metode substitusi, tentunya kita harus sudah menguasai konsep-konsep turunan, dimana \(\mathrm{\frac{du}{dx}}\) adalah turunan u terhadap x..
Misalkan u = 2x + 1, turunan u terhadap x ditulis :
\(\mathrm{\frac{du}{dx}}\) = 2 ⇔ du = 2 dx
B. Contoh Soal Integral Substitusi
Untuk memahami proses pengintegralan dengan metode substitusi, simaklah contoh-contoh berikut.
Contoh 1
∫ x2 (x3 + 5)7 dx = ...
Jawaban :
Misalkan : u = x3 + 5
\(\mathrm{\frac{du}{dx}}\) = 3x2 ⇔ \(\mathrm{\frac{du}{3}}\) = x2 dx
\(\begin{align}
\mathrm{\int x^{2}(x^{3}+5)^{7}\,dx} & = \mathrm{\int (x^{3}+5)^{7}\,x^{2}dx} \\
& = \mathrm{\int u^{7}\,\frac{du}{3}} \\
& = \mathrm{\frac{1}{3}\int u^{7}\,du} \\
& = \mathrm{\frac{1}{3}\cdot \frac{1}{8}u^{8}+C} \\
& = \mathrm{\frac{1}{24}u^{8}+C} \\
& = \mathrm{\frac{1}{24}(x^{3}+5)^{8}+C}
\end{align}\)
Contoh 2
\(\mathrm{\int \frac{4x}{\sqrt{x^{2}-2}}}\) dx = ...
Jawaban :
Misalkan : u = x2 - 2
\(\mathrm{\frac{du}{dx}}\) = 2x ⇔ \(\mathrm{\frac{du}{2}}\) = x dx
\(\begin{align}
\mathrm{\int \frac{4x}{\sqrt{x^{2}-2}}\,dx} & = \mathrm{4\int \frac{1}{\sqrt{x^{2}-2}}\cdot x\,dx} \\
& = \mathrm{4\int \frac{1}{\sqrt{u}}\cdot \frac{du}{2}} \\
& = \mathrm{\frac{4}{2}\int u^{-\frac{1}{2}}\,du} \\
& = \mathrm{2\cdot 2u^{\frac{1}{2}}+C} \\
& = \mathrm{4\sqrt{u}+C} \\
& = \mathrm{4\sqrt{x^{2}-2}+C}
\end{align}\)
Contoh 3
∫ x(x + 4)7 dx = ...
Jawaban :
Misalkan : u = x + 4 maka x = u - 4
\(\mathrm{\frac{du}{dx}}\) = 1 ⇔ du = dx
\(\begin{align}
\mathrm{\int x(x+4)^{7}\,dx} & = \mathrm{\int (u-4)u^{7}\,du} \\
& = \mathrm{\int \left (u^{8}-4u^{7} \right )du} \\
& = \mathrm{\frac{1}{9}u^{9}-\frac{1}{2}u^{8}+C} \\
& = \mathrm{\frac{1}{18}\left (2u^{9}-9u^{8} \right )+C} \\
& = \mathrm{\frac{1}{18}\left (2u-9 \right )u^{8}+C} \\
& = \mathrm{\frac{1}{18}\left (2(x+4)-9 \right )(x+4)^{8}+C} \\
& = \mathrm{\frac{1}{18}(2x-1)(x+4)^{8}+C}
\end{align}\)
Untuk fungsi-fungsi trigonometri, langkah-langkah pengintegralannya sama saja dengan fungsi aljabar diatas, tetapi untuk kasus-kasus tertentu kita harus mengubah terlebih dahulu fungsi yang akan diintegralkan sebelum melakukan pemisalan.
Contoh 4
∫ cos4x sin x dx = ...
Jawaban :
Misalkan : u = cos x
\(\mathrm{\frac{du}{dx}}\) = -sin x ⇔ -du = sin x dx
\(\begin{align}
\mathrm{\int cos^{4}x\,sin\,x\,dx} & = \mathrm{\int u^{4}\,(-du)} \\
& = \mathrm{-\int u^{4}\,du} \\
& = \mathrm{-\frac{1}{5}u^{5}+C} \\
& = \mathrm{-\frac{1}{5}cos^{5}x+C}
\end{align}\)
Contoh 5
∫ cos5x dx = ...
Jawaban :
\(\begin{align} \mathrm{\int cos^{5}x\,dx} & = \mathrm{\int \left (cos^{2}x \right )^{2}cos\,x\,dx} \\
& = \mathrm{\int \left (1-sin^{2}x \right )^{2}cos\,x\,dx} \\
& = \mathrm{\int \left (1-2\,sin^{2}x+sin^{4}x \right )cos\,x\,dx}
\end{align}\)
Misalkan : u = sin x
\(\mathrm{\frac{du}{dx}}\) = cos x ⇔ du = cos x dx
\(\begin{align}
\mathrm{\int cos^{5}x\,dx} & = \mathrm{\int \left (1-2\,sin^{2}x+sin^{4}x \right )cos\,x\,dx} \\
& = \mathrm{\int \left ( 1-2u^{2}+u^{4} \right )du} \\
& = \mathrm{u-\frac{2}{3}u^{3}+\frac{1}{5}u^{5}+C} \\
& = \mathrm{sin\,x-\frac{2}{3}sin^{3}x+\frac{1}{5}sin^{5}x+C}
\end{align}\)
Latihan 1
Selesaikan integral berikut!
a. \(\mathrm{\int x\sqrt{3x^{2}+1}\:dx=\,...}\)
Jawaban :
Misalkan : u = 3x2 + 1
\(\mathrm{\frac{du}{dx}}\) = 6x ⇔ \(\mathrm{\frac{du}{6}}\) = x dx
\(\begin{align}
\mathrm{\int x\sqrt{3x^{2}+1}\,dx }
& = \mathrm{\int (3x^{2}+1)^{\frac{1}{2}}\,x\,dx} \\
& = \mathrm{\int u^{\frac{1}{2}}\,\frac{du}{6}} \\
& = \mathrm{\frac{1}{6}\int u^{\frac{1}{2}}\,du} \\
& = \mathrm{\frac{1}{6}\cdot \frac{2}{3}u^{\frac{3}{2}}+C} \\
& = \mathrm{\frac{1}{9}u\cdot u^{\frac{1}{2}}+C} \\
& = \mathrm{\frac{1}{9}u\sqrt{u}+C} \\
& = \mathrm{\frac{1}{9}(3x^{2}+1)\sqrt{3x^{2}+1}+C}
\end{align}\)
b. \(\mathrm{\int \frac{4x+2}{\sqrt{x^{2}+x-1}}\;dx=\,...}\)
Jawaban :
Misalkan : u = x2 + x - 1
\(\mathrm{\frac{du}{dx}}\) = 2x + 1 ⇔ du = (2x + 1) dx
\(\begin{align}
\mathrm{\int \frac{4x+2}{\sqrt{x^{2}+x-1}}\,dx }
& = \mathrm{2\int \frac{1}{\sqrt{x^{2}+x-1}}\,(2x+1)\,dx} \\
& = \mathrm{2\int \frac{1}{\sqrt{u}}\,du} \\
& = \mathrm{2\int u^{-\frac{1}{2}}\,du} \\
& = \mathrm{2\cdot 2u^{\frac{1}{2}}+C} \\
& = \mathrm{4\sqrt{u}+C} \\
& = \mathrm{4\sqrt{x^{2}+x-1}+C} \\
\end{align}\)
c. \(\mathrm{\int x\left ( x-1 \right )^{3}\;dx}=\,...\)
Jawaban :
Misalkan : u = x - 1 maka x = u + 1
\(\mathrm{\frac{du}{dx}}\) = 1 ⇔ du = dx
\(\begin{align}
\mathrm{\int x(x-1)^{3}\,dx }
& = \mathrm{\int (u+1)u^{3}\,du} \\
& = \mathrm{\int (u^{4}+u^{3})\,du} \\
& = \mathrm{\frac{1}{5}u^{5}+\frac{1}{4}u^{4}+C} \\
& = \mathrm{\frac{1}{20}(4u^{5}+5u^{4})+C} \\
& = \mathrm{\frac{1}{20}(4u+5)u^{4}+C} \\
& = \mathrm{\frac{1}{20}(4(x-1)+5)(x-1)^{4}+C} \\
& = \mathrm{\frac{1}{20}(4x+1)(x-1)^{4}+C}
\end{align}\)
Latihan 2
Contoh 1
∫ x2 (x3 + 5)7 dx = ...
Jawaban :
Misalkan : u = x3 + 5
\(\mathrm{\frac{du}{dx}}\) = 3x2 ⇔ \(\mathrm{\frac{du}{3}}\) = x2 dx
\(\begin{align}
\mathrm{\int x^{2}(x^{3}+5)^{7}\,dx} & = \mathrm{\int (x^{3}+5)^{7}\,x^{2}dx} \\
& = \mathrm{\int u^{7}\,\frac{du}{3}} \\
& = \mathrm{\frac{1}{3}\int u^{7}\,du} \\
& = \mathrm{\frac{1}{3}\cdot \frac{1}{8}u^{8}+C} \\
& = \mathrm{\frac{1}{24}u^{8}+C} \\
& = \mathrm{\frac{1}{24}(x^{3}+5)^{8}+C}
\end{align}\)
Contoh 2
\(\mathrm{\int \frac{4x}{\sqrt{x^{2}-2}}}\) dx = ...
Jawaban :
Misalkan : u = x2 - 2
\(\mathrm{\frac{du}{dx}}\) = 2x ⇔ \(\mathrm{\frac{du}{2}}\) = x dx
\(\begin{align}
\mathrm{\int \frac{4x}{\sqrt{x^{2}-2}}\,dx} & = \mathrm{4\int \frac{1}{\sqrt{x^{2}-2}}\cdot x\,dx} \\
& = \mathrm{4\int \frac{1}{\sqrt{u}}\cdot \frac{du}{2}} \\
& = \mathrm{\frac{4}{2}\int u^{-\frac{1}{2}}\,du} \\
& = \mathrm{2\cdot 2u^{\frac{1}{2}}+C} \\
& = \mathrm{4\sqrt{u}+C} \\
& = \mathrm{4\sqrt{x^{2}-2}+C}
\end{align}\)
Contoh 3
∫ x(x + 4)7 dx = ...
Jawaban :
Misalkan : u = x + 4 maka x = u - 4
\(\mathrm{\frac{du}{dx}}\) = 1 ⇔ du = dx
\(\begin{align}
\mathrm{\int x(x+4)^{7}\,dx} & = \mathrm{\int (u-4)u^{7}\,du} \\
& = \mathrm{\int \left (u^{8}-4u^{7} \right )du} \\
& = \mathrm{\frac{1}{9}u^{9}-\frac{1}{2}u^{8}+C} \\
& = \mathrm{\frac{1}{18}\left (2u^{9}-9u^{8} \right )+C} \\
& = \mathrm{\frac{1}{18}\left (2u-9 \right )u^{8}+C} \\
& = \mathrm{\frac{1}{18}\left (2(x+4)-9 \right )(x+4)^{8}+C} \\
& = \mathrm{\frac{1}{18}(2x-1)(x+4)^{8}+C}
\end{align}\)
Untuk fungsi-fungsi trigonometri, langkah-langkah pengintegralannya sama saja dengan fungsi aljabar diatas, tetapi untuk kasus-kasus tertentu kita harus mengubah terlebih dahulu fungsi yang akan diintegralkan sebelum melakukan pemisalan.
Contoh 4
∫ cos4x sin x dx = ...
Jawaban :
Misalkan : u = cos x
\(\mathrm{\frac{du}{dx}}\) = -sin x ⇔ -du = sin x dx
\(\begin{align}
\mathrm{\int cos^{4}x\,sin\,x\,dx} & = \mathrm{\int u^{4}\,(-du)} \\
& = \mathrm{-\int u^{4}\,du} \\
& = \mathrm{-\frac{1}{5}u^{5}+C} \\
& = \mathrm{-\frac{1}{5}cos^{5}x+C}
\end{align}\)
Contoh 5
∫ cos5x dx = ...
Jawaban :
\(\begin{align} \mathrm{\int cos^{5}x\,dx} & = \mathrm{\int \left (cos^{2}x \right )^{2}cos\,x\,dx} \\
& = \mathrm{\int \left (1-sin^{2}x \right )^{2}cos\,x\,dx} \\
& = \mathrm{\int \left (1-2\,sin^{2}x+sin^{4}x \right )cos\,x\,dx}
\end{align}\)
Misalkan : u = sin x
\(\mathrm{\frac{du}{dx}}\) = cos x ⇔ du = cos x dx
\(\begin{align}
\mathrm{\int cos^{5}x\,dx} & = \mathrm{\int \left (1-2\,sin^{2}x+sin^{4}x \right )cos\,x\,dx} \\
& = \mathrm{\int \left ( 1-2u^{2}+u^{4} \right )du} \\
& = \mathrm{u-\frac{2}{3}u^{3}+\frac{1}{5}u^{5}+C} \\
& = \mathrm{sin\,x-\frac{2}{3}sin^{3}x+\frac{1}{5}sin^{5}x+C}
\end{align}\)
C. Latihan Soal Integral Substitusi
Berikut beberapa contoh soal yang dapat dijadikan latihan untuk menambah pemahaman menyangkut integral dengan metode/teknik substitusi.Latihan 1
Selesaikan integral berikut!
a. \(\mathrm{\int x\sqrt{3x^{2}+1}\:dx=\,...}\)
Jawaban :
Misalkan : u = 3x2 + 1
\(\mathrm{\frac{du}{dx}}\) = 6x ⇔ \(\mathrm{\frac{du}{6}}\) = x dx
\(\begin{align}
\mathrm{\int x\sqrt{3x^{2}+1}\,dx }
& = \mathrm{\int (3x^{2}+1)^{\frac{1}{2}}\,x\,dx} \\
& = \mathrm{\int u^{\frac{1}{2}}\,\frac{du}{6}} \\
& = \mathrm{\frac{1}{6}\int u^{\frac{1}{2}}\,du} \\
& = \mathrm{\frac{1}{6}\cdot \frac{2}{3}u^{\frac{3}{2}}+C} \\
& = \mathrm{\frac{1}{9}u\cdot u^{\frac{1}{2}}+C} \\
& = \mathrm{\frac{1}{9}u\sqrt{u}+C} \\
& = \mathrm{\frac{1}{9}(3x^{2}+1)\sqrt{3x^{2}+1}+C}
\end{align}\)
b. \(\mathrm{\int \frac{4x+2}{\sqrt{x^{2}+x-1}}\;dx=\,...}\)
Jawaban :
Misalkan : u = x2 + x - 1
\(\mathrm{\frac{du}{dx}}\) = 2x + 1 ⇔ du = (2x + 1) dx
\(\begin{align}
\mathrm{\int \frac{4x+2}{\sqrt{x^{2}+x-1}}\,dx }
& = \mathrm{2\int \frac{1}{\sqrt{x^{2}+x-1}}\,(2x+1)\,dx} \\
& = \mathrm{2\int \frac{1}{\sqrt{u}}\,du} \\
& = \mathrm{2\int u^{-\frac{1}{2}}\,du} \\
& = \mathrm{2\cdot 2u^{\frac{1}{2}}+C} \\
& = \mathrm{4\sqrt{u}+C} \\
& = \mathrm{4\sqrt{x^{2}+x-1}+C} \\
\end{align}\)
c. \(\mathrm{\int x\left ( x-1 \right )^{3}\;dx}=\,...\)
Jawaban :
Misalkan : u = x - 1 maka x = u + 1
\(\mathrm{\frac{du}{dx}}\) = 1 ⇔ du = dx
\(\begin{align}
\mathrm{\int x(x-1)^{3}\,dx }
& = \mathrm{\int (u+1)u^{3}\,du} \\
& = \mathrm{\int (u^{4}+u^{3})\,du} \\
& = \mathrm{\frac{1}{5}u^{5}+\frac{1}{4}u^{4}+C} \\
& = \mathrm{\frac{1}{20}(4u^{5}+5u^{4})+C} \\
& = \mathrm{\frac{1}{20}(4u+5)u^{4}+C} \\
& = \mathrm{\frac{1}{20}(4(x-1)+5)(x-1)^{4}+C} \\
& = \mathrm{\frac{1}{20}(4x+1)(x-1)^{4}+C}
\end{align}\)
Latihan 2
Tunjukkan bahwa :
a. \(\mathrm{\int sin^{n}ax\:cos\,ax\;dx=\frac{1}{a(n+1)}sin^{n+1}x+C}\)
a. \(\mathrm{\int sin^{n}ax\:cos\,ax\;dx=\frac{1}{a(n+1)}sin^{n+1}x+C}\)
Jawaban :
Misalkan : u = sin ax
\(\mathrm{\frac{du}{dx}}\) = a cos ax ⇔ \(\mathrm{\frac{du}{a}}\) = cos ax dx
\(\begin{align}
\mathrm{\int sin^{n}ax\,cos\,ax\,dx }
& = \mathrm{\int u^{n}\cdot \frac{du}{a}} \\
& = \mathrm{\frac{1}{a}\int u^{n}\,du} \\
& = \mathrm{\frac{1}{a}\cdot \frac{1}{n+1}u^{n+1}+C} \\
& = \mathrm{\frac{1}{a(n+1)}sin^{n+1}ax+C} \\
\end{align}\)
b. \(\mathrm{\int cos^{n}ax\:sin\,ax\;dx=-\frac{1}{a(n+1)}cos^{n+1}x+C}\)
Jawaban :
Misalkan : u = cos ax
\(\mathrm{\frac{du}{dx}}\) = -a sin ax ⇔ \(\mathrm{-\frac{du}{a}}\) = sin ax dx
\(\begin{align}
\mathrm{\int cos^{n}ax\,sin\,ax\,dx }
& = \mathrm{\int u^{n}\cdot \left (-\frac{du}{a} \right )} \\
& = \mathrm{-\frac{1}{a}\int u^{n}\,du} \\
& = \mathrm{-\frac{1}{a}\cdot \frac{1}{n+1}u^{n+1}+C} \\
& = \mathrm{-\frac{1}{a(n+1)}cos^{n+1}ax+C} \\
\end{align}\)
Latihan 3
Dengan menggunakan rumus pada contoh 2, selesaikan integral berikut!
a. ∫ sin43x cos 3x dx = ...
Jawaban :
∫ sin43x cos 3x dx = \(\frac{1}{3(4+1)}\)sin4+13x + C
∫ sin43x cos 3x dx = \(\frac{1}{15}\)sin53x + C
b. ∫ cos52x sin 2x dx = ...
Jawaban :
∫ cos52x sin 2x dx = -\(\frac{1}{2(5+1)}\)cos5+12x + C
∫ cos52x sin 2x dx = -\(\frac{1}{12}\)cos62x + C
Latihan 4
Selesaikan integral berikut!
a. ∫ sin2x cos3x dx = ...
Jawaban :
Misalkan : u = sin x
\(\mathrm{\frac{du}{dx}}\) = cos x ⇔ du = cos x dx
∫ sin2x cos3x dx = ∫ sin2x cos2x cos x dx
∫ sin2x cos3x dx = ∫ sin2x (1 - sin2x) cos x dx
∫ sin2x cos3x dx = ∫ u2 (1 - u2) du
∫ sin2x cos3x dx = ∫ (u2 - u4) du
∫ sin2x cos3x dx = \(\frac{1}{3}\)u3 - \(\frac{1}{5}\)u5 + C
∫ sin2x cos3x dx = \(\frac{1}{3}\)sin3x - \(\frac{1}{5}\)sin5x + C
b. ∫ tan x sec3x dx = ...
Jawaban :
Misalkan : u = sec x
\(\mathrm{\frac{du}{dx}}\) = sec x tan x ⇔ du = sec x tan x dx
∫ tan x sec3x dx = ∫ sec2x sec x tan x dx
∫ tan x sec3x dx = ∫ u2 du
∫ tan x sec3x dx = \(\frac{1}{3}\)u3 + C
∫ tan x sec3x dx = \(\frac{1}{3}\)sec3x + C
\(\mathrm{\frac{du}{dx}}\) = a cos ax ⇔ \(\mathrm{\frac{du}{a}}\) = cos ax dx
\(\begin{align}
\mathrm{\int sin^{n}ax\,cos\,ax\,dx }
& = \mathrm{\int u^{n}\cdot \frac{du}{a}} \\
& = \mathrm{\frac{1}{a}\int u^{n}\,du} \\
& = \mathrm{\frac{1}{a}\cdot \frac{1}{n+1}u^{n+1}+C} \\
& = \mathrm{\frac{1}{a(n+1)}sin^{n+1}ax+C} \\
\end{align}\)
b. \(\mathrm{\int cos^{n}ax\:sin\,ax\;dx=-\frac{1}{a(n+1)}cos^{n+1}x+C}\)
Jawaban :
Misalkan : u = cos ax
\(\mathrm{\frac{du}{dx}}\) = -a sin ax ⇔ \(\mathrm{-\frac{du}{a}}\) = sin ax dx
\(\begin{align}
\mathrm{\int cos^{n}ax\,sin\,ax\,dx }
& = \mathrm{\int u^{n}\cdot \left (-\frac{du}{a} \right )} \\
& = \mathrm{-\frac{1}{a}\int u^{n}\,du} \\
& = \mathrm{-\frac{1}{a}\cdot \frac{1}{n+1}u^{n+1}+C} \\
& = \mathrm{-\frac{1}{a(n+1)}cos^{n+1}ax+C} \\
\end{align}\)
Latihan 3
Dengan menggunakan rumus pada contoh 2, selesaikan integral berikut!
a. ∫ sin43x cos 3x dx = ...
Jawaban :
∫ sin43x cos 3x dx = \(\frac{1}{3(4+1)}\)sin4+13x + C
∫ sin43x cos 3x dx = \(\frac{1}{15}\)sin53x + C
b. ∫ cos52x sin 2x dx = ...
Jawaban :
∫ cos52x sin 2x dx = -\(\frac{1}{2(5+1)}\)cos5+12x + C
∫ cos52x sin 2x dx = -\(\frac{1}{12}\)cos62x + C
Selesaikan integral berikut!
a. ∫ sin2x cos3x dx = ...
Jawaban :
Misalkan : u = sin x
\(\mathrm{\frac{du}{dx}}\) = cos x ⇔ du = cos x dx
∫ sin2x cos3x dx = ∫ sin2x cos2x cos x dx
∫ sin2x cos3x dx = ∫ sin2x (1 - sin2x) cos x dx
∫ sin2x cos3x dx = ∫ u2 (1 - u2) du
∫ sin2x cos3x dx = ∫ (u2 - u4) du
∫ sin2x cos3x dx = \(\frac{1}{3}\)u3 - \(\frac{1}{5}\)u5 + C
∫ sin2x cos3x dx = \(\frac{1}{3}\)sin3x - \(\frac{1}{5}\)sin5x + C
b. ∫ tan x sec3x dx = ...
Misalkan : u = sec x
\(\mathrm{\frac{du}{dx}}\) = sec x tan x ⇔ du = sec x tan x dx
∫ tan x sec3x dx = ∫ sec2x sec x tan x dx
∫ tan x sec3x dx = ∫ u2 du
∫ tan x sec3x dx = \(\frac{1}{3}\)u3 + C
∫ tan x sec3x dx = \(\frac{1}{3}\)sec3x + C
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