Integral Substitusi

Integral dengan metode/metode substitusi dipakai dikala proses pengintegralan tidak sanggup diselesaikan dengan rumus-rumus dasar integral, atau seandainya sanggup diselesaikan namun akan memerlukan proses yang cukup panjang.

Dalam pengintegralan dengan metode substitusi, tentunya kita harus sudah menguasai konsep-konsep turunan, dimana \(\mathrm{\frac{du}{dx}}\) yaitu turunan u terhadap x..
Misalkan u = 2x + 1, turunan u terhadap x ditulis :
\(\mathrm{\frac{du}{dx}}\) = 2  ⇔  du = 2 dx

Untuk memahami proses pengintegralan dengan metode substitusi, simaklah contoh-contoh diberikut.

misal 1
∫ x² (x³ + 5)⁷ dx = ...

Jawab :
Misalkan : u = x³ + 5
\(\mathrm{\frac{du}{dx}}\) = 3x²   ⇔   \(\mathrm{\frac{du}{3}}\) = x² dx

\(\begin{align}
\mathrm{\int x^{2}(x^{3}+5)^{7}\,dx} & = \mathrm{\int (x^{3}+5)^{7}\,x^{2}dx} \\
& = \mathrm{\int u^{7}\,\frac{du}{3}} \\
& = \mathrm{\frac{1}{3}\int u^{7}\,du} \\
& = \mathrm{\frac{1}{3}\cdot \frac{1}{8}u^{8}+C} \\
& = \mathrm{\frac{1}{24}u^{8}+C} \\
& = \mathrm{\frac{1}{24}(x^{3}+5)^{8}+C}
\end{align}\)


misal 2
\(\mathrm{\int \frac{4x}{\sqrt{x^{2}-2}}}\) dx = ...

Jawab :
Misalkan : u = x² - 2
\(\mathrm{\frac{du}{dx}}\) = 2x   ⇔   \(\mathrm{\frac{du}{2}}\) = x dx

\(\begin{align}
\mathrm{\int \frac{4x}{\sqrt{x^{2}-2}}\,dx} & = \mathrm{4\int \frac{1}{\sqrt{x^{2}-2}}\cdot x\,dx} \\
& = \mathrm{4\int \frac{1}{\sqrt{u}}\cdot \frac{du}{2}} \\
& = \mathrm{\frac{4}{2}\int u^{-\frac{1}{2}}\,du} \\
& = \mathrm{2\cdot 2u^{\frac{1}{2}}+C} \\
& = \mathrm{4\sqrt{u}+C} \\
& = \mathrm{4\sqrt{x^{2}-2}+C}
\end{align}\)


misal 3
 ∫ x(x + 4)⁷ dx = ...

Jawab :
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 lampau fungsi yang akan diintegralkan sebelum melaksanakan pemisalan.

misal 4
∫ cos⁴x sin x dx = ...

Jawab :
Misalkan : u = cos x
\(\mathrm{\frac{du}{dx}}\) = -sin x   ⇔   -du = sin x dx

\(\begin{align}
\mathrm{\int cos^{4}x\,sin\,x\,dx} & = \mathrm{\int u^{4}\,(-du)} \\
& = \mathrm{-\int u^{4}\,du} \\
& = \mathrm{-\frac{1}{5}u^{5}+C} \\
& = \mathrm{-\frac{1}{5}cos^{5}x+C}
\end{align}\)


misal 5
∫ cos⁵x dx = ...

Jawab :
\(\begin{align} \mathrm{\int cos^{5}x\,dx} & = \mathrm{\int \left (cos^{2}x  \right )^{2}cos\,x\,dx} \\
& = \mathrm{\int \left (1-sin^{2}x  \right )^{2}cos\,x\,dx} \\
& = \mathrm{\int \left (1-2\,sin^{2}x+sin^{4}x  \right )cos\,x\,dx}
\end{align}\)

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 Soal Integral Substitusi

Berikut beberapa rujukan soal yang sanggup dijadikan tes untuk menambah pemahaman menyangkut integral dengan metode/metode substitusi.

Latihan 1
Selesaikan integral diberikut!

a.  \(\mathrm{\int x\sqrt{3x^{2}+1}\:dx=\,...}\)

Jawab :
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.  \(\mathrm{\int \frac{4x+2}{\sqrt{x^{2}+x-1}}\;dx=\,...}\)

Jawab :
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.  \(\mathrm{\int x\left ( x-1 \right )^{3}\;dx}=\,...\)

Jawab :
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.  \(\mathrm{\int sin^{n}ax\:cos\,ax\;dx=\frac{1}{a(n+1)}sin^{n+1}x+C}\)

Jawab :
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.  \(\mathrm{\int cos^{n}ax\:sin\,ax\;dx=-\frac{1}{a(n+1)}cos^{n+1}x+C}\)

Jawab :
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
melaluiataubersamaini memakai rumus pada rujukan 2, selesaikan integral diberikut!

a.  ∫ sin⁴3x cos 3x dx = ...

Jawab :
∫ 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 = ...

Jawab :
∫ 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 diberikut!

a.  ∫ sin²x cos³x dx = ...

Jawab :
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 = ...

Jawab :
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