Integral- Substitusi Trigonometri ~ NADIA MUTMAINNAH

Integral- Substitusi Trigonometri

Terkadang kita bertemu dengan tipe soal integral yang memuat bentuk akar yang sulit untuk diselesaikan sehingga dibutuhkan teknik substitusi trigonometri agar bentuk akarnya hilang. Setelah variabelnya diganti dengan fungsi trigonometri yang sesuai, maka bentuknya menjadi fungsi trigonometri yang dapat diselesaikan dengan rumus reduksi atau rumus yang sebelumnya sudah dipelajari.

TABEL KESAMAAN TRIGONOMETRI

Bentuk penyederhanaannya;

\sqrt{a^{2} - x^{2}}

$\sqrt{a^2-x^2}$

Gunakan substitusi

x = a sin θ

$x=asin\theta$

\sqrt{a^{2} - {(a sin θ)}^{2}} = \sqrt{a^{2} - a^{2} {sin}^{2} θ}

$\sqrt{a^2-(asin\theta)^2}=\sqrt{a^2-a^2sin^2\theta}$

= \sqrt{a^{2} (1 - {sin}^{2} θ)}

$=\sqrt{a^2(1-sin^2\theta)}$

= a cos θ

$=a cos\theta$

\sqrt{a^{2} + x^{2}}

$\sqrt{a^2+x^2}$

Gunakan substitusi

x = a tan θ

$x=atan\theta$

\sqrt{a^{2} + {(a tan θ)}^{2}} = \sqrt{a^{2} + a^{2} {tan}^{2} θ}

$\sqrt{a^2+(atan\theta)^2}=\sqrt{a^2+a^2tan^2\theta}$

= \sqrt{a^{2} (1 + {tan}^{2} θ)}

$=\sqrt{a^2(1+tan^2\theta)}$

= a sec θ

$=asec\theta$

\sqrt{x^{2} - a^{2}}

$\sqrt{x^2-a^2}$

Gunakan substitusi

x = a sec θ

$x=asec\theta$

\sqrt{{(a sec θ)}^{2} - a^{2}} = \sqrt{(a^{2} {sec}^{2} θ) - a^{2}}

$\sqrt{(asec\theta)^2-a^2}= \sqrt{(a^2sec^2\theta)-a^2}$

= \sqrt{(sec θ - 1) a^{2}}

$=\sqrt{(sec\theta-1)a^2}$

= a tan θ

$=atan\theta$

Note!

Pembatasan nilai

θ

$\theta$ pada interval tertentu pada tabel diatas dimaksudkan agar sinus, tangen dan secan menjadi fungsi yang dapat di inverskan.

Contoh:

(1) Tentukanlah

\int \frac{\sqrt{9 - x^{2}}}{x^{2}} d x

$\int\frac{\sqrt{9-x^2}}{x^2}dx$

Penyelesaian

\int \frac{\sqrt{9 - x^{2}}}{x^{2}} d x = \int \frac{\sqrt{3^{3} - x^{2}}}{x^{2}} d x

$\int\frac{\sqrt{9-x^2}}{x^2}dx=\int\frac{\sqrt{3^3-x^2}}{x^2}dx$

Substitusikan

x = 3 sin θ

$x= 3 sin\theta$ berdasarkan

x = a sin θ

$x= a sin\theta$ ,

maka

d x = 3 cos θ d θ

$dx=3cos\theta d\theta$ sehingga,

\int \frac{\sqrt{9 - x^{2}}}{x^{2}} d x = \int \frac{\sqrt{9 - {(3 sin θ)}^{2}}}{{(3 sin θ)}^{2}} d x

$\int\frac{\sqrt{9-x^2}}{x^2}dx= \int\frac{\sqrt{9-(3sin\theta)^2}}{(3sin\theta)^2}dx$

= \int \frac{\sqrt{9 - 9 {sin}^{2} θ}}{9 {sin}^{2} θ} d x

$=\int\frac{\sqrt{9-9sin^2\theta}}{9sin^2\theta} dx$

= \int \frac{\sqrt{3^{2} - 3^{2} {sin}^{2} θ}}{9 {sin}^{2} θ} 3 cos θ d θ

$=\int\frac{\sqrt{3^2-3^2sin^2\theta}}{9sin^2\theta} 3cos\thetad\theta$

= \int \frac{\sqrt{3^{2} (1 - {sin}^{2} θ)}}{9 {sin}^{2} θ} 3 cos θ d θ

$=\int\frac{\sqrt{3^2(1-sin^2\theta)}}{9sin^2\theta}3cos\thetad\theta$

= \int \frac{3 cos θ}{9 {sin}^{2} θ} 3 cos θ d θ

$=\int\frac{3cos\theta}{9sin^2\theta}3cos\thetad\theta$

= \int \frac{{cos}^{2} θ}{{sin}^{2} θ} d θ

$=\int\frac{cos^2\theta}{sin^2\theta}d\theta$

= \int {cot}^{2} θ d θ

$=\intcot^2\theta d\theta$

= \int ({csc}^{2} θ - 1) d θ

$=\int(csc^2\theta-1)d\theta$

= - cot θ - θ + C

$=-cot\theta -\theta +C$

(2) Tentukanlah

\int \frac{x}{\sqrt{x^{2} + 4}} d x

$\int\frac{x}{\sqrt{x^2+4}}dx$

Penyelesaian:

\int \frac{x}{\sqrt{x^{2} + 4}} d x = \int \frac{x}{\sqrt{x^{2} + 2^{2}}} d x

$\int\frac{x}{\sqrt{x^2+4}}dx=\int\frac{x}{\sqrt{x^2+2^2}}dx$

Substitusikan

x = 2 tan θ

$x=2tan\theta$ maka

d x = 2 {sec}^{2} θ d θ

$dx=2sec^2\thetad\theta$ sehingga,

\int \frac{x}{\sqrt{x^{2} + 4}} d x = \int \frac{2 tan θ}{\sqrt{{(2 tan θ)}^{2} + 4}} 2 {sec}^{2} d θ

$\int\frac{x}{\sqrt{x^2+4}}dx=\int\frac{2tan\theta}{\sqrt{(2tan\theta)^2+4}}2sec^2d\theta$

= \int \frac{2 tan θ}{\sqrt{4 {tan}^{2} θ + 4}} 2 {sec}^{2} d θ

$=\int\frac{2tan\theta}{\sqrt{4tan^2\theta+4}}2sec^2d\theta$

$=\int\frac{2tan\theta}{\sqrt{4 (tan^2\theta+1)}}2sec^2d\theta$

$=\int\frac{2tan\theta}{\sqrt{2^2sec^2\theta}}2sec^2d\theta$

$=\int\frac{2tan\theta2sec^2\theta}{2sec\theta}d\theta$

$=\int\frac{2tan\theta2(sec\theta.sec\theta)}{2sec\theta}d\theta$

$=\int2tan\thetasec\thetad\theta$

$=2\int\frac{sin\theta}{cos\theta}\frac{1}{cos\theta}d\theta$

$=2\int\frac{sin\theta}{cos^2\theta}d\theta$

$=-2\int\frac{d cos\theta}{cos^2\theta}$

$=2cos\theta+C$

(3) Tentukanlah

$\int\frac{x}{\sqrt{3-2x-x^2}}dx$

Penyelesaian:

Terlebih dahulu kita nyatakan

$3-2x-x^2$ menjadi bentuk kuadrat

$3+1-1-2x-x^2=4-(1+2x+x^2)=4-(1+x)^2$

lalu misalkan

$u=x+1$ atau

$x=u-1$ sehingga,

$\int\frac{x}{\sqrt{3-2x-x^2}}dx=\int\frac{u-1]{\sqrt{4-u^2}}dx=\int\frac{u-1]{\sqrt{2^2-u^2}}dx$

Substitusi `u=2sin\theta berdasarkan tabel kesamaan trigonometri diatas.

Maka

$du=2cos\theta$ dan

$\sqrt{4-u^2}=2cos\theta$ sehingga,

$\int\frac{x}{\sqrt{3-2x-x^2}}dx=\int\frac{2sin\theta-1}{2cos\theta}2cos\thetad\theta$

$=\int(2sin\theta-1)d\theta$

$=-2cos\theta-\theta+C$

$=-\sqrt{4-u^2}-sin^{-1}(\frac{u}{2})+C$

$=-\sqrt{3-2x-x^2}-sin^{-1}(\frac{x+1}{2})+C$

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Selasa, 14 Maret 2023