BAB 1 : FUNGSI TRANSENDEN

1.3 Fungsi Hiperbolik

A) Rangkuman Materi

1 Definisi Fungsi Hiperbolik

Definisi 1.1

Fungsi sinus hiperbolik dan cosinus hiperbolik, masing-masing dinotasikan dengan \(\sinh\) dan \(\cosh\), dan didefinisikan dengan

\( \sinh x= \frac{e^x-e^{-x}}{2} \) \(\cosh x = \frac{e^x+e^{-x}}{2}\)

2 Kesamaan Hiperbolik

\(\cosh (-x)=\cosh x\)

\(\sinh (-x)=-\cosh x\)

\(\cosh^2 x - \sinh^2 x =1 \)

\(\tanh^2 x + \mathrm{sech}^2 {x} =1 \)

\(\coth^2 x - \mathrm{csch}^2 {x} =1 \)

\(\cosh x + \sinh x =e^x \)

\(\cosh x - \sinh x =e^{-x} \)

\(2\sinh x \cosh x =\sinh 2x \)

\(\cosh^2 x + \sinh^2 x =\cosh 2x \)

\(\sinh^2 x + 1 =\cosh 2x \)

\(\cosh^2 x - 1 =\cosh 2x \)

\(\sinh (x+y)=\sinh x \cosh y + \cosh x \sinh y\)

\(\sinh(x-y)=\sinh x \cosh y - \cosh x \sinh y\)

\(\cosh (x+y)=\cosh x \cosh y + \sinh x \sinh y\)

\(\cosh(x-y)=\cosh x \cosh y - \sinh x \sinh y\)

3 Rumus Turunan Fungsi Hiperbolik

\(\frac{d}{dx}(\sinh u) = \cosh u \frac{du}{dx}\)

\(\frac{d}{dx}(\cosh u) = \sinh u \frac{du}{dx}\)

\(\frac{d}{dx}(\tanh u) = \mathrm{sech}^2 {u} \frac{du}{dx}\)

\(\frac{d}{dx}(\coth u) = -\mathrm{csch}^2 {u} \frac{du}{dx}\)

\(\frac{d}{dx}(\mathrm{sech} {u}) = -\mathrm{sech} {u} \tanh u \frac{du}{dx}\)

\(\frac{d}{dx}(\mathrm{csch} {u}) = -\mathrm{csch} {u} \coth u \frac{du}{dx}\)

4 Rumus Integral Fungsi Hiperbolik

\(\int \sinh u \, du = \cosh u + C\)

\(\int \cosh u \, du = \sinh u + C\)

\(\int \tanh u \, du = \ln |\cosh u| + C\)

\(\int \coth u \, du = \ln |\sinh u| + C\)

\(\int \mathrm{sech}^2 {u} \, du = \tanh u + C\)

\(\int \mathrm{csch}^2 {u} \, du = -\coth u + C\)

\(\int (\mathrm{sech} {u}\) \(\tanh u ) du = -\mathrm{sech} {u} + C\)

\(\int (\mathrm{csch} {u}\) \(\mathrm{coth} {u}) du = -\mathrm{csch} {u} + C\)

5 Invers Fungsi Hiperbolik

Fungsi	Domain	Range	Hubungan Dasar
\(\sinh^{-1} x\)	\((-\infty, +\infty)\)	\((-\infty, +\infty)\)	\(\sinh^{-1}(\sinh x) = x \text{ jika } -\infty < x < +\infty\) \(\sinh(\sinh^{-1} x) = x \text{ jika } -\infty < x < +\infty\)
\(\cosh^{-1} x\)	\([1, +\infty)\)	\([0, +\infty)\)	\(\cosh^{-1}(\cosh x) = x \text{ jika } x \ge 0\) \(\cosh(\cosh^{-1} x) = x \text{ jika } x \ge 1\)
\(\tanh^{-1} x\)	\((-1, 1)\)	\((-\infty, +\infty)\)	\(\tanh^{-1}(\tanh x) = x \text{ jika } -\infty < x < +\infty\) \(\tanh(\tanh^{-1} x) = x \text{ jika } -1 < x < 1\)
\(\coth^{-1} x\)	\((-\infty, -1) \cup (1, +\infty)\)	\((-\infty, 0) \cup (0, +\infty)\)	\(\coth^{-1}(\coth x) = x \text{ jika } x < 0 \text{ atau } x > 0\) \(\coth(\coth^{-1} x) = x \text{ jika } x < -1 \text{ atau } x > 1\)
\(\mathrm{sech}^{-1} {x}\)	\((0, 1]\)	\([0, +\infty)\)	\(\mathrm{sech}^{-1}{\mathrm{(sech)} {x}} = x \text{ jika } x \ge 0\) \(\mathrm{sech}{\mathrm{(sech)}^{-1} {x}} = x \text{ jika } 0 < x \le 1\)
\(\mathrm{csch}^{-1} x\)	\((-\infty, 0) \cup (0, +\infty)\)	\((-\infty, 0) \cup (0, +\infty)\)	\(\mathrm{csch}^{-1}(\mathrm{csch} {x}) = x \text{ jika } x < 0 \text{ atau } x > 0\) \(\mathrm{csch}(\mathrm{csch}^{-1} {x}) = x \text{ jika } x < 0 \text{ atau } x > 0\)

Tabel 5.1 Hubungan Dasar Fungsi Invers Hiperbolik

Teorema 5.1

\(y=\sinh^{-1}x \) ekuivalen dengan \(\sinh y = x\) untuk semua \(x,y\)

\(y=\cosh^{-1}x \) ekuivalen dengan \(\cosh y = x\) jika \(\begin{cases} x \ge 1 \\ y \ge 0. \end{cases}\)

\(y=\tanh^{-1}x \) ekuivalen dengan \(\tanh y = x\) jika \(\begin{cases} -1 < x < 1 \\ -\infty < y < +\infty. \end{cases}\)

\(y=\coth^{-1}x \) ekuivalen dengan \(\coth y = x\) jika \(\begin{cases} |x| > 1 \\ y ≠ 0 \end{cases}\)

\(y=\mathrm{sech}^{-1}x \) ekuivalen dengan \(\mathrm{sech} {y} = x\) jika \(\begin{cases} 0 < x \le 1 \\ y \ge 0. \end{cases}\)

\(y=\mathrm{csch}^{-1}x \) ekuivalen dengan \(\mathrm{csch} {y} = x\) jika \(\begin{cases} x ≠ 0 \\ y ≠ 0. \end{cases}\)

6 Bentuk Logaritma Fungsi Invers Hiperbolik

Teorema 6.1

Hubungan berikut ini berlaku, dengan pembatasan pada \(x\) dalam (1.5.1)

\(\sinh^{-1}x = \ln\left(x + \sqrt{x^2 + 1}\right)\)

\(\cosh^{-1}x = \ln\left(x + \sqrt{x^2 - 1}\right)\)

\(\tanh^{-1}x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right)\)

\(\coth^{-1}x = \frac{1}{2} \ln\left(\frac{x + 1}{x - 1}\right)\)

\(\mathrm{sech}^{-1}x = \ln\left(\frac{1 + \sqrt{1 - x^2}}{x}\right)\)

\(\mathrm{csch}^{-1}x = \ln\left(\frac{1}{x} + \frac{\sqrt{1 + x^2}}{|x|}\right)\)

7 Rumus Turunan Fungsi Invers Hiperbolik

Teorema 7.1

Hubungan berikut ini berlaku, dengan pembatasan pada \(x\) dalam (1.5.1)

\(\frac{d}{dx}[\sinh^{-1}x] = \frac{1}{\sqrt{x^2 + 1}}\)

\(\frac{d}{dx}[\cosh^{-1}x] = \frac{1}{\sqrt{x^2 - 1}}\)

\(\frac{d}{dx}[\tanh^{-1}x] = \frac{1}{1 - x^2}\)

\(\frac{d}{dx}[\coth^{-1}x] = -\frac{1}{1 - x^2}\)

\(\frac{d}{dx}[\mathrm{sech}^{-1}x] = -\frac{1}{x\sqrt{1 - x^2}}\)

\(\frac{d}{dx}[\mathrm{csch}^{-1}x] = -\frac{1}{|x|\sqrt{1 + x^2}}\)

(7.1)

Teorema 7.2

\(\int \frac{dx}{\sqrt{x^2 + 1}} = \sinh^{-1}x + C\)

\(\int \frac{dx}{\sqrt{x^2 - 1}} = \cosh^{-1}x + C,\) untuk \(u > 1\)

\(\int \frac{dx}{1 - x^2} = \tanh^{-1}x + C,\) untuk \(|u| < 1\)

\(\int \frac{dx}{1 - x^2} = -\coth^{-1}x + C,\) untuk \(|u| > 1\)

\(\int \frac{dx}{x\sqrt{1 - x^2}} = -\mathrm{sech}^{-1}x + C\)

\(\int \frac{dx}{|x|\sqrt{1 + x^2}} = -\mathrm{csch}^{-1}x + C\)

B) Contoh Soal

1. Dapatkan \(\frac{dy}{dx}\) dari

\(y=\sinh (cos3x)\)

Jawab:

Diketahui bahwa \(\frac {d}{dx}[\sinh u]= \cosh u \frac{du}{dx}\), sehingga

\(\frac{dy}{dx}=\frac{d}{dx}[\sinh (cos3x)]\)

\(= \cosh (cos3x) \frac{d}{dx}(cos3x)\)

\(= \cosh (cos3x) (-3 \sin 3x)\)

\(= -3 \sin 3x \cosh (cos3x))\)

2. Dapatkan \(\frac{dy}{dx}\) dari

\(y=\sinh^{-1} (\tan x)\)

Jawab:

Diketahui bahwa \(\frac {d}{dx}[\sinh^{-1} u]= \frac{1}{\sqrt{u^2 + 1}} \frac{du}{dx}\), sehingga

\(\frac{dy}{dx}=\frac{d}{dx}[\sinh^{-1} (\tan x)]\)

\(= \frac{1}{\sqrt{(\tan x)^2 + 1}} \frac{d}{dx}(\tan x)\)

\(= \frac{1}{\sqrt{\sec^2 x}} \sec^2 x\)

\(= \sec x\)

3. Hitung integral

\(\int \tanh 2x, dx\)

Jawab::
Misalkan \(u=sech 2x,\) sehingga \(du=-2 sech 2x \tanh 2x dx\) ⇔ \(\frac{du}{-2 sech 2x} = \frac{du}{-2u} = \tanh 2x dx\)

\(\int \tanh 2x dx = \int \frac{du}{2u}\)

\(=-\frac{1}{2} ln u + C\)

\(-\frac{1}{2} \ln(\mathrm{sech} 2x) + C\)

4. Hitung integral

\(\frac{dx}{\sqrt{1-e^{2x}}}\)

Jawab::
Misalkan \(u=e^x,\) sehingga \(du=e^xdx ⇔ dx=\frac{du}{e^x}=\frac{du}{u}\)

\(\int \frac{dx}{\sqrt{1-e^{2x}}} = \int \frac{1}{\sqrt{1-u^2}} \frac{du}{u}\)

\(= \int \frac{1}{u \sqrt{1-u^2}} \frac{du}{u}\)

\(= -\mathrm{sech}^{-1}|u| + C\)

\(= -\mathrm{sech}^{-1}|e^x| + C\)

C) Latihan Soal

1. Soal ETS 2024
Dapatkan turunan dari

\(f(x) = \frac{e^x+ \ln x}{\sinh 3x}\)

Pembahasan

Misalkan \(u=e^{x}+ \ln x\) dan \(v= \sinh 3x\), sehingga \(u'=e^x + \frac{1}{x} dan v'=3\cosh 3x\)
Dengan rumus turunan fungsi pecahan, diperoleh

\(f'(x)=\frac{u'v-uv'}{v^2}\)

\(=\frac{(e^x+\frac{1}{x})\sinh 3x- (e^x+ \ln x)3\cosh 3x}{\sinh^2 3x}\)

\(=\frac{e^x \sinh 3x+\frac{1}{x}\sinh 3x-3e^x\cosh 3x-3\ln x \cosh 3x}{\sinh^2 3x}\)

Jadi, turunan dari fungsi tersebut adalah

\(f'(x)=\frac{e^x \sinh 3x+\frac{1}{x}\sinh 3x-3e^x\cosh 3x-3\ln x \cosh 3x}{\sinh^2 3x}\)

atau

\(= \frac{xe^x \sinh 3x + \sinh 3x - 3xe^x \cosh 3x - 3x \ln x cosh 3x}{x \sinh^2 3x} \)

2. Soal ETS 2021
Dapatkan \(\frac{dy}{dx}\) dari

\(y=\ln(\cosh^{-1}x)\)

Pembahasan

\(\frac{dy}{dx}=\frac{d}{dx}[\ln(\cosh^{-1}x)]\)

=\(\frac{1}{\cosh^{-1}x}\frac{d}{dx}[\cosh^{-1}x]\)

=\(\frac{1}{\cosh^{-1}x}\cdot\frac{1}{\sqrt{x^2-1}}\)

3. Soal kuis Dapatkan bahwa

\(\tanh^{-1}x = \frac{1}{2} \ln (\frac{1+x}{1-x}); -1 < x < 1 \)

Pembahasan:

Dapatkan \(y = \tanh^{-1}x, \) sehingga \(x = \tanh y\)

\( x = \frac{\sinh y}{\cosh y} \)
\(= \frac{\frac{e^y - e^{-y}}{2}}{\frac{e^y+e^{-y}}{2}}\)
\(= \frac{e^y - e^{-y}}{e^y + e^{-y}}\)
\(= \frac{e^{2y} - 1}{e^{2y} + 1}\)

\(xe^{2y}+x = e^{2y}- 1 \)
\(e^{2y} - xe^{2y} = 1 + x\)
\(e^{2y}(1-x) = 1+x\)
\(e^{2y} = \frac{1+x}{1-x}\)
\(2y = \ln(\frac{1+x}{1-x})\)
\(y = \frac{1}{2} \ln(\frac{1+x}{1-x})\)

4. Soal Kuis Selesaikan \(\int e^{-x} \tanh(2-e^{-x});dx \)
Hint 1: Substitusi \(u=2-e^{-x}\)

Pembahasan:

Misalkan \(u=2-e{-x}\), sehingga \(du=e^{-x} dx\)

\(\int e^{-x} \tanh(2-e^{-x});dx = \int \tanh(u) du\)

Misalkan \(a=\mathrm{sech} {u},\) sehingga \(da=-\mathrm{sech} {u} \tanh u du\) ⇔ \(\frac{da}{-\mathrm{sech}{u}} = \tanh u du\)

\(\int \tanh(u) du = \int \frac{da}{-a}\)

\(= -\ln |a| + C\)

\(= -\ln |\mathrm{sech} {u}| + C\)

\(= -\ln |\mathrm{sech}(2-e^{-x})| + C\)

\(= -\ln \left(\frac{1}{\sqrt{1-(2-e^{-x})^2}}\right) + C\)

\(= \frac{1}{2} \ln(1-(2-e^{-x})^2) + C\)

\(= \frac{1}{2} \ln(1-(4-4e^{-x}+e^{-2x})) + C\)

\(= \frac{1}{2} \ln(e^{-2x}-4e^{-x}+3) + C\)

\(= \frac{1}{2} \ln((e^{-x}-3)(e^{-x}-1)) + C\)

\(= \frac{1}{2} \ln(e^{-x}-3) + \frac{1}{2} \ln(e^{-x}-1) + C\)

5. Hitung integral

\(\int \frac{1}{\sqrt{9x^2-25}};dx \)

Hint 1: Substitusi \(x=5u\)

Pembahasan:

Misalkan \(x=5u,\) sehingga \(dx=5du\)

\(\int \frac{1}{\sqrt{9x^2-25}};dx = 5\int \frac{1}{\sqrt{9(5u)^2-25}};du\)

\(= 5\int \frac{1}{\sqrt{225u^2-25}};du\)

\(= 5\int \frac{1}{\sqrt{25(9u^2-1)}};du\)

\(= \int \frac{1}{\sqrt{9u^2-1}};du\)

Misalkan \(u=\frac{1}{3} \sec v,\) sehingga \(du=\frac{1}{3} \sec v \tan v dv\)

\(= \int \frac{1}{\sqrt{9(\frac{1}{3} \sec v)^2-1}} \cdot \frac{1}{3} \sec v \tan v dv\)

\(= \int \frac{1}{\sqrt{\sec^2 v-1}} \cdot \frac{1}{3} \sec v \tan v dv\)

\(= \int \frac{1}{\tan v} \cdot \frac{1}{3} \sec v \tan v dv\)

\(= \frac{1}{3} \int \sec v dv\)

\(= \frac{1}{3} \ln |\sec v + \tan v| + C\)

\(= \frac{1}{3} \ln |\sec(\frac{1}{3} \sec^{-1}(3u)) + \tan(\frac{1}{3} \sec^{-1}(3u))| + C\)

\(= \frac{1}{3} \ln |\sec(\sec^{-1}(3u)) + \tan(\sec^{-1}(3u))| + C\)

\(= \frac{1}{3} \ln |3u + \sqrt{9u^2-1}| + C\)

\(= \frac{1}{3} \ln |3(\frac{x}{5}) + \sqrt{9(\frac{x}{5})^2-1}| + C\)

\(= \frac{1}{3} \ln |3(\frac{x}{5}) + \sqrt{\frac{9x^2}{25}-1}| + C\)

\(= \frac{1}{3} \ln |3(\frac{x}{5}) + \sqrt{\frac{9x^2-25}{25}}| + C\)

\(= \frac{1}{3} \ln |3x + \sqrt{9x^2-25}| - \frac{1}{3} \ln 5 + C\)

6. Hitung integral

\(\int \sqrt{\tanh x} \mathrm{sech}^2{x};dx \)

Hint 1: Substitusi \(u = \tanh x\)

Pembahasan:

Misalkan \(u= \tanh x,\) sehingga \(du= \mathrm{sech}^2x ;dx\)

\(\int \sqrt{\tanh x}\mathrm{sech}^2x;dx = \int \sqrt{u} du\)

\(= \frac{2}{3} u^{\frac{3}{2}} + C\)

\(= \frac{2}{3} (\tanh x)^{\frac{3}{2}} + C\)