Integral dari $$$\frac{\sin{\left(x \right)}}{\cos{\left(2 x \right)}}$$$

Temukan $$$\int \frac{\sin{\left(x \right)}}{\cos{\left(2 x \right)}}\, dx$$$.

Tulis ulang kosinus menggunakan rumus sudut ganda $$$\cos{\left(2 x \right)} = 2 \cos^{2}{\left(x \right)} - 1$$$:

$${\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(2 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(x \right)}}{2 \cos^{2}{\left(x \right)} - 1} d x}}}$$

Misalkan $$$u=\cos{\left(x \right)}$$$.

Kemudian $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\sin{\left(x \right)} dx = - du$$$.

Integralnya menjadi

$${\color{red}{\int{\frac{\sin{\left(x \right)}}{2 \cos^{2}{\left(x \right)} - 1} d x}}} = {\color{red}{\int{\left(- \frac{1}{2 u^{2} - 1}\right)d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = \frac{1}{2 u^{2} - 1}$$$:

$${\color{red}{\int{\left(- \frac{1}{2 u^{2} - 1}\right)d u}}} = {\color{red}{\left(- \int{\frac{1}{2 u^{2} - 1} d u}\right)}}$$

Lakukan dekomposisi pecahan parsial (langkah-langkah dapat dilihat di »):

$$- {\color{red}{\int{\frac{1}{2 u^{2} - 1} d u}}} = - {\color{red}{\int{\left(- \frac{1}{2 \left(\sqrt{2} u + 1\right)} + \frac{1}{2 \left(\sqrt{2} u - 1\right)}\right)d u}}}$$

Integralkan suku demi suku:

$$- {\color{red}{\int{\left(- \frac{1}{2 \left(\sqrt{2} u + 1\right)} + \frac{1}{2 \left(\sqrt{2} u - 1\right)}\right)d u}}} = - {\color{red}{\left(\int{\frac{1}{2 \left(\sqrt{2} u - 1\right)} d u} - \int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \frac{1}{\sqrt{2} u - 1}$$$:

$$\int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - {\color{red}{\int{\frac{1}{2 \left(\sqrt{2} u - 1\right)} d u}}} = \int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{2} u - 1} d u}}{2}\right)}}$$

Misalkan $$$v=\sqrt{2} u - 1$$$.

Kemudian $$$dv=\left(\sqrt{2} u - 1\right)^{\prime }du = \sqrt{2} du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = \frac{\sqrt{2} dv}{2}$$$.

Oleh karena itu,

$$\int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{\sqrt{2} u - 1} d u}}}}{2} = \int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - \frac{{\color{red}{\int{\frac{\sqrt{2}}{2 v} d v}}}}{2}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{\sqrt{2}}{2}$$$ dan $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$\int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - \frac{{\color{red}{\int{\frac{\sqrt{2}}{2 v} d v}}}}{2} = \int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - \frac{{\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{v} d v}}{2}\right)}}}{2}$$

Integral dari $$$\frac{1}{v}$$$ adalah $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{v} d v}}}}{4} = \int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} - \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$

Ingat bahwa $$$v=\sqrt{2} u - 1$$$:

$$- \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} + \int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u} = - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(\sqrt{2} u - 1\right)}}}\right| \right)}}{4} + \int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \frac{1}{\sqrt{2} u + 1}$$$:

$$- \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + {\color{red}{\int{\frac{1}{2 \left(\sqrt{2} u + 1\right)} d u}}} = - \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{2} u + 1} d u}}{2}\right)}}$$

Misalkan $$$v=\sqrt{2} u + 1$$$.

Kemudian $$$dv=\left(\sqrt{2} u + 1\right)^{\prime }du = \sqrt{2} du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = \frac{\sqrt{2} dv}{2}$$$.

Integral tersebut dapat ditulis ulang sebagai

$$- \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{{\color{red}{\int{\frac{1}{\sqrt{2} u + 1} d u}}}}{2} = - \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{{\color{red}{\int{\frac{\sqrt{2}}{2 v} d v}}}}{2}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{\sqrt{2}}{2}$$$ dan $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$- \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{{\color{red}{\int{\frac{\sqrt{2}}{2 v} d v}}}}{2} = - \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{{\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{v} d v}}{2}\right)}}}{2}$$

Integral dari $$$\frac{1}{v}$$$ adalah $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{v} d v}}}}{4} = - \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$

Ingat bahwa $$$v=\sqrt{2} u + 1$$$:

$$- \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = - \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} u - 1}\right| \right)}}{4} + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(\sqrt{2} u + 1\right)}}}\right| \right)}}{4}$$

Ingat bahwa $$$u=\cos{\left(x \right)}$$$:

$$- \frac{\sqrt{2} \ln{\left(\left|{-1 + \sqrt{2} {\color{red}{u}}}\right| \right)}}{4} + \frac{\sqrt{2} \ln{\left(\left|{1 + \sqrt{2} {\color{red}{u}}}\right| \right)}}{4} = - \frac{\sqrt{2} \ln{\left(\left|{-1 + \sqrt{2} {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{4} + \frac{\sqrt{2} \ln{\left(\left|{1 + \sqrt{2} {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{4}$$

Oleh karena itu,

$$\int{\frac{\sin{\left(x \right)}}{\cos{\left(2 x \right)}} d x} = - \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} \cos{\left(x \right)} - 1}\right| \right)}}{4} + \frac{\sqrt{2} \ln{\left(\left|{\sqrt{2} \cos{\left(x \right)} + 1}\right| \right)}}{4}$$

Sederhanakan:

$$\int{\frac{\sin{\left(x \right)}}{\cos{\left(2 x \right)}} d x} = \frac{\sqrt{2} \left(- \ln{\left(\left|{\sqrt{2} \cos{\left(x \right)} - 1}\right| \right)} + \ln{\left(\left|{\sqrt{2} \cos{\left(x \right)} + 1}\right| \right)}\right)}{4}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\sin{\left(x \right)}}{\cos{\left(2 x \right)}} d x} = \frac{\sqrt{2} \left(- \ln{\left(\left|{\sqrt{2} \cos{\left(x \right)} - 1}\right| \right)} + \ln{\left(\left|{\sqrt{2} \cos{\left(x \right)} + 1}\right| \right)}\right)}{4}+C$$

$$$\int \frac{\sin{\left(x \right)}}{\cos{\left(2 x \right)}}\, dx = \frac{\sqrt{2} \left(- \ln\left(\left|{\sqrt{2} \cos{\left(x \right)} - 1}\right|\right) + \ln\left(\left|{\sqrt{2} \cos{\left(x \right)} + 1}\right|\right)\right)}{4} + C$$$A