Integral dari $$$\frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}}\, dx$$$.
Solusi
Misalkan $$$u=2 x$$$.
Kemudian $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{2}$$$.
Oleh karena itu,
$${\color{red}{\int{\frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{2 \sin{\left(2 u \right)}} d u}}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \frac{\sin{\left(u \right)}}{\sin{\left(2 u \right)}}$$$:
$${\color{red}{\int{\frac{\sin{\left(u \right)}}{2 \sin{\left(2 u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{\sin{\left(u \right)}}{\sin{\left(2 u \right)}} d u}}{2}\right)}}$$
Tulis ulang integran:
$$\frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{\sin{\left(2 u \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\frac{1}{2 \cos{\left(u \right)}} d u}}}}{2}$$
Tulis ulang kosinus dalam bentuk sinus menggunakan rumus $$$\cos\left( u \right)=\sin\left( u + \frac{\pi}{2}\right)$$$ dan kemudian tulis ulang sinus menggunakan rumus sudut rangkap $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$:
$$\frac{{\color{red}{\int{\frac{1}{2 \cos{\left(u \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\frac{1}{4 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2}$$
Kalikan pembilang dan penyebut dengan $$$\sec^2\left(\frac{ u }{2} + \frac{\pi}{4} \right)$$$:
$$\frac{{\color{red}{\int{\frac{1}{4 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{4 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2}$$
Misalkan $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$.
Kemudian $$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$.
Integral tersebut dapat ditulis ulang sebagai
$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{4 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{2}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(v \right)} = \frac{1}{v}$$$:
$$\frac{{\color{red}{\int{\frac{1}{2 v} d v}}}}{2} = \frac{{\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}}{2}$$
Integral dari $$$\frac{1}{v}$$$ adalah $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$
Ingat bahwa $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{4}$$
Ingat bahwa $$$u=2 x$$$:
$$\frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{4} = \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{\left(2 x\right)}}}{2} \right)}}\right| \right)}}{4}$$
Oleh karena itu,
$$\int{\frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}} d x} = \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{4}$$
Tambahkan konstanta integrasi:
$$\int{\frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}} d x} = \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{4}+C$$
Jawaban
$$$\int \frac{\sin{\left(2 x \right)}}{\sin{\left(4 x \right)}}\, dx = \frac{\ln\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right|\right)}{4} + C$$$A