Integral dari $$$\frac{1}{\cos{\left(3 x \right)} + 1}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \frac{1}{\cos{\left(3 x \right)} + 1}\, dx$$$.
Solusi
Misalkan $$$u=3 x$$$.
Kemudian $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{3}$$$.
Integral tersebut dapat ditulis ulang sebagai
$${\color{red}{\int{\frac{1}{\cos{\left(3 x \right)} + 1} d x}}} = {\color{red}{\int{\frac{1}{3 \left(\cos{\left(u \right)} + 1\right)} d u}}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{3}$$$ dan $$$f{\left(u \right)} = \frac{1}{\cos{\left(u \right)} + 1}$$$:
$${\color{red}{\int{\frac{1}{3 \left(\cos{\left(u \right)} + 1\right)} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{\cos{\left(u \right)} + 1} d u}}{3}\right)}}$$
Tulis ulang kosinus menggunakan rumus sudut ganda $$$\cos\left( u \right)=2\cos^2\left(\frac{ u }{2}\right)-1$$$ dan sederhanakan:
$$\frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)} + 1} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{2 \cos^{2}{\left(\frac{u}{2} \right)}} d u}}}}{3}$$
Misalkan $$$v=\frac{u}{2}$$$.
Kemudian $$$dv=\left(\frac{u}{2}\right)^{\prime }du = \frac{du}{2}$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = 2 dv$$$.
Oleh karena itu,
$$\frac{{\color{red}{\int{\frac{1}{2 \cos^{2}{\left(\frac{u}{2} \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{\cos^{2}{\left(v \right)}} d v}}}}{3}$$
Tulis ulang integran dalam bentuk fungsi sekan.:
$$\frac{{\color{red}{\int{\frac{1}{\cos^{2}{\left(v \right)}} d v}}}}{3} = \frac{{\color{red}{\int{\sec^{2}{\left(v \right)} d v}}}}{3}$$
Integral dari $$$\sec^{2}{\left(v \right)}$$$ adalah $$$\int{\sec^{2}{\left(v \right)} d v} = \tan{\left(v \right)}$$$:
$$\frac{{\color{red}{\int{\sec^{2}{\left(v \right)} d v}}}}{3} = \frac{{\color{red}{\tan{\left(v \right)}}}}{3}$$
Ingat bahwa $$$v=\frac{u}{2}$$$:
$$\frac{\tan{\left({\color{red}{v}} \right)}}{3} = \frac{\tan{\left({\color{red}{\left(\frac{u}{2}\right)}} \right)}}{3}$$
Ingat bahwa $$$u=3 x$$$:
$$\frac{\tan{\left(\frac{{\color{red}{u}}}{2} \right)}}{3} = \frac{\tan{\left(\frac{{\color{red}{\left(3 x\right)}}}{2} \right)}}{3}$$
Oleh karena itu,
$$\int{\frac{1}{\cos{\left(3 x \right)} + 1} d x} = \frac{\tan{\left(\frac{3 x}{2} \right)}}{3}$$
Tambahkan konstanta integrasi:
$$\int{\frac{1}{\cos{\left(3 x \right)} + 1} d x} = \frac{\tan{\left(\frac{3 x}{2} \right)}}{3}+C$$
Jawaban
$$$\int \frac{1}{\cos{\left(3 x \right)} + 1}\, dx = \frac{\tan{\left(\frac{3 x}{2} \right)}}{3} + C$$$A