Integral dari $$$\cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)}\, dx$$$.
Solusi
Misalkan $$$u=\cot{\left(\frac{x}{2} \right)}$$$.
Kemudian $$$du=\left(\cot{\left(\frac{x}{2} \right)}\right)^{\prime }dx = - \frac{\csc^{2}{\left(\frac{x}{2} \right)}}{2} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\csc^{2}{\left(\frac{x}{2} \right)} dx = - 2 du$$$.
Integralnya menjadi
$${\color{red}{\int{\cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)} d x}}} = {\color{red}{\int{\left(- 2 u^{2}\right)d u}}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-2$$$ dan $$$f{\left(u \right)} = u^{2}$$$:
$${\color{red}{\int{\left(- 2 u^{2}\right)d u}}} = {\color{red}{\left(- 2 \int{u^{2} d u}\right)}}$$
Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=2$$$:
$$- 2 {\color{red}{\int{u^{2} d u}}}=- 2 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- 2 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Ingat bahwa $$$u=\cot{\left(\frac{x}{2} \right)}$$$:
$$- \frac{2 {\color{red}{u}}^{3}}{3} = - \frac{2 {\color{red}{\cot{\left(\frac{x}{2} \right)}}}^{3}}{3}$$
Oleh karena itu,
$$\int{\cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)} d x} = - \frac{2 \cot^{3}{\left(\frac{x}{2} \right)}}{3}$$
Tambahkan konstanta integrasi:
$$\int{\cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)} d x} = - \frac{2 \cot^{3}{\left(\frac{x}{2} \right)}}{3}+C$$
Jawaban
$$$\int \cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)}\, dx = - \frac{2 \cot^{3}{\left(\frac{x}{2} \right)}}{3} + C$$$A