Integral dari $$$7 \tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int 7 \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx$$$.
Solusi
Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=7$$$ dan $$$f{\left(x \right)} = \tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$:
$${\color{red}{\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\left(7 \int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}\right)}}$$
Keluarkan satu faktor tangen dan nyatakan sisanya dalam bentuk sekan, menggunakan rumus $$$\tan^2\left(x \right)=\sec^2\left(x \right)-1$$$:
$$7 {\color{red}{\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = 7 {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}}$$
Misalkan $$$u=\sec{\left(x \right)}$$$.
Kemudian $$$du=\left(\sec{\left(x \right)}\right)^{\prime }dx = \tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\tan{\left(x \right)} \sec{\left(x \right)} dx = du$$$.
Integral tersebut dapat ditulis ulang sebagai
$$7 {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}} = 7 {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$
Integralkan suku demi suku:
$$7 {\color{red}{\int{\left(u^{2} - 1\right)d u}}} = 7 {\color{red}{\left(- \int{1 d u} + \int{u^{2} d u}\right)}}$$
Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:
$$7 \int{u^{2} d u} - 7 {\color{red}{\int{1 d u}}} = 7 \int{u^{2} d u} - 7 {\color{red}{u}}$$
Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=2$$$:
$$- 7 u + 7 {\color{red}{\int{u^{2} d u}}}=- 7 u + 7 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- 7 u + 7 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Ingat bahwa $$$u=\sec{\left(x \right)}$$$:
$$- 7 {\color{red}{u}} + \frac{7 {\color{red}{u}}^{3}}{3} = - 7 {\color{red}{\sec{\left(x \right)}}} + \frac{7 {\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$
Oleh karena itu,
$$\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{7 \sec^{3}{\left(x \right)}}{3} - 7 \sec{\left(x \right)}$$
Sederhanakan:
$$\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{7 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3}$$
Tambahkan konstanta integrasi:
$$\int{7 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{7 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3}+C$$
Jawaban
$$$\int 7 \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx = \frac{7 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3} + C$$$A