Integral dari $$$\sin{\left(x \right)} \cos^{3}{\left(x \right)}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \sin{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$.
Solusi
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
$${\color{red}{\int{\sin{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- u^{3}\right)d u}}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = u^{3}$$$:
$${\color{red}{\int{\left(- u^{3}\right)d u}}} = {\color{red}{\left(- \int{u^{3} d u}\right)}}$$
Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=3$$$:
$$- {\color{red}{\int{u^{3} d u}}}=- {\color{red}{\frac{u^{1 + 3}}{1 + 3}}}=- {\color{red}{\left(\frac{u^{4}}{4}\right)}}$$
Ingat bahwa $$$u=\cos{\left(x \right)}$$$:
$$- \frac{{\color{red}{u}}^{4}}{4} = - \frac{{\color{red}{\cos{\left(x \right)}}}^{4}}{4}$$
Oleh karena itu,
$$\int{\sin{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\cos^{4}{\left(x \right)}}{4}$$
Tambahkan konstanta integrasi:
$$\int{\sin{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\cos^{4}{\left(x \right)}}{4}+C$$
Jawaban
$$$\int \sin{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = - \frac{\cos^{4}{\left(x \right)}}{4} + C$$$A