Integral dari $$$\frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}}\, dx$$$.
Solusi
Kalikan pembilang dan penyebut dengan $$$\cos^{2}{\left(x \right)}$$$ dan ubah $$$\frac{\sin^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}$$$ menjadi $$$\tan^{2}{\left(x \right)}$$$:
$${\color{red}{\int{\frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\tan^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}}$$
Konversikan $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ menjadi $$$\sec^{2}{\left(x \right)}$$$:
$${\color{red}{\int{\frac{\tan^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\tan^{2}{\left(x \right)} \sec^{2}{\left(x \right)} d x}}}$$
Misalkan $$$u=\tan{\left(x \right)}$$$.
Kemudian $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\sec^{2}{\left(x \right)} dx = du$$$.
Integralnya menjadi
$${\color{red}{\int{\tan^{2}{\left(x \right)} \sec^{2}{\left(x \right)} d x}}} = {\color{red}{\int{u^{2} d u}}}$$
Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=2$$$:
$${\color{red}{\int{u^{2} d u}}}={\color{red}{\frac{u^{1 + 2}}{1 + 2}}}={\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Ingat bahwa $$$u=\tan{\left(x \right)}$$$:
$$\frac{{\color{red}{u}}^{3}}{3} = \frac{{\color{red}{\tan{\left(x \right)}}}^{3}}{3}$$
Oleh karena itu,
$$\int{\frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}} d x} = \frac{\tan^{3}{\left(x \right)}}{3}$$
Tambahkan konstanta integrasi:
$$\int{\frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}} d x} = \frac{\tan^{3}{\left(x \right)}}{3}+C$$
Jawaban
$$$\int \frac{\sin^{2}{\left(x \right)}}{\cos^{4}{\left(x \right)}}\, dx = \frac{\tan^{3}{\left(x \right)}}{3} + C$$$A