Integral dari $$$\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$

Temukan $$$\int \frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx$$$.

Kalikan pembilang dan penyebut dengan $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ dan ubah $$$\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$ menjadi $$$\frac{1}{\tan^{2}{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}}$$

Kalikan pembilang dan penyebut dengan $$$\cos^{2}{\left(x \right)}$$$ dan ubah $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ menjadi $$$\sec^{2}{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\cos^{4}{\left(x \right)} \sec^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}}$$

Tulis ulang kosinus dalam bentuk tangen menggunakan rumus $$$\cos^{2}{\left(x \right)}=\frac{1}{\tan^{2}{\left(x \right)} + 1}$$$:

$${\color{red}{\int{\frac{\cos^{4}{\left(x \right)} \sec^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\left(\tan^{2}{\left(x \right)} + 1\right)^{2} \tan^{2}{\left(x \right)}} d x}}}$$

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 $$$\sec^{2}{\left(x \right)} dx = du$$$.

Jadi,

$${\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\left(\tan^{2}{\left(x \right)} + 1\right)^{2} \tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{u^{2} \left(u^{2} + 1\right)^{2}} d u}}}$$

Lakukan dekomposisi pecahan parsial (langkah-langkah dapat dilihat di »):

$${\color{red}{\int{\frac{1}{u^{2} \left(u^{2} + 1\right)^{2}} d u}}} = {\color{red}{\int{\left(- \frac{1}{u^{2} + 1} - \frac{1}{\left(u^{2} + 1\right)^{2}} + \frac{1}{u^{2}}\right)d u}}}$$

Integralkan suku demi suku:

$${\color{red}{\int{\left(- \frac{1}{u^{2} + 1} - \frac{1}{\left(u^{2} + 1\right)^{2}} + \frac{1}{u^{2}}\right)d u}}} = {\color{red}{\left(\int{\frac{1}{u^{2}} d u} - \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-2$$$:

$$- \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{u^{-2} d u}}}=- \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\left(- u^{-1}\right)}}=- \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\left(- \frac{1}{u}\right)}}$$

Integral dari $$$\frac{1}{u^{2} + 1}$$$ adalah $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$- \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} - \frac{1}{u} = - \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} - {\color{red}{\operatorname{atan}{\left(u \right)}}} - \frac{1}{u}$$

Untuk menghitung integral $$$\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}$$$, terapkan integrasi parsial $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$ pada integral $$$\int{\frac{1}{u^{2} + 1} d u}$$$.

Misalkan $$$\operatorname{c}=\frac{1}{u^{2} + 1}$$$ dan $$$\operatorname{dv}=du$$$.

Maka $$$\operatorname{dc}=\left(\frac{1}{u^{2} + 1}\right)^{\prime }du=- \frac{2 u}{\left(u^{2} + 1\right)^{2}} du$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{1 d u}=u$$$ (langkah-langkah dapat dilihat di »).

Integral tersebut dapat ditulis ulang sebagai

$$\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}=\frac{1}{u^{2} + 1} \cdot u-\int{u \cdot \left(- \frac{2 u}{\left(u^{2} + 1\right)^{2}}\right) d u}=\frac{u}{u^{2} + 1} - \int{\left(- \frac{2 u^{2}}{\left(u^{2} + 1\right)^{2}}\right)d u}$$

Keluarkan konstanta:

$$\frac{u}{u^{2} + 1} - \int{\left(- \frac{2 u^{2}}{\left(u^{2} + 1\right)^{2}}\right)d u}=\frac{u}{u^{2} + 1} + 2 \int{\frac{u^{2}}{\left(u^{2} + 1\right)^{2}} d u}$$

Tulis ulang pembilang integran sebagai $$$u^{2}=u^{2}{\color{red}{+1}}{\color{red}{-1}}$$$ dan pisahkan:

$$\frac{u}{u^{2} + 1} + 2 \int{\frac{u^{2}}{\left(u^{2} + 1\right)^{2}} d u}=\frac{u}{u^{2} + 1} + 2 \int{\left(- \frac{1}{\left(u^{2} + 1\right)^{2}} + \frac{u^{2} + 1}{\left(u^{2} + 1\right)^{2}}\right)d u}=\frac{u}{u^{2} + 1} + 2 \int{\left(\frac{1}{u^{2} + 1} - \frac{1}{\left(u^{2} + 1\right)^{2}}\right)d u}$$

Pisahkan integral-integral tersebut:

$$\frac{u}{u^{2} + 1} + 2 \int{\left(\frac{1}{u^{2} + 1} - \frac{1}{\left(u^{2} + 1\right)^{2}}\right)d u}=\frac{u}{u^{2} + 1} - 2 \int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u} + 2 \int{\frac{1}{u^{2} + 1} d u}$$

Dengan demikian, kita memperoleh persamaan linear sederhana berikut terhadap integral:

$$\int{\frac{1}{u^{2} + 1} d u}=\frac{u}{u^{2} + 1} + 2 \int{\frac{1}{u^{2} + 1} d u} - 2 {\color{red}{\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}}}$$

Dengan menyelesaikannya, kita memperoleh bahwa

$$\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}=\frac{u}{2 \left(u^{2} + 1\right)} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{2}$$

Oleh karena itu,

$$- \operatorname{atan}{\left(u \right)} - {\color{red}{\int{\frac{1}{\left(u^{2} + 1\right)^{2}} d u}}} - \frac{1}{u} = - \operatorname{atan}{\left(u \right)} - {\color{red}{\left(\frac{u}{2 \left(u^{2} + 1\right)} + \frac{\int{\frac{1}{u^{2} + 1} d u}}{2}\right)}} - \frac{1}{u}$$

Integral dari $$$\frac{1}{u^{2} + 1}$$$ adalah $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$- \frac{u}{2 \left(u^{2} + 1\right)} - \operatorname{atan}{\left(u \right)} - \frac{{\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{2} - \frac{1}{u} = - \frac{u}{2 \left(u^{2} + 1\right)} - \operatorname{atan}{\left(u \right)} - \frac{{\color{red}{\operatorname{atan}{\left(u \right)}}}}{2} - \frac{1}{u}$$

Ingat bahwa $$$u=\tan{\left(x \right)}$$$:

$$- \frac{3 \operatorname{atan}{\left({\color{red}{u}} \right)}}{2} - {\color{red}{u}}^{-1} - \frac{{\color{red}{u}} \left(1 + {\color{red}{u}}^{2}\right)^{-1}}{2} = - \frac{3 \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)}}{2} - {\color{red}{\tan{\left(x \right)}}}^{-1} - \frac{{\color{red}{\tan{\left(x \right)}}} \left(1 + {\color{red}{\tan{\left(x \right)}}}^{2}\right)^{-1}}{2}$$

Oleh karena itu,

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 \operatorname{atan}{\left(\tan{\left(x \right)} \right)}}{2} - \frac{1}{\tan{\left(x \right)}} - \frac{\tan{\left(x \right)}}{2 \left(\tan^{2}{\left(x \right)} + 1\right)}$$

Sederhanakan:

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\tan{\left(x \right)}}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x} = - \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\tan{\left(x \right)}}+C$$

$$$\int \frac{\cos^{4}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx = \left(- \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\tan{\left(x \right)}}\right) + C$$$A