Integral dari $$$\sec^{5}{\left(u \right)}$$$

Temukan $$$\int \sec^{5}{\left(u \right)}\, du$$$.

Untuk integral $$$\int{\sec^{5}{\left(u \right)} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{\mu} \operatorname{dv} = \operatorname{\mu}\operatorname{v} - \int \operatorname{v} \operatorname{d\mu}$$$.

Misalkan $$$\operatorname{\mu}=\sec^{3}{\left(u \right)}$$$ dan $$$\operatorname{dv}=\sec^{2}{\left(u \right)} du$$$.

Maka $$$\operatorname{d\mu}=\left(\sec^{3}{\left(u \right)}\right)^{\prime }du=3 \tan{\left(u \right)} \sec^{3}{\left(u \right)} du$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{\sec^{2}{\left(u \right)} d u}=\tan{\left(u \right)}$$$ (langkah-langkah dapat dilihat di »).

Oleh karena itu,

$$\int{\sec^{5}{\left(u \right)} d u}=\sec^{3}{\left(u \right)} \cdot \tan{\left(u \right)}-\int{\tan{\left(u \right)} \cdot 3 \tan{\left(u \right)} \sec^{3}{\left(u \right)} d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} - \int{3 \tan^{2}{\left(u \right)} \sec^{3}{\left(u \right)} d u}$$

Keluarkan konstanta:

$$\tan{\left(u \right)} \sec^{3}{\left(u \right)} - \int{3 \tan^{2}{\left(u \right)} \sec^{3}{\left(u \right)} d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\tan^{2}{\left(u \right)} \sec^{3}{\left(u \right)} d u}$$

Terapkan rumus $$$\tan^{2}{\left(u \right)} = \sec^{2}{\left(u \right)} - 1$$$:

$$\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\tan^{2}{\left(u \right)} \sec^{3}{\left(u \right)} d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\left(\sec^{2}{\left(u \right)} - 1\right) \sec^{3}{\left(u \right)} d u}$$

Kembangkan:

$$\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\left(\sec^{2}{\left(u \right)} - 1\right) \sec^{3}{\left(u \right)} d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\left(\sec^{5}{\left(u \right)} - \sec^{3}{\left(u \right)}\right)d u}$$

Integral dari selisih adalah selisih dari integral:

$$\tan{\left(u \right)} \sec^{3}{\left(u \right)} - 3 \int{\left(\sec^{5}{\left(u \right)} - \sec^{3}{\left(u \right)}\right)d u}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} + 3 \int{\sec^{3}{\left(u \right)} d u} - 3 \int{\sec^{5}{\left(u \right)} d u}$$

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

$${\color{red}{\int{\sec^{5}{\left(u \right)} d u}}}=\tan{\left(u \right)} \sec^{3}{\left(u \right)} + 3 \int{\sec^{3}{\left(u \right)} d u} - 3 {\color{red}{\int{\sec^{5}{\left(u \right)} d u}}}$$

Dengan menyelesaikannya, kita memperoleh bahwa

$$\int{\sec^{5}{\left(u \right)} d u}=\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \int{\sec^{3}{\left(u \right)} d u}}{4}$$

Untuk integral $$$\int{\sec^{3}{\left(u \right)} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{\mu} \operatorname{dv} = \operatorname{\mu}\operatorname{v} - \int \operatorname{v} \operatorname{d\mu}$$$.

Misalkan $$$\operatorname{\mu}=\sec{\left(u \right)}$$$ dan $$$\operatorname{dv}=\sec^{2}{\left(u \right)} du$$$.

Maka $$$\operatorname{d\mu}=\left(\sec{\left(u \right)}\right)^{\prime }du=\tan{\left(u \right)} \sec{\left(u \right)} du$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{\sec^{2}{\left(u \right)} d u}=\tan{\left(u \right)}$$$ (langkah-langkah dapat dilihat di »).

Jadi,

$$\int{\sec^{3}{\left(u \right)} d u}=\sec{\left(u \right)} \cdot \tan{\left(u \right)}-\int{\tan{\left(u \right)} \cdot \tan{\left(u \right)} \sec{\left(u \right)} d u}=\tan{\left(u \right)} \sec{\left(u \right)} - \int{\tan^{2}{\left(u \right)} \sec{\left(u \right)} d u}$$

Terapkan rumus $$$\tan^{2}{\left(u \right)} = \sec^{2}{\left(u \right)} - 1$$$:

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

Kembangkan:

$$\tan{\left(u \right)} \sec{\left(u \right)} - \int{\left(\sec^{2}{\left(u \right)} - 1\right) \sec{\left(u \right)} d u}=\tan{\left(u \right)} \sec{\left(u \right)} - \int{\left(\sec^{3}{\left(u \right)} - \sec{\left(u \right)}\right)d u}$$

Integral dari selisih adalah selisih dari integral:

$$\tan{\left(u \right)} \sec{\left(u \right)} - \int{\left(\sec^{3}{\left(u \right)} - \sec{\left(u \right)}\right)d u}=\tan{\left(u \right)} \sec{\left(u \right)} + \int{\sec{\left(u \right)} d u} - \int{\sec^{3}{\left(u \right)} d u}$$

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

$${\color{red}{\int{\sec^{3}{\left(u \right)} d u}}}=\tan{\left(u \right)} \sec{\left(u \right)} + \int{\sec{\left(u \right)} d u} - {\color{red}{\int{\sec^{3}{\left(u \right)} d u}}}$$

Dengan menyelesaikannya, kita memperoleh bahwa

$$\int{\sec^{3}{\left(u \right)} d u}=\frac{\tan{\left(u \right)} \sec{\left(u \right)}}{2} + \frac{\int{\sec{\left(u \right)} d u}}{2}$$

Oleh karena itu,

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 {\color{red}{\int{\sec^{3}{\left(u \right)} d u}}}}{4} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 {\color{red}{\left(\frac{\tan{\left(u \right)} \sec{\left(u \right)}}{2} + \frac{\int{\sec{\left(u \right)} d u}}{2}\right)}}}{4}$$

Tulis ulang sekan sebagai $$$\sec\left(u\right)=\frac{1}{\cos\left(u\right)}$$$:

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\sec{\left(u \right)} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{8}$$

Tulis ulang kosinus dalam bentuk sinus menggunakan rumus $$$\cos\left(u\right)=\sin\left(u + \frac{\pi}{2}\right)$$$ dan kemudian tulis ulang sinus menggunakan rumus sudut rangkap $$$\sin\left(u\right)=2\sin\left(\frac{u}{2}\right)\cos\left(\frac{u}{2}\right)$$$:

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8}$$

Kalikan pembilang dan penyebut dengan $$$\sec^2\left(\frac{u}{2} + \frac{\pi}{4} \right)$$$:

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8}$$

Misalkan $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$.

Kemudian $$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$.

Dengan demikian,

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{v} d v}}}}{8}$$

Integral dari $$$\frac{1}{v}$$$ adalah $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{v} d v}}}}{8} = \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} + \frac{3 {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{8}$$

Ingat bahwa $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{3 \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8} = \frac{3 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}$$

Oleh karena itu,

$$\int{\sec^{5}{\left(u \right)} d u} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}$$

Tambahkan konstanta integrasi:

$$\int{\sec^{5}{\left(u \right)} d u} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}+C$$

$$$\int \sec^{5}{\left(u \right)}\, du = \left(\frac{3 \ln\left(\left|{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}\right|\right)}{8} + \frac{\tan{\left(u \right)} \sec^{3}{\left(u \right)}}{4} + \frac{3 \tan{\left(u \right)} \sec{\left(u \right)}}{8}\right) + C$$$A