7.1 Integration By Parts/50

Prove $\int _{}^{}\sec ^{n}x={\frac {\tan(x)}{n-1}}$

$\int _{}^{}\left(\ln(x)^{n}\right)dx$

${\begin{aligned}&u=\ln(x)^{n}\quad dv=1dx\\[2ex]&du=1dx\quad v=x\\[2ex]\end{aligned}}$

<math> \begin{align}

\int_{}^{} \left(\ln(x)^{n}\right)dx &= x \ln(x)^{n} - \int_{}^{} \left((x \frac{n \ln(x)^{n-1}}{x}) \right)dx \\[2ex] &= x \ln(x)^{n} - \int_{}^{} \left(n \ln(x)^{n-1} \right)dx \\[2ex] &= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex]

\int_{}^{} \sec^{n}x = \frac{\tanx \sec^{n-2}x}{n-1} + \frac{n-2}{n-1} \int_{}^{} \sec^{n-2}xdx