7.1 Integration By Parts/50

Prove $\int _{}^{}\sec ^{n}(x)dx={\frac {\tan(x)\cdot \sec ^{n-2}(x)}{n-1}}+{\frac {n-2}{n-1}}\int _{}^{}\sec ^{n-2}(x)dx$

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{}^{} \sec^{n}(x)dx = \int_{}^{} \sec^{2}(x) \cdot \sec^{n-2}(x) \dx }

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

${\begin{aligned}\int _{}^{}\left(\ln(x)^{n}\right)dx&=x\ln(x)^{n}-\int _{}^{}\left((x{\frac {n\ln(x)^{n-1}}{x}})\right)dx\\[2ex]&=x\ln(x)^{n}-\int _{}^{}\left(n\ln(x)^{n-1}\right)dx\\[2ex]&=x\ln(x)^{n}-n\int _{}^{}\left(\ln(x)^{n-1}\right)dx\\[2ex]\end{aligned}}$