7.1 Integration By Parts/50

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

$\int _{}^{}\sec ^{n}(x)dx=\int _{}^{}\sec ^{2}(x)\cdot \sec ^{n-2}(x)dx$

${\begin{aligned}&u=\sec ^{n-2}(x)\quad dv=\sec ^{2}(x)dx\\[2ex]&du=(n-2)\sec ^{n-3}\cdot \sec(x)\cdot \tan(x)dx\quad v=\tan(x)\\[2ex]\end{aligned}}$

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