7.1 Integration By Parts/50

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

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

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

${\begin{aligned}\int _{}^{}\sec ^{2}(x)\cdot \sec ^{n-2}(x)dx&=\sec ^{n-2}(x)\cdot \tan(x)-\int _{}^{}\left[(n-2)\sec ^{n-3}(x)\cdot \sec(x)\tan(x)\right]\cdot \tan(x)dx\\[2ex]&=sec^{n-2}(x)\cdot \tan(x)-(n-2)\int _{}^{}\left[\sec ^{n-2}(x)\cdot \tan ^{2}(x)\right]dx\\[2ex]&=sec^{n-2}(x)\cdot \tan(x)-(n-2)\int _{}^{}\sec ^{n-2}(x)\cdot [\sec ^{2}(x)-1]dx\\[2ex]&=x\ln(x)^{n}-n\int _{}^{}\left(\ln(x)^{n-1}\right)dx\\[2ex]\end{aligned}}$

sec^{n-2}(x) \cdot \tan(x) - (n-2)\int_{}^{} \left[\sec^{n-2}(x) \cdot [\sec^{2}(x)-1]dx