7.1 Integration By Parts/49

Prove $\int _{}^{}\left(\tan ^{n}(x)\right)dx={\frac {\tan ^{n-1}x}{n-1}}-\int _{}^{}\left(\tan ^{n-2}x\right)dx$

Failed to parse (syntax error): {\displaystyle \int_{}^{} \left(\tan^{n}(x)\right)dx = \int_{}^{} \left((\tan^{2}x)(\tan^{n-2}x)\right)dx = \int_{}^{} (\sec^{2}(x)-1)\tan^{n-2}x dx &= \int_{}^{} (\sec^{2}x)(\tan^{n-2}x)-\tan^{n-2}x dx }

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

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