7.1 Integration By Parts/49

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

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

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

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