7.1 Integration By Parts/49

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

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

Solving for $\int _{}^{}(\sec ^{2}x)(\tan ^{n-2}x)$

Failed to parse (unknown function "\qqquad"): {\displaystyle \begin{align} &u = \tan^{n-2}x \qqquad dv= 1dx \\[2ex] &du = (n-2)\tan^{n-3}x \cdot \sec(x)\tan(x)dx \quad v=x \\[2ex] \end{align} }

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