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

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

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int_{}^{} (\sec^{2}x)(\tan^{n-2}x)dx = \tan^{n-2}(x) \cdot \tan(x) - \int_{}^{} (n-2)\tan^{n-3}(x)\sec^{2} \cdot \tan(x)dx \\[2ex] = \tan^{n-1}(x) - \int_{}^{} (n-2)\tan^{n-2}(x)\sec^{2}dx \\[2ex] \tan^{n-1}(x) - \int_{}^{} (n-2)\tan^{n-2}(x)\sec^{2}dx = \int_{}^{} (\sec^{2}x)(\tan^{n-2}x)dx \\[2ex] <math> \begin{align} &+(n-2)\int_{}^{} \sec^{2}(x)dx \quad &&&+(n-2)\int_{}^{} \sec^{2}(x)dx \end{align} }

\end{align} </math>

Note: ${\begin{aligned}\tan ^{2}(x)=\sec ^{2}(x)-1\end{aligned}}$