7.1 Integration By Parts/50

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

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

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

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \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_{}^{} \left[\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{align} }