Prove ∫ sec n ( x ) d x = tan ( x ) ⋅ sec n − 2 ( x ) n − 1 + n − 2 n − 1 ∫ sec n − 2 ( x ) d x {\displaystyle \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}
∫ sec n ( x ) d x = ∫ sec 2 ( x ) ⋅ sec n − 2 ( x ) d x {\displaystyle \int _{}^{}\sec ^{n}(x)dx=\int _{}^{}\sec ^{2}(x)\cdot \sec ^{n-2}(x)dx}
u = ln ( x ) n d v = 1 d x d u = 1 d x v = x {\displaystyle {\begin{aligned}&u=\ln(x)^{n}\quad dv=1dx\\[2ex]&du=1dx\quad v=x\\[2ex]\end{aligned}}}
∫ ( ln ( x ) n ) d x = x ln ( x ) n − ∫ ( ( x n ln ( x ) n − 1 x ) ) d x = x ln ( x ) n − ∫ ( n ln ( x ) n − 1 ) d x = x ln ( x ) n − n ∫ ( ln ( x ) n − 1 ) d x {\displaystyle {\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}}}
![{\displaystyle {\begin{aligned}&u=\ln(x)^{n}\quad dv=1dx\\[2ex]&du=1dx\quad v=x\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5fd34bb0856f51b99669b80b874ddcf377bc75f4)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/19be1d5890fa5a4661b92c41d3728d2463eb1083)