7.1 Integration By Parts/49

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

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

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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{align} }