7.1 Integration By Parts/50
< 7.1 Integration By Parts
Jump to navigation
Jump to search
Revision as of 18:23, 29 November 2022 by Ricardom59381@students.laalliance.org (talk | contribs) (Created page with "Prove <math> \int_{}^{} \sec^{n}x = \frac{\tanx \cdot \sec^{n-2}x}{n-1} + \frac{n-2}{n-1} \int_{}^{} \sec^{n-2}xdx </math> <math> \int_{}^{} \left(\ln(x)^{n}\right)dx </math> <math> \begin{align} &u = \ln(x)^{n} \quad dv= 1dx \\[2ex] &du =1dx \quad v=x \\[2ex] \end{align} </math> <math> \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(...")
Prove Failed to parse (unknown function "\tanx"): {\displaystyle \int_{}^{} \sec^{n}x = \frac{\tanx \cdot \sec^{n-2}x}{n-1} + \frac{n-2}{n-1} \int_{}^{} \sec^{n-2}xdx }
![{\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)
<math> \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]