7.1 Integration By Parts/13: Difference between revisions
Jump to navigation
Jump to search
Andy Arevalo (talk | contribs) No edit summary |
Andy Arevalo (talk | contribs) No edit summary |
||
| Line 2: | Line 2: | ||
<math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex] | <math>u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex] | ||
&= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex] | &= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex] </math> <br><br> | ||
</math> <br><br> | |||
<math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math> | <math>du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)</math> | ||
Revision as of 03:01, 29 November 2022
Failed to parse (syntax error): {\displaystyle u = t \qquad dv = \sec^2\left(2t\right) \qquad & = \int\sec^2\left(2t\right)dt \\[2ex] &= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex] }
