7.1 Integration By Parts/13: Difference between revisions

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<math>\int t\sec^2\left(2t\right) dt </math>
<math>\int t\sec^2\left(2t\right) dt </math>


<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] </math>
&= \frac{1}{2}\int\sec^2\left(u\right)du \\[2ex] </math>
<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 03:02, 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] }