7.1 Integration By Parts/13: Difference between revisions
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<math> | <math> | ||
= \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c | \int t\sec^2\left(2t\right) dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c | ||
</math> | </math> | ||
Revision as of 05:40, 16 December 2022
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 du = dt \qquad v = \frac{1}{2}\tan\left(2t\right)}
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 \int t\sec^2\left(2t\right) dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{2}\int\tan\left(2t\right)dt = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\int\tan\left(u\right)du = \frac{1}{2}\tan\left(2t\right)-\frac{1}{4}\ln|\sec2t|+c }
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} & u=2t \\[2ex] & du=2dt \\[2ex] & \frac{du}{2}=dt \\[2ex] \end{align} }