7.1 Integration By Parts/11: Difference between revisions
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u=1+16t^{2} \\[1ex] | u=1+16t^{2} \\[1ex] | ||
du= | du=32t dt \\[1ex] | ||
\frac{1}{32}du= | \frac{1}{32}du=t dt | ||
\end{align} | \end{align} | ||
Revision as of 05:04, 29 November 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 \int \text {arctan(4t)}dt }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int {\text{arctan(4t)}}dt={\text{tarctan(4t)}}-4\int {\frac {t}{1+16t^{2}}}dt\\[1ex]u=1+16t^{2}\\[1ex]du=32tdt\\[1ex]{\frac {1}{32}}du=tdt\end{aligned}}}
&= \text {tarctan(4t)} \\[1ex]
&= \pi\left[\left(4(2)-(2)^2+\frac{1}{12}(2)^3\right)-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[2ex]
&= \pi\left[\left(8-4+\frac{8}{12}\right)-\left(4-1+\frac{1}{12}\right)\right] \\[2ex]
&= \pi\left[4+\frac{8}{12}-3-\frac{1}{12}\right]= \pi\left[1+\frac{7}{12}\right] \\[2ex]
&= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex]
&= \frac{19\pi}{12}
\end{align} </math>