5.4 Indefinite Integrals and the Net Change Theorem/41: Difference between revisions

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&=\left[\arctan\left(\frac{1}{\sqrt{3}}\right)\right]-[\arctan(0)] \\[2ex]
&=\left[\arctan\left(\frac{1}{\sqrt{3}}\right)\right]-[\arctan(0)] \\[2ex]


&=\frac{\pi}{6}-0=\frac{\pi}{6}
&=\frac{\pi}{6}-0 \\[2ex]
 
&=\frac{\pi}{6}


\end{align}
\end{align}
</math>
</math>

Revision as of 16:33, 21 September 2022