5.4 Indefinite Integrals and the Net Change Theorem/41: Difference between revisions
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<math>\begin{align}\int_{0}^\frac{1}\sqrt{3}\frac{t^2-1}{t^4-1} dt&=\int_{0}^\frac{1}\sqrt{3} \frac{(t^2-1)}{(t^2-1)(t^2+1)} dt=\int_{0}^\frac{1}\sqrt{3} \frac{1}{(t^2+1)}dt\\[2ex]&=\ | <math> | ||
\end{align}</math> | \begin{align} | ||
\int_{0}^\frac{1}\sqrt{3}\frac{t^2-1}{t^4-1} dt &= \int_{0}^\frac{1}\sqrt{3} \frac{(t^2-1)}{(t^2-1)(t^2+1)} dt=\int_{0}^\frac{1}\sqrt{3} \frac{1}{(t^2+1)}dt \\[2ex] | |||
&=\arctan{(t)}\bigg|_{0}^{\frac{1}{\sqrt{3}}} = \arctan(\frac{1}{\sqrt{3}})-[\arctan(0)^{-1}]\\[2ex] | |||
&=\frac{\pi}{6}-0=\frac{\pi}{6} | |||
\end{align} | |||
</math> | |||
Revision as of 16:32, 21 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {t^{2}-1}{t^{4}-1}}dt&=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {(t^{2}-1)}{(t^{2}-1)(t^{2}+1)}}dt=\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {1}{(t^{2}+1)}}dt\\[2ex]&=\arctan {(t)}{\bigg |}_{0}^{\frac {1}{\sqrt {3}}}=\arctan({\frac {1}{\sqrt {3}}})-[\arctan(0)^{-1}]\\[2ex]&={\frac {\pi }{6}}-0={\frac {\pi }{6}}\end{aligned}}}