5.4 Indefinite Integrals and the Net Change Theorem/41: Difference between revisions
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\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] | \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{(t)}\bigg|_{0}^{\frac{1}{\sqrt{3}}} \\[2ex] | ||
&=\arctan\left(\frac{1}{\sqrt{3}}\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> | ||
Latest revision as of 19:41, 21 September 2022
![{\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}}}\\[2ex]&=\arctan \left({\frac {1}{\sqrt {3}}}\right)-\arctan(0)\\[2ex]&={\frac {\pi }{6}}-0\\[2ex]&={\frac {\pi }{6}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/094d08285abbfc9a04df77ce69f091702bf79da4)