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(0) | &=\arctan{(t)}\bigg|_{0}^{\frac{1}{\sqrt{3}}} = \arctan\left(\frac{1}{\sqrt{3}}\right)-[\arctan(0)] \\[2ex] | ||
&=\frac{\pi}{6}-0=\frac{\pi}{6} | &=\frac{\pi}{6}-0=\frac{\pi}{6} | ||
Revision as of 16:32, 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}}}=\arctan \left({\frac {1}{\sqrt {3}}}\right)-[\arctan(0)]\\[2ex]&={\frac {\pi }{6}}-0={\frac {\pi }{6}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5e244893b0876e47813f37bc3f84b73caa71248f)