5.4 Indefinite Integrals and the Net Change Theorem/41 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/41" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:41:51Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/41" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

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Revision as of 19:41, 21 September 2022

2022-09-21T16:33:57Z

← Older revision		Revision as of 16:33, 21 September 2022
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	&=\arctan{(t)}\bigg\|_{0}^{\frac{1}{\sqrt{3}}} \\[2ex]		&=\arctan{(t)}\bigg\|_{0}^{\frac{1}{\sqrt{3}}} \\[2ex]
	&=~~\left[~~\arctan\left(\frac{1}{\sqrt{3}}\right)~~\right]~~-[\arctan(0)] \\[2ex]		&=\arctan\left(\frac{1}{\sqrt{3}}\right)-\arctan(0) \\[2ex]

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

2022-09-21T16:33:35Z

← Older revision		Revision as of 16:33, 21 September 2022
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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>

2022-09-21T16:33:15Z

← Older revision		Revision as of 16:33, 21 September 2022
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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\left(\frac{1}{\sqrt{3}}\right)-[\arctan(0)] \\[2ex]		&=\arctan{(t)}\bigg\|_{0}^{\frac{1}{\sqrt{3}}} \\[2ex]
			&=\left[\arctan\left(\frac{1}{\sqrt{3}}\right)\right]-[\arctan(0)] \\[2ex]

	&=\frac{\pi}{6}-0=\frac{\pi}{6}		&=\frac{\pi}{6}-0=\frac{\pi}{6}

2022-09-21T16:32:45Z

← Older revision		Revision as of 16:32, 21 September 2022
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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)~~^{-1}~~]\\[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}		&=\frac{\pi}{6}-0=\frac{\pi}{6}

2022-09-21T16:32:17Z

← Older revision		Revision as of 16:32, 21 September 2022
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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]&=\~~tan^{-1}~~{(t)}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\~~tan^{-1}~~(\frac{1}{\sqrt{3}})~~^{-1}~~-[~~tan~~(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}		<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>

2022-09-21T16:31:09Z

← Older revision		Revision as of 16:31, 21 September 2022
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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]&=\tan^{-1}{(t)}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\tan(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}		<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]&=\tan^{-1}{(t)}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\tan^{-1}(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}
	\end{align}</math>		\end{align}</math>

2022-09-21T16:30:51Z

← Older revision		Revision as of 16:30, 21 September 2022
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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]&=\tan^{-1}{t}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\tan(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}		<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]&=\tan^{-1}{(t)}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\tan(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}
	\end{align}</math>		\end{align}</math>

2022-09-21T16:30:42Z

← Older revision		Revision as of 16:30, 21 September 2022
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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]&=\tan^{-1}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\tan(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}		<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]&=\tan^{-1}{t}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\tan(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}
	\end{align}</math>		\end{align}</math>

2022-09-21T16:30:26Z

← Older revision		Revision as of 16:30, 21 September 2022
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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]&=tan^{-1}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}= tan(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}		<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]&=\tan^{-1}\bigg\|_{0}^{\frac{1}{\sqrt{3}}}=\tan(\frac{1}{\sqrt{3}})^{-1}-[tan(0)^{-1}]\\[2ex]&=\frac{\pi}{6}-0=\frac{\pi}{6}
	\end{align}</math>		\end{align}</math>