5.4 Indefinite Integrals and the Net Change Theorem/17 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/17" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:39:59Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/17" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 19:39, 21 September 2022
(No difference)

2022-09-13T17:51:23Z

← Older revision		Revision as of 17:51, 13 September 2022
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	Or,		Or,


	<math>		<math>
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	\int\sec^2\alpha \,d\alpha = \tan{\alpha}+C		\int\sec^2\alpha \,d\alpha = \tan{\alpha}+C
	</math>		</math>


	Note: <math>\cos^2\alpha+sin^2\alpha=1</math>		Note: <math>\cos^2\alpha+sin^2\alpha=1</math>

2022-09-13T17:51:11Z

← Older revision		Revision as of 17:51, 13 September 2022
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	<math>		<math>

	\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{1}{cos^2\alpha}d\alpha =		\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{1}{cos^2\alpha}\,d\alpha =

	\int\sec^2\alpha \,d\alpha = \tan{alpha}+C		\int\sec^2\alpha \,d\alpha = \tan{\alpha}+C
	</math>		</math>

	Note: <math>\cos^2\alpha+sin^2\alpha=1</math>		Note: <math>\cos^2\alpha+sin^2\alpha=1</math>

2022-09-13T17:50:47Z

← Older revision		Revision as of 17:50, 13 September 2022
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	\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{1}{cos^2\alpha}d\alpha =		\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{1}{cos^2\alpha}d\alpha =

	\int\sec^2\alpha \,d\alpha =		\int\sec^2\alpha \,d\alpha = \tan{alpha}+C

	\tan{x}+C
	</math>		</math>

	Note: <math>\cos^2\alpha+sin^2\alpha=1</math>		Note: <math>\cos^2\alpha+sin^2\alpha=1</math>

2022-09-13T17:49:51Z

← Older revision		Revision as of 17:49, 13 September 2022
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	<math>		<math>

	\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha		\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{1}{cos^2\alpha}d\alpha =

	~~\cos^2x+sin^2x=1~~		\int\sec^2\alpha \,d\alpha =

	\int\~~frac{1}{cos~~^~~2x}dx =~~

	\~~int~~\~~sec^2xdx~~ =

	\tan{x}+C		\tan{x}+C

2022-09-13T17:49:14Z

← Older revision		Revision as of 17:49, 13 September 2022
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	</math>		</math>

	Note:<math>\cos^2\alpha+sin^2\alpha=1</math>		Note: <math>\cos^2\alpha+sin^2\alpha=1</math>

2022-09-13T17:49:08Z

← Older revision		Revision as of 17:49, 13 September 2022
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	\tan{x}+C		\tan{x}+C
	</math>		</math>

			Note:<math>\cos^2\alpha+sin^2\alpha=1</math>

2022-09-13T17:48:36Z

← Older revision		Revision as of 17:48, 13 September 2022
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	<math>		<math>

	\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}d\alpha		\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha

	\cos^2x+sin^2x=1		\cos^2x+sin^2x=1

2022-09-13T17:48:17Z

← Older revision		Revision as of 17:48, 13 September 2022
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	<math>		<math>

	\int(1+\tan^2{\alpha})\,d\alpha = \int 1+\frac{sin^2\alpha}{cos^2\alpha}d\alpha = \int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}d\alpha		\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}d\alpha

	\cos^2x+sin^2x=1		\cos^2x+sin^2x=1

2022-09-13T17:48:03Z

← Older revision		Revision as of 17:48, 13 September 2022
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	<math>		<math>

	\int(1+\tan^2{\alpha})\,d\alpha		\int(1+\tan^2{\alpha})\,d\alpha = \int 1+\frac{sin^2\alpha}{cos^2\alpha}d\alpha = \int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}d\alpha

	\int 1+\frac{sin^2\alpha}{cos^2\alpha}d\alpha =

	\int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}d\alpha

	\cos^2x+sin^2x=1		\cos^2x+sin^2x=1