5.4 Indefinite Integrals and the Net Change Theorem/3 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/3" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:38:48Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/3" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 19:38, 21 September 2022
(No difference)

2022-09-13T17:15:42Z

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← Older revision		Revision as of 17:15, 13 September 2022
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	\begin{align}		\begin{align}

	\frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} ~~\\[2ex]~~		\frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} &= {\cos{x} - \frac{1}{3}\cdot 3\sin^2{x} \cos{x} +0} \\[2ex]

	& ={\cos{x} - \frac{1}{3}\cdot 3\sin^2{x} \cos{x} +0} \\[2ex]

	& =\cos{x} - \sin^2{x}\cos{x} \\[2ex]		& =\cos{x} - \sin^2{x}\cos{x} \\[2ex]

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← Older revision		Revision as of 17:14, 13 September 2022
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	\begin{align}		\begin{align}

	\int\cos^{3}xdx = \sin{x}-\frac{1}{3}\sin^{3}x+C &= \frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} \\[2ex]		\int\cos^{3}xdx = \sin{x}-\frac{1}{3}\sin^{3}x+C \\[2ex]

			\frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} \\[2ex]

	& ={\cos{x} - \frac{1}{3}\cdot 3\sin^2{x} \cos{x} +0} \\[2ex]		& ={\cos{x} - \frac{1}{3}\cdot 3\sin^2{x} \cos{x} +0} \\[2ex]

2022-09-13T17:14:04Z

← Older revision		Revision as of 17:14, 13 September 2022
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	\begin{align}		\begin{align}

	& \int\cos^{3}xdx = \sin{x}-\frac{1}{3}\sin^{3}x+C ~~\\[2ex]~~		\int\cos^{3}xdx = \sin{x}-\frac{1}{3}\sin^{3}x+C &= \frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} \\[2ex]

	& \frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} \\[2ex]

	& ={\cos{x} - \frac{1}{3}\cdot 3\sin^2{x} \cos{x} +0} \\[2ex]		& ={\cos{x} - \frac{1}{3}\cdot 3\sin^2{x} \cos{x} +0} \\[2ex]

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