Secant-cubed function - Revision history

Vipul: /* First antiderivative */

2012-04-09T20:08:17Z

First antiderivative

← Older revision		Revision as of 20:08, 9 April 2012
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	We now use the fact that <math>\tan^2 x + 1 = \sec^2 x</math>, or more explicitly, <math>\tan^2 x = \sec^2 x - 1</math>, to rewrite this as:		We now use the fact that <math>\tan^2 x + 1 = \sec^2 x</math>, or more explicitly, <math>\tan^2 x = \sec^2 x - 1</math>, to rewrite this as:

	<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \int \sec x \, dx - \int \sec^3 x \, dx</math>		<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x x - \int \sec^3 x \, dx + \int \sec x \, d</math>

	We now use the [[secant function#Integration\|integration of the secant function]] to simplify this as:		We now use the [[secant function#Integration\|integration of the secant function]] to simplify this as:

	<math>\! \int \sec^3 x \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| ~~- \int \sec^3 x \, dx~~</math>		<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec^3 x \, dx + \ln\|\sec x + \tan x\| </math>

	We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:		We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:

	<math>\! I = \sec x \tan x + \ln\|\sec x + \tan x\| ~~- I~~</math>		<math>\! I = \sec x \tan x - I + \ln\|\sec x + \tan x\| </math>

	We rearrange and obtain:		We rearrange and obtain:

Vipul: /* First antiderivative */

2012-04-09T20:06:30Z

First antiderivative

← Older revision		Revision as of 20:06, 9 April 2012
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	We now use the fact that <math>\tan^2 x + 1 = \sec^2 x</math>, or more explicitly, <math>\tan^2 x = \sec^2 x - 1</math>, to rewrite this as:		We now use the fact that <math>\tan^2 x + 1 = \sec^2 x</math>, or more explicitly, <math>\tan^2 x = \sec^2 x - 1</math>, to rewrite this as:

	<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \int \sec x \, dx - \int \sec^3 x \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>		<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \int \sec x \, dx - \int \sec^3 x \, dx</math>

			We now use the [[secant function#Integration\|integration of the secant function]] to simplify this as:

			<math>\! \int \sec^3 x \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>

	We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:		We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:

Vipul: /* First antiderivative */

2012-04-09T20:05:30Z

First antiderivative

← Older revision		Revision as of 20:05, 9 April 2012
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	<math>\int \sec^3 x \, dx = \int \sec x (\sec^2 x) \,dx = \sec x \tan x - \int (\sec x \tan x) \tan x \, dx = \sec x \tan x - \int \sec x \tan^2x \, dx</math>		<math>\int \sec^3 x \, dx = \int \sec x (\sec^2 x) \,dx = \sec x \tan x - \int (\sec x \tan x) \tan x \, dx = \sec x \tan x - \int \sec x \tan^2x \, dx</math>

	We now use the fact that <math>\tan^2 x + 1 = \sec^2 x</math> to rewrite this as:		We now use the fact that <math>\tan^2 x + 1 = \sec^2 x</math>, or more explicitly, <math>\tan^2 x = \sec^2 x - 1</math>, to rewrite this as:

	<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \int \sec x \, dx - \int \sec^3 x \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>		<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \int \sec x \, dx - \int \sec^3 x \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>

Vipul: /* Integration */

2012-04-09T20:03:09Z

Integration

← Older revision		Revision as of 20:03, 9 April 2012
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	<section begin="integration by parts"/>		<section begin="integration by parts"/>

	We rewrite <math>\sec^3x = \sec x \cdot \sec^2x</math> and perform integration by parts, taking <math>\sec^2</math> as the part to integrate:		We rewrite <math>\sec^3x = \sec x \cdot \sec^2x</math> and perform integration by parts, taking <math>\sec^2</math> as the part to integrate. We use that an antiderivative of <math>\sec^2 x</math> is <math>\tan x</math> whereas the derivative of <math>\sec x</math> is <math>\sec x \tan x</math>:

	<math>\int \sec^3 x \, dx = \sec x \tan x - \int (\sec x \tan x) \tan x \, dx = \sec x \tan x - \int \sec x \tan^2x \, dx</math>		<math>\int \sec^3 x \, dx = \int \sec x (\sec^2 x) \,dx = \sec x \tan x - \int (\sec x \tan x) \tan x \, dx = \sec x \tan x - \int \sec x \tan^2x \, dx</math>

	<math>\! = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>		We now use the fact that <math>\tan^2 x + 1 = \sec^2 x</math> to rewrite this as:

			<math>\! \int \sec^3 x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \int \sec x \, dx - \int \sec^3 x \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>

	We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:		We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:

Vipul: /* First antiderivative */

2011-09-19T21:18:26Z

First antiderivative

← Older revision		Revision as of 21:18, 19 September 2011
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	<math>\! = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>		<math>\! = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>

	We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above holds without any additive constant adjustment, and we get:		We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above equality (between the left-most and right-most expression) holds without any additive constant adjustment, and we get:

	<math>I = \sec x \tan x + \ln\|\sec x + \tan x\| - I</math>		<math>\! I = \sec x \tan x + \ln\|\sec x + \tan x\| - I</math>

	We rearrange and obtain:		We rearrange and obtain:

Vipul: /* First antiderivative */

2011-09-19T21:17:46Z

First antiderivative

← Older revision		Revision as of 21:17, 19 September 2011
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	We use the [[recursive version of integration by parts]].		We use the [[recursive version of integration by parts]].

			<section begin="integration by parts"/>

	We rewrite <math>\sec^3x = \sec x \cdot \sec^2x</math> and perform integration by parts, taking <math>\sec^2</math> as the part to integrate:		We rewrite <math>\sec^3x = \sec x \cdot \sec^2x</math> and perform integration by parts, taking <math>\sec^2</math> as the part to integrate:

	<math>\int \sec^3 x \, dx = \sec x \tan x - \int (\sec x \tan x) \tan x \, dx = \sec x \tan x - \int \sec x \tan^2x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>		<math>\int \sec^3 x \, dx = \sec x \tan x - \int (\sec x \tan x) \tan x \, dx = \sec x \tan x - \int \sec x \tan^2x \, dx</math>

			<math>\! = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>

	We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above holds without any additive constant adjustment, and we get:		We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above holds without any additive constant adjustment, and we get:
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	<math>\frac{\sec x \tan x + \ln\|\sec x + \tan x\|}{2} + C</math>		<math>\frac{\sec x \tan x + \ln\|\sec x + \tan x\|}{2} + C</math>

			<section end="integration by parts"/>

Vipul at 12:34, 28 August 2011

2011-08-28T12:34:58Z

← Older revision		Revision as of 12:34, 28 August 2011
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	<math>x \mapsto (\sec x)^3 = \frac{1}{(\cos x)^3}</math>		<math>x \mapsto (\sec x)^3 = \frac{1}{(\cos x)^3}</math>

			==Differentiation==

			===First derivative===

			We can differentiate this function as a composite function, using the [[chain rule for differentiation]]. We get:

			<math>\! \frac{d}{dx}(\sec^3x) = [3(\sec x)^2](\sec x \tan x) = 3 \sec^3x \tan x</math>

			Alternatively, we could consider it as <math>1/(\cos x)^3</math>, and hence as a composite of <math>t \mapsto 1/t^3</math> and <math>\cos</math>. With that interpretation, the [[chain rule for differentiation]] yields:

			<math>\! \frac{-3}{\cos^4x} \cdot (-\sin x) = \frac{3\sin x}{\cos^4 x}</math>

			The two answers can be verified to be equivalent.

			==Integration==

			===First antiderivative===

			We use the [[recursive version of integration by parts]].

			We rewrite <math>\sec^3x = \sec x \cdot \sec^2x</math> and perform integration by parts, taking <math>\sec^2</math> as the part to integrate:

			<math>\int \sec^3 x \, dx = \sec x \tan x - \int (\sec x \tan x) \tan x \, dx = \sec x \tan x - \int \sec x \tan^2x \, dx = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx = \sec x \tan x + \ln\|\sec x + \tan x\| - \int \sec^3 x \, dx</math>

			We can choose an antiderivative <math>I</math> of <math>\sec^3</math> so that the above holds without any additive constant adjustment, and we get:

			<math>I = \sec x \tan x + \ln\|\sec x + \tan x\| - I</math>

			We rearrange and obtain:

			<math>2I = \sec x \tan x + \ln\|\sec x + \tan x\|</math>

			Dividing by 2, we get:

			<math>\! I = \frac{\sec x \tan x + \ln\|\sec x + \tan x\|}{2}</math>

			The general antiderivative expression is thus:

			<math>\frac{\sec x \tan x + \ln\|\sec x + \tan x\|}{2} + C</math>

Vipul: /* Definition */

2011-08-28T12:27:12Z

Definition

← Older revision		Revision as of 12:27, 28 August 2011
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	==Definition==		==Definition==

	The function, denoted <math>\sec^3</math>, is defined as the [[defining ingredient::composite of two functions\|composite]] of the [[defining ingredient::cube function]] and the [[defining ingredient::secant function]] (which in turn is the composite of the [[reciprocal function]] and the cosine function). Explicitly, it is given as:		The function, denoted <math>\sec^3</math>, is defined as the [[defining ingredient::composite of two functions\|composite]] of the [[defining ingredient::cube function]] and the [[defining ingredient::secant function]] (which in turn is the composite of the [[reciprocal function]] and the [[cosine function]]). Explicitly, it is given as:

	<math>x \mapsto (\sec x)^3 = \frac{1}{(\cos x)^3}</math>		<math>x \mapsto (\sec x)^3 = \frac{1}{(\cos x)^3}</math>

Vipul: Created page with "{{particular function}} {{angular function radian convention}} ==Definition== The function, denoted $\sec^3$ , is defined as the [[defining ingredient::composite of ..."

2011-08-28T12:26:34Z

Created page with "{{particular function}} {{angular function radian convention}} ==Definition== The function, denoted <math>\sec^3</math>, is defined as the [[defining ingredient::composite of ..."

New page

{{particular function}}

{{angular function radian convention}}

==Definition==

The function, denoted <math>\sec^3</math>, is defined as the [[defining ingredient::composite of two functions|composite]] of the [[defining ingredient::cube function]] and the [[defining ingredient::secant function]] (which in turn is the composite of the [[reciprocal function]] and the cosine function). Explicitly, it is given as:

<math>x \mapsto (\sec x)^3 = \frac{1}{(\cos x)^3}</math>

Secant-cubed function - Revision history

Vipul: /* First antiderivative */

Vipul: /* First antiderivative */

Vipul: /* First antiderivative */

Vipul: /* Integration */

Vipul: /* First antiderivative */

Vipul: /* First antiderivative */

Vipul at 12:34, 28 August 2011

Vipul: /* Definition */

Vipul: Created page with "{{particular function}} {{angular function radian convention}} ==Definition== The function, denoted \sec^3, is defined as the [[defining ingredient::composite of ..."

Vipul: Created page with "{{particular function}} {{angular function radian convention}} ==Definition== The function, denoted $\sec^3$ , is defined as the [[defining ingredient::composite of ..."