Practical:Integration by parts - Revision history

Vipul at 20:52, 14 April 2012

2012-04-14T20:52:54Z

Vipul: /* Products and typical strategies */

2012-04-14T17:13:18Z

Products and typical strategies

← Older revision		Revision as of 17:13, 14 April 2012
Line 36:		Line 36:
	The examples here are done quickly, without the explicit function/letter naming procedure.		The examples here are done quickly, without the explicit function/letter naming procedure.

	===~~Simple examples of polynomial~~ times trigonometric===		===Polynomial times basic trigonometric===

	In all these, we choose to differentiate the polynomial and integrate the trigonometric function. We use the following formulas:		In all these, we choose to differentiate the polynomial and integrate the trigonometric function. We use the following formulas:
Line 58:		Line 58:
	\|}		\|}

	===~~Simple examples of polynomial~~ times exponential===		===Polynomial times exponential===

	In all these, we choose to differentiate the polynomial and integrate the exponential function. We use that the exponential function is its own antiderivative.		In all these, we choose to differentiate the polynomial and integrate the exponential function. We use that the exponential function is its own antiderivative.
Line 70:		Line 70:
	\|}		\|}

	===~~Simple examples of polynomial~~ times logarithmic===		===Polynomial times logarithmic===

	In all these, we choose to integrate the polynomial and differentiate the logarithmic function. We use that <math>d(\ln x)/dx = 1/x</math>.		In all these, we choose to integrate the polynomial and differentiate the logarithmic function. We use that <math>d(\ln x)/dx = 1/x</math>.
Line 84:		Line 84:
	Embedding of Part 2 video {{fillin}}		Embedding of Part 2 video {{fillin}}

	===~~Simple examples of polynomial~~ times inverse trigonometric===		===Polynomial times inverse trigonometric===

	In all these, we choose to integrate the polynomial and differentiate the inverse trigonometric function. We use that <math>\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}</math>.		In all these, we choose to integrate the polynomial and differentiate the inverse trigonometric function. We use that <math>\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}</math>.

Vipul: /* Explicit procedure */

2012-04-10T18:43:39Z

Explicit procedure

← Older revision		Revision as of 18:43, 10 April 2012
Line 12:		Line 12:
	# '''Apply integration by parts'''		# '''Apply integration by parts'''
	# '''Now, do the second integral''': This may require integration by parts again, or we may do it some other way, or we may use the [[recursive version of integration by parts]].		# '''Now, do the second integral''': This may require integration by parts again, or we may do it some other way, or we may use the [[recursive version of integration by parts]].
			# (Optional) '''Verify your answer by differentiating the antiderivative obtained''': You should find yourself using the [[product rule for differentiation]], and further, you should find some of the terms canceling and the left-over term should be the integrand.
	===Example of <math>x \cos x</math>===		===Example of <math>x \cos x</math>===

Vipul: /* Simple examples of polynomial times inverse trigonometric */

2012-04-09T19:30:10Z

Simple examples of polynomial times inverse trigonometric

← Older revision		Revision as of 19:30, 9 April 2012
Line 87:		Line 87:

	In all these, we choose to integrate the polynomial and differentiate the inverse trigonometric function. We use that <math>\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}</math>.		In all these, we choose to integrate the polynomial and differentiate the inverse trigonometric function. We use that <math>\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}</math>.

			For many of these examples, we need to use some tricks/techniques related to the integration of rational functions to simplify the new rational function or algebraic expressions we obtain.

	{\| class="sortable" border="1"		{\| class="sortable" border="1"

Vipul: /* Simple examples of polynomial times inverse trigonometric */

2012-04-09T19:28:40Z

Simple examples of polynomial times inverse trigonometric

← Older revision		Revision as of 19:28, 9 April 2012
Line 91:		Line 91:
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 2 video (embedded above)?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 2 video (embedded above)?
	\|-		\|-
	\| <math>\int \arctan x \, dx</math> \|\| Think of integrand as <math>(\arctan x)(1)</math>, integrate 1 to get <math>x</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math> \|\| <math>\int \arctan x \, dx = (\arctan x)(x) - \int \frac{1}{1 + x^2}x \, dx = x \arctan x - \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx = x \arctan x - \frac{1}{2}\ln(1 + x^2) + C</math><br>Note that for the latter integration, we use the <math>\int g'/g = \ln\|g\|</math> rule. \|\| Yes		\| <math>\int \arctan x \, dx</math> \|\| Think of integrand as <math>(\arctan x)(1)</math>, integrate 1 to get <math>x</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math> \|\| <math>\int \arctan x \, dx = (\arctan x)(x) - \int \frac{1}{1 + x^2}x \, dx = x \arctan x - \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx</math><br>This simplifies to <math>x \arctan x - \frac{1}{2}\ln(1 + x^2) + C</math><br>Note that for the latter integration, we use the <math>\int g'/g = \ln\|g\|</math> rule. \|\| Yes
	\|-		\|-
	\| <math>\int x \arctan x \, dx</math> \|\| integrate <math>x</math> to get <math>x^2/2</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math>. \|\| <math>\int x \arctan x \, dx = (\arctan x)(x^2/2) - \int \frac{1}{1 + x^2} \frac{x^2}{2} \, dx = \frac{x^2\arctan x}{2} - \frac{1}{2} \int \frac{x^2}{1 + x^2} \, dx</math><br>We now use that the fraction is an improper fraction to convert it to the mixed fraction <matH>1 - \frac{1}{1 + x^2}</math>, and get: <math>\frac{x^2\arctan x}{2} - \frac{1}{2} \int [1 - \frac{1}{1 + x^2}] \, dx</math>.<br> Simplfying gives <math>\frac{x^2 \arctan x}{2} - \frac{1}{2}(x-\arctan x) + C = \frac{(x^2 + 1)\arctan x}{2} - \frac{x}{2} + C</math> \|\| Yes		\| <math>\int x \arctan x \, dx</math> \|\| integrate <math>x</math> to get <math>x^2/2</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math>. \|\| <math>\int x \arctan x \, dx = (\arctan x)(x^2/2) - \int \frac{1}{1 + x^2} \frac{x^2}{2} \, dx = \frac{x^2\arctan x}{2} - \frac{1}{2} \int \frac{x^2}{1 + x^2} \, dx</math><br>We now use that the fraction is an improper fraction to convert it to the mixed fraction <matH>1 - \frac{1}{1 + x^2}</math>, and get: <math>\frac{x^2\arctan x}{2} - \frac{1}{2} \int [1 - \frac{1}{1 + x^2}] \, dx</math>.<br> Simplfying gives <math>\frac{x^2 \arctan x}{2} - \frac{1}{2}(x-\arctan x) + C = \frac{(x^2 + 1)\arctan x}{2} - \frac{x}{2} + C</math> \|\| Yes
	\|}		\|}

Vipul: /* Products and typical strategies */

2012-04-09T19:27:32Z

Products and typical strategies

← Older revision		Revision as of 19:27, 9 April 2012
Line 38:		Line 38:
	===Simple examples of polynomial times trigonometric===		===Simple examples of polynomial times trigonometric===

	In all these, we choose to differentiate the polynomial and integrate the trigonometric function.		In all these, we choose to differentiate the polynomial and integrate the trigonometric function. We use the following formulas:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video?		! Function !! Antiderivative
			\|-
			\| <math>\sin</math> \|\| <math>-\cos</math>
			\|-
			\| <math>\cos</math> \|\| <math>\sin</math>
			\|}

			{\| class="sortable" border="1"
			! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video (embedded above)?
	\|-		\|-
	\| <math>\int x\cos x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\cos x</math> to get <math>\sin x</math> \|\| <math>\int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C</math> \|\| Yes		\| <math>\int x\cos x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\cos x</math> to get <math>\sin x</math> \|\| <math>\int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C</math> \|\| Yes
Line 52:		Line 60:
	===Simple examples of polynomial times exponential===		===Simple examples of polynomial times exponential===

	In all these, we choose to differentiate the polynomial and integrate the exponential function.		In all these, we choose to differentiate the polynomial and integrate the exponential function. We use that the exponential function is its own antiderivative.

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video (embedded above)?
	\|-		\|-
	\| <math>\int xe^x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>e^x</math> to get <math>e^x</math> \|\| <math>\int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C</math> \|\| Yes		\| <math>\int xe^x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>e^x</math> to get <math>e^x</math> \|\| <math>\int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C</math> \|\| Yes
Line 64:		Line 72:
	===Simple examples of polynomial times logarithmic===		===Simple examples of polynomial times logarithmic===

	In all these, we choose to integrate the polynomial and differentiate the logarithmic function.		In all these, we choose to integrate the polynomial and differentiate the logarithmic function. We use that <math>d(\ln x)/dx = 1/x</math>.

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video (embedded above)?
	\|-		\|-
	\| <math>\int \ln x \, dx</math> \|\| Think of integrand as <math>(\ln x)(1)</math>, differentiate <math>\ln x</math> to get <math>1/x</math>, integrate 1 to get <math>x</math> \|\| <math>\int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C</math> \|\| Yes		\| <math>\int \ln x \, dx</math> \|\| Think of integrand as <math>(\ln x)(1)</math>, differentiate <math>\ln x</math> to get <math>1/x</math>, integrate 1 to get <math>x</math> \|\| <math>\int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C</math> \|\| Yes
Line 73:		Line 81:
	\| <math>\int x \ln x \, dx</math> \|\| Differentiate <math>\ln x</math> to get <matH>1/x</math>, integrate <math>x</math> to get <math>x^2/2</math> \|\| <math>\int x \ln x \, dx = (\ln x)(x^2/2) - \int \frac{1}{x} \frac{x^2}{2} \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C</math> \|\| Yes		\| <math>\int x \ln x \, dx</math> \|\| Differentiate <math>\ln x</math> to get <matH>1/x</math>, integrate <math>x</math> to get <math>x^2/2</math> \|\| <math>\int x \ln x \, dx = (\ln x)(x^2/2) - \int \frac{1}{x} \frac{x^2}{2} \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C</math> \|\| Yes
	\|}		\|}

			Embedding of Part 2 video {{fillin}}

	===Simple examples of polynomial times inverse trigonometric===		===Simple examples of polynomial times inverse trigonometric===

			In all these, we choose to integrate the polynomial and differentiate the inverse trigonometric function. We use that <math>\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}</math>.

			{\| class="sortable" border="1"
			! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 2 video (embedded above)?
			\|-
			\| <math>\int \arctan x \, dx</math> \|\| Think of integrand as <math>(\arctan x)(1)</math>, integrate 1 to get <math>x</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math> \|\| <math>\int \arctan x \, dx = (\arctan x)(x) - \int \frac{1}{1 + x^2}x \, dx = x \arctan x - \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx = x \arctan x - \frac{1}{2}\ln(1 + x^2) + C</math><br>Note that for the latter integration, we use the <math>\int g'/g = \ln\|g\|</math> rule. \|\| Yes
			\|-
			\| <math>\int x \arctan x \, dx</math> \|\| integrate <math>x</math> to get <math>x^2/2</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math>. \|\| <math>\int x \arctan x \, dx = (\arctan x)(x^2/2) - \int \frac{1}{1 + x^2} \frac{x^2}{2} \, dx = \frac{x^2\arctan x}{2} - \frac{1}{2} \int \frac{x^2}{1 + x^2} \, dx</math><br>We now use that the fraction is an improper fraction to convert it to the mixed fraction <matH>1 - \frac{1}{1 + x^2}</math>, and get: <math>\frac{x^2\arctan x}{2} - \frac{1}{2} \int [1 - \frac{1}{1 + x^2}] \, dx</math>.<br> Simplfying gives <math>\frac{x^2 \arctan x}{2} - \frac{1}{2}(x-\arctan x) + C = \frac{(x^2 + 1)\arctan x}{2} - \frac{x}{2} + C</math> \|\| Yes
			\|}

Vipul: /* Simple examples of polynomial times logarithmic */

2012-04-09T19:17:32Z

Simple examples of polynomial times logarithmic

← Older revision		Revision as of 19:17, 9 April 2012
Line 71:		Line 71:
	\| <math>\int \ln x \, dx</math> \|\| Think of integrand as <math>(\ln x)(1)</math>, differentiate <math>\ln x</math> to get <math>1/x</math>, integrate 1 to get <math>x</math> \|\| <math>\int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C</math> \|\| Yes		\| <math>\int \ln x \, dx</math> \|\| Think of integrand as <math>(\ln x)(1)</math>, differentiate <math>\ln x</math> to get <math>1/x</math>, integrate 1 to get <math>x</math> \|\| <math>\int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C</math> \|\| Yes
	\|-		\|-
	\| <math>\int x \ln x \, dx</math> \|\| Differentiate <math>\ln x</math> to get <matH>1/x</math>, integrate <math>x</math> to get <math>x^2/2</math> \|\| <math>\int x \ln x \, dx = (\ln x)(x^2/2) - \int \frac{1}{x} \frac{x^2}{2} \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C</math>		\| <math>\int x \ln x \, dx</math> \|\| Differentiate <math>\ln x</math> to get <matH>1/x</math>, integrate <math>x</math> to get <math>x^2/2</math> \|\| <math>\int x \ln x \, dx = (\ln x)(x^2/2) - \int \frac{1}{x} \frac{x^2}{2} \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C</math> \|\| Yes
	\|}		\|}

	===Simple examples of polynomial times inverse trigonometric===		===Simple examples of polynomial times inverse trigonometric===

Vipul: /* Products and typical strategies */

2012-04-09T19:16:46Z

Products and typical strategies

← Older revision		Revision as of 19:16, 9 April 2012
Line 41:		Line 41:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video?
	\|-		\|-
	\| <math>\int x\cos x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\cos x</math> to get <math>\sin x</math> \|\| <math>\int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C</math> \|\| Yes		\| <math>\int x\cos x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\cos x</math> to get <math>\sin x</math> \|\| <math>\int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C</math> \|\| Yes
Line 55:		Line 55:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video?
	\|-		\|-
	\| <math>\int xe^x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>e^x</math> to get <math>e^x</math> \|\| <math>\int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C</math> \|\| Yes		\| <math>\int xe^x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>e^x</math> to get <math>e^x</math> \|\| <math>\int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C</math> \|\| Yes
Line 61:		Line 61:
	\| <math>\int x^2e^x \, dx</math> \|\| Differentiate <math>x^2</math> to get <math>2x</math>, integrate <matH>e^x</math> to get <math>e^x</math>, then again differentiate the polynomial and integrate the exponential \|\| <math>\int x^2e^x \, dx = x^2e^x - \int 2x e^x \, dx = x^2e^x - 2[xe^x - \int (1)(e^x) \, dx] = x^2e^x - 2xe^x + 2e^x + C</math> \|\| No		\| <math>\int x^2e^x \, dx</math> \|\| Differentiate <math>x^2</math> to get <math>2x</math>, integrate <matH>e^x</math> to get <math>e^x</math>, then again differentiate the polynomial and integrate the exponential \|\| <math>\int x^2e^x \, dx = x^2e^x - \int 2x e^x \, dx = x^2e^x - 2[xe^x - \int (1)(e^x) \, dx] = x^2e^x - 2xe^x + 2e^x + C</math> \|\| No
	\|}		\|}

			===Simple examples of polynomial times logarithmic===

			In all these, we choose to integrate the polynomial and differentiate the logarithmic function.

			{\| class="sortable" border="1"
			! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In part 1 video?
			\|-
			\| <math>\int \ln x \, dx</math> \|\| Think of integrand as <math>(\ln x)(1)</math>, differentiate <math>\ln x</math> to get <math>1/x</math>, integrate 1 to get <math>x</math> \|\| <math>\int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C</math> \|\| Yes
			\|-
			\| <math>\int x \ln x \, dx</math> \|\| Differentiate <math>\ln x</math> to get <matH>1/x</math>, integrate <math>x</math> to get <math>x^2/2</math> \|\| <math>\int x \ln x \, dx = (\ln x)(x^2/2) - \int \frac{1}{x} \frac{x^2}{2} \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C</math>
			\|}

			===Simple examples of polynomial times inverse trigonometric===

Vipul: Created page with "{{perspectives}} {{#lst:Integration by parts|statement}} ==Explicit procedure== The procedure for applying integration by parts, explicitly, is as follows: # '''Identify t..."

2012-04-09T19:11:11Z

Created page with "{{perspectives}} {{#lst:Integration by parts|statement}} ==Explicit procedure== The procedure for applying integration by parts, explicitly, is as follows: # '''Identify t..."

New page

{{perspectives}}

{{#lst:Integration by parts|statement}}

==Explicit procedure==

The procedure for applying integration by parts, explicitly, is as follows:

# '''Identify the function being integrated as a product of two functions''': In some cases,we may take one of the factors to be the function 1, and the other one to be the whole integrand.
# '''Figure out which is the part to differentiate and which is the part to integrate''': In the explicit function notation, the part to differentiate is <math>F</math> and the part to integrate is <math>g</math>. In the <math>u-v</math> notation, the part to differentiate is <math>u</math> and the part to integrate is what we'll eventually call <math>dv/dx</math>. This is the step where we need to use explicit heuristics. The general heuristic used is the ILATE or LIATE rule. It says that inverse trigonometric and logarithmic functions are most preferable for differentiation, algebraic functions (polynomials, rational functions, and radicals) are in between, and exponential and trigonometric functions are most preferable for integration.
# '''Determine the antiderivative of the part to integrate and the derivative of the part to differentiate''': In the function notation, these are <math>F'</math> and <math>G</math>. In the <math>u-v</math> notation, these are <math>du/dx</math> and <math>v</math> respectively.
# '''Apply integration by parts'''
# '''Now, do the second integral''': This may require integration by parts again, or we may do it some other way, or we may use the [[recursive version of integration by parts]].

===Example of <math>x \cos x</math>===

Consider the integration problem:

{{quotation|Compute the antiderivative <math>\int x \cos x \, dx</math>}}

# '''Identify the function being integrated as the product of two functions''': Here, the two factor functions are <math>x</math> and <math>\cos x</math> respectively.
# '''Figure out which is the part to differentiate and which is the part to integrate''': Using the ILATE rule or through direct strategic thinking, we see that <math>x</math> (the polynomial, algebraic part) should be the part to differentiate and <math>\cos x</math> should be the part to integrate. Explicitly:
#* In the function notation, we have <math>F(x) = x</math> and <math>g(x) = \cos x</math>.
#* In the <math>u-v</math> notation, we have <math>u = x</math> and <math>dv/dx = \cos x</math>.
# '''Determine the antiderivative of the part to integrate and the derivative of the part to differentiate''': The antiderivative of <math>\cos x</math> is <math>\sin x</math> and the derivative of <math>x</math> is <math>1</math>. Explicitly:
#* In the function notation, we have <math>F'(x) = 1</math> and <math>G(x) = \sin x</math>.
#* In the <math>u-v</math> notation, we have <math>du/dx = 1</math> and <math>v = \sin x</math>.
# '''Apply integration by parts''': We get<br><math>\int x\cos x \, dx = x\sin x - \int (1)(\sin x) \, dx = x \sin x - \int \sin x \, dx</math>
# '''Now, do the second integral''': We have <math>\int \sin x = -\cos x</math>, so plugging it back in, we get <math>x \sin x -
(- \cos x) + C = x \sin x + \cos x + C</math>.

==Products and typical strategies==

{{#lst:integration by parts|product strategies}}

The examples here are done quickly, without the explicit function/letter naming procedure.

===Simple examples of polynomial times trigonometric===

In all these, we choose to differentiate the polynomial and integrate the trigonometric function.

{| class="sortable" border="1"
! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video?
|-
| <math>\int x\cos x \, dx</math> || Differentiate <math>x</math> to get 1, integrate <math>\cos x</math> to get <math>\sin x</math> || <math>\int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C</math> || Yes
|-
| <math>\int x^2 \cos x \, dx</math> || Differentiate <math>x^2</math> to get <math>2x</math>, integrate <math>\cos x</math> to get <math>\sin x</math>, then again differentiate the polynomial and integrate the trigonometric function || <math>\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2\sin x - 2[x(-\cos x) - \int (1)(-\cos x) \, dx]</math><br> This simplifies to <math>x^2 \sin x + 2x \cos x - 2 \sin x + C</math> || Yes
|-
| <math>\int x \sin x \, dx</math> || Differentiate <math>x</math> to get 1, integrate <math>\sin x</math> to get <math>-\cos x</math> || <math>\int x \sin x \, dx = x(-\cos x) - \int (1)(-\cos x) \, dx = -x\cos x + \sin x + C</math> || No
|}

===Simple examples of polynomial times exponential===

In all these, we choose to differentiate the polynomial and integrate the exponential function.

{| class="sortable" border="1"
! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video?
|-
| <math>\int xe^x \, dx</math> || Differentiate <math>x</math> to get 1, integrate <math>e^x</math> to get <math>e^x</math> || <math>\int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C</math> || Yes
|-
| <math>\int x^2e^x \, dx</math> || Differentiate <math>x^2</math> to get <math>2x</math>, integrate <matH>e^x</math> to get <math>e^x</math>, then again differentiate the polynomial and integrate the exponential || <math>\int x^2e^x \, dx = x^2e^x - \int 2x e^x \, dx = x^2e^x - 2[xe^x - \int (1)(e^x) \, dx] = x^2e^x - 2xe^x + 2e^x + C</math> || No
|}

← Older revision		Revision as of 20:52, 14 April 2012
Line 29:		Line 29:
	# '''Now, do the second integral''': We have <math>\int \sin x = -\cos x</math>, so plugging it back in, we get <math>x \sin x -		# '''Now, do the second integral''': We have <math>\int \sin x = -\cos x</math>, so plugging it back in, we get <math>x \sin x -
	(- \cos x) + C = x \sin x + \cos x + C</math>.		(- \cos x) + C = x \sin x + \cos x + C</math>.

			<center>{{#widget:YouTube\|id=ir4wy8NCBDU}}</center>

	==Products and typical strategies==		==Products and typical strategies==
Line 49:		Line 51:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In ~~part 1~~ video (embedded ~~above~~)?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video (embedded below)?
	\|-		\|-
	\| <math>\int x\cos x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\cos x</math> to get <math>\sin x</math> \|\| <math>\int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C</math> \|\| Yes		\| <math>\int x\cos x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\cos x</math> to get <math>\sin x</math> \|\| <math>\int x\cos x \, dx = x \sin x - \int (1) (\sin x) \, dx = x\sin x + \cos x + C</math> \|\| Yes
			\|-
			\| <math>\int x \sin x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\sin x</math> to get <math>-\cos x</math> \|\| <math>\int x \sin x \, dx = x(-\cos x) - \int (1)(-\cos x) \, dx = -x\cos x + \sin x + C</math> \|\| Yes
	\|-		\|-
	\| <math>\int x^2 \cos x \, dx</math> \|\| Differentiate <math>x^2</math> to get <math>2x</math>, integrate <math>\cos x</math> to get <math>\sin x</math>, then again differentiate the polynomial and integrate the trigonometric function \|\| <math>\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2\sin x - 2[x(-\cos x) - \int (1)(-\cos x) \, dx]</math><br> This simplifies to <math>x^2 \sin x + 2x \cos x - 2 \sin x + C</math> \|\| Yes		\| <math>\int x^2 \cos x \, dx</math> \|\| Differentiate <math>x^2</math> to get <math>2x</math>, integrate <math>\cos x</math> to get <math>\sin x</math>, then again differentiate the polynomial and integrate the trigonometric function \|\| <math>\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2\sin x - 2[x(-\cos x) - \int (1)(-\cos x) \, dx]</math><br> This simplifies to <math>x^2 \sin x + 2x \cos x - 2 \sin x + C</math> \|\| Yes
	\|-
	\| <math>\int x \sin x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>\sin x</math> to get <math>-\cos x</math> \|\| <math>\int x \sin x \, dx = x(-\cos x) - \int (1)(-\cos x) \, dx = -x\cos x + \sin x + C</math> \|\| No
	\|}		\|}

			<center>{{#widget:YouTube\|id=BxxwiDcAHo8}}</center>

	===Polynomial times exponential===		===Polynomial times exponential===
Line 63:		Line 67:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In ~~part 1~~ video (embedded above)?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video (embedded above)?
	\|-		\|-
	\| <math>\int xe^x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>e^x</math> to get <math>e^x</math> \|\| <math>\int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C</math> \|\| Yes		\| <math>\int xe^x \, dx</math> \|\| Differentiate <math>x</math> to get 1, integrate <math>e^x</math> to get <math>e^x</math> \|\| <math>\int xe^x \, dx = xe^x - \int (1)(e^x) \, dx = xe^x - e^x + C</math> \|\| Yes
	\|-		\|-
	\| <math>\int x^2e^x \, dx</math> \|\| Differentiate <math>x^2</math> to get <math>2x</math>, integrate <matH>e^x</math> to get <math>e^x</math>, then again differentiate the polynomial and integrate the exponential \|\| <math>\int x^2e^x \, dx = x^2e^x - \int 2x e^x \, dx = x^2e^x - 2[xe^x - \int (1)(e^x) \, dx] = x^2e^x - 2xe^x + 2e^x + C</math> \|\| No		\| <math>\int x^2e^x \, dx</math> \|\| Differentiate <math>x^2</math> to get <math>2x</math>, integrate <matH>e^x</math> to get <math>e^x</math>, then again differentiate the polynomial and integrate the exponential \|\| <math>\int x^2e^x \, dx = x^2e^x - \int 2x e^x \, dx = x^2e^x - 2[xe^x - \int (1)(e^x) \, dx] = x^2e^x - 2xe^x + 2e^x + C</math> \|\| Yes
	\|}		\|}

Line 75:		Line 79:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In ~~part 1~~ video (embedded ~~above~~)?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video (embedded below)?
	\|-		\|-
	\| <math>\int \ln x \, dx</math> \|\| Think of integrand as <math>(\ln x)(1)</math>, differentiate <math>\ln x</math> to get <math>1/x</math>, integrate 1 to get <math>x</math> \|\| <math>\int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C</math> \|\| Yes		\| <math>\int \ln x \, dx</math> \|\| Think of integrand as <math>(\ln x)(1)</math>, differentiate <math>\ln x</math> to get <math>1/x</math>, integrate 1 to get <math>x</math> \|\| <math>\int \ln x \, dx = \int (\ln x)(1) \, dx = (\ln x)(x) - \int (1/x)(x) \, dx = x\ln x - x + C</math> \|\| Yes
Line 82:		Line 86:
	\|}		\|}

	~~Embedding of Part 2 video~~ {{~~fillin~~}}		<center>{{#widget:YouTube\|id=TuZNwRxuDIw}}</center>

	===Polynomial times inverse trigonometric===		===Polynomial times inverse trigonometric===
Line 91:		Line 95:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In ~~part 2~~ video (embedded above)?		! Integration problem !! What we choose to differentiate and integrate !! Full solution (quick version) !! In video (embedded above)
	\|-		\|-
	\| <math>\int \arctan x \, dx</math> \|\| Think of integrand as <math>(\arctan x)(1)</math>, integrate 1 to get <math>x</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math> \|\| <math>\int \arctan x \, dx = (\arctan x)(x) - \int \frac{1}{1 + x^2}x \, dx = x \arctan x - \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx</math><br>This simplifies to <math>x \arctan x - \frac{1}{2}\ln(1 + x^2) + C</math><br>Note that for the latter integration, we use the <math>\int g'/g = \ln\|g\|</math> rule. \|\| Yes		\| <math>\int \arctan x \, dx</math> \|\| Think of integrand as <math>(\arctan x)(1)</math>, integrate 1 to get <math>x</math>, differentiate <math>\arctan x</math> to get <math>1/(1 + x^2)</math> \|\| <math>\int \arctan x \, dx = (\arctan x)(x) - \int \frac{1}{1 + x^2}x \, dx = x \arctan x - \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx</math><br>This simplifies to <math>x \arctan x - \frac{1}{2}\ln(1 + x^2) + C</math><br>Note that for the latter integration, we use the <math>\int g'/g = \ln\|g\|</math> rule. \|\| Yes