Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Calculus BC

Calculus BC Integration By Parts

Section 3: Integration: Lecture 1 | 19:04 min

Lecture Description

Next Lecture

Previous Lecture

Discussion
Answer Engine
Study Guides
Practice Questions
Download Lecture Slides
Table of Contents
Related Books

Lecture Comments (7)

	0 answers Post by Dawud Muhammad on January 18, 2015 hey prof...? what if i had the intergal of (sqrtof 3x+4)dx by parts??
	1 answer Last reply by: Thomas Zhang Sun Mar 16, 2014 10:11 PM Post by Hongxin Wan on March 9, 2014 For example 3, do we add "C" at the end?
	1 answer Last reply by: Becky Liao Sun Sep 22, 2013 7:51 PM Post by chetachi ohagi on June 27, 2013 Where does the "C" come from in the answer of example 1?
	1 answer Last reply by: Akshay Tiwary Sat Feb 23, 2013 6:55 AM Post by Shehan Gunasekara on May 26, 2012 Why did you take the exponential as u, shouldnt it be cos x, using the liate memonic????

Answer EngineGet answers to any question!Ask any question related to Calculus BC

Working on the solution...

Integration By Parts

Know by heart:
Choosing uand dv:

dvcan be integrated
is at most “just as difficult” as original integral

If the first integration is not successful, apply integration by parts again

Integration By Parts

∫ 2xcosxdx

Let u = 2x and dv = cosx
du = 2dx
dv = cosxdx
v = sinx + C

Apply Integration by Parts
∫ 2xcosxdx = 2x(sinx) − ∫ sinx(2dx) + C = 2xsinx − 2∫ sinxdx + C = 2xsinx + 2cosx + C

∫ xsin3xdx

Let u = x and dv = sin3xdx
u = x
du = dx
dv = sin3xdx
v = [( − cos3x)/3] + C
Apply Integration by Parts
∫ xsin3xdx = [( − cos3x)/3]( x ) − ∫ [( − cos3x)/3]dx + C
= [( − xcos3x)/3] + [1/3]∫ cos3xdx + C
= [( − xcos3x)/3] + [1/3]( [sin3x/3] ) + C
= [( − xcos3x)/3] + [sin3x/9] + C

= [( − 3xcos3x + sin3x)/9] + C

∫ lnx²dx

Use logarithm properties to simplify
∫ lnx²dx = ∫ 2lnxdx
= 2∫ lnxdx
Let u = lnx and dv = dx
du = [dx/x]
dv = dx
v = x + C
Apply Integration by Parts
∫ lnx²dx = 2∫ lnxdx

= 2( xlnx − ∫ dx + C ) = 2( xlnx − x + C ) = 2xlnx − 2x + C

∫ ln(x + 1)dx

Let u = x + 1 and dv = dx
u = ln(x + 1)
du = [1/(x + 1)]dx
dv = dx
v = x + C
Apply Integration by Parts
∫ ln(x + 1)dx = x( ln(x + 1) ) − ∫ x( [dx/(3x + 1)] ) + C
= x( ln(x + 1) ) − ∫ [x/(x + 1)] dx + C
Note the alternative form of [x/(x + 1)]
[x/(x + 1)] = 1 − [1/(x + 1)]
x( ln(x + 1) ) − ∫ [x/(x + 1)] dx = x( ln(x + 1) ) − ∫ 1 − [1/(x + 1)] dx + C
= xln(x + 1) − ∫ dx + ∫ [1/(x + 1)] dx + C

= xln(x + 1) − x + ln(x + 1) + C

∫ 3x² e^xdx

Let u = x² and dv = e^xdx
du = 2x dx
dv = e^xdx
v = e^x + C
Apply Integration by Parts
∫ 3x² e^xdx = 3∫ x² e^xdx
= 3( e^xx² − ∫ e^x( 2x dx ) )
Apply Integration by Parts again on ∫ e^x( 2x dx )
∫ e^x( 2x dx ) = e^x(2x) + ∫ e^x(2dx)
= 2xe^x + 2e^x + C

Apply substitution
3( e^xx² − ∫ e^x( 2x dx ) ) = 3( e^xx² − 2xe^x + 2e^x + C )

∫₀^π (x − 1)sinxdx

Let u = x − 1 and dv = sinxdx
du = dx
dv = sinxdx
v = − cosx
Apply Integration by Parts
∫₀^π (x − 1)sinxdx = − cosx(x − 1) − ∫₀^π− cosx(dx)
= (cosx − xcosx + sinx)₀^π
= cosπ− πcos π+ sinπ− (cos0 − (0)cos0 + sin0)
= − 1 + π+ 0 − (1)

= π− 2

∫₀¹⁰ x e^xdx

Let u = x and dv = e^xdx
du = dx
dv = e^xdx
v = e^x
Apply Integration by Parts
∫₀¹⁰ x e^xdx = [ e^x(x) ]₀¹⁰ + ∫₀¹⁰ e^xdx

= e¹⁰(10) + e¹⁰ − (e⁰(0) + e⁰) = e¹⁰(10) + e¹⁰ − 1 = 11e¹⁰ − 1

∫₀^e [x/(x + 1)]dx

Let u = x and dv = [dx/(x + 1)]
du = dx
dv = [1/(x + 1)]dx
v = ln(x + 1)
Apply Integration by Parts
∫₀^e [x/(x + 1)]dx = xln(x + 1) − ∫₀^e ln(x + 1) dx
From problem 4, we have solved ∫ ln(x + 1) dx
∫ ln(x + 1) dx = xln(x + 1) − x + ln(x + 1) + C
xln(x + 1) − ∫₀^e ln(x + 1) dx = xln(x + 1) − (xln(x + 1) − x + ln(x + 1))
= [ xln(x + 1) − xln(x + 1) + x − ln(x + 1) ]₀^e
= eln(e + 1) − eln(e + 1) + e − ln(e + 1) − ((0)ln(0 + 1) − (0)ln(0 + 1) + 0 − ln(0 + 1)
= e − ln(e + 1) + ln1

= e − ln(e + 1)

∫₂³ ln(3x − 5)³dx

Use logarithm properties to change form
∫₂³ ln(3x − 5)³dx = ∫₂³ 3ln(3x − 5)dx
= ∫₀² 3n(3x − 5)3dx
Let u = ln(3x − 5) and dv = 3dx
du = [3/(3x − 5)]dx
dv = 3dx
v = 3x
Apply Integration by Parts
∫₂³ ln(3x − 5)dx = xln(3x − 5) − ∫₂³ [3x/(3x − 5)]dx
Find the alternative form of [3x/(3x − 5)]
[3x/(3x − 5)] = [(3x − 5)/(3x − 5)] + [5/(3x − 5)] = 1 + [5/(3x − 5)]
xln(3x − 5) − ∫₀² [3x/(3x − 5)]dx = xln(3x − 5) − ∫₂³ 1 + [5/(3x − 5)] dx

= 3ln4 − 3 − [5/3]ln4 − (2ln1 − 2 − [5/3]ln1) = 3ln4 − [5/3]ln4 − 2ln1 + [5/3]ln1 − 1 = [4/3]ln4 − 1

Prove ∫ sinxcosx = [(sin²x)/2] + C

Let u = sinx and dv = cosx
du = cosx
dv = cosx
v = sinx
Apply Integration by Parts
∫ sinxcosx = sinx(sinx) − ∫ sinxcosx + C
= sin²x − ∫ sinxcosx + C
Isolate sin²x
2∫ sinxcosx = sin²x + C

∫ sinxcosx = [(sin²x)/2] + C

« Previous Question | Next Question »

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Integration By Parts

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Integration by Parts

Integration by Parts: Rules to Follow

Recap of the Rules

Example 1

Example 2

Example 3

Example 4

Example 5

Intro 0:00
Integration by Parts 0:12

Formula
How It's Derived From Product Rule

Integration by Parts: Rules to Follow 3:24

Recap of the Rules

Example 1 4:29
Example 2 7:04
Example 3 8:40
Example 4 13:48
Example 5 15:36

Mathematics: AP Calculus BC

Section 1: Prerequisites
	Parametric Curves	10:10
	Polar Coordinates	14:54
	Vector Functions	12:05
Section 2: Differentiation
	Parametric Differentiation	13:15
	Polar Differentiation	10:14
	L'Hopital's Rule	6:43
Section 3: Integration
	Integration By Parts	19:04
	Integration By Partial Fractions	22:04
	Improper Integrals	19:01
Section 4: Applications of Integration
	Logistic Growth	19:16
	Arc Length for Parametric & Polar Curves	17:40
	Area for Parametric & Polar Curves	4:33
Section 5: Sequences, Series, & Approximations
	Definition & Convergence	7:10
	Geometric Series	10:59
	Harmonic & P Series	3:58
	Integral Test	11:31
	Comparison Test	7:47
	Limit Comparison Test	9:40
	Ratio Test	11:20
	Alternating Series	9:11
	Absolute Convergence	7:13
	Power Series Convergence	12:52
	Taylor Series	15:11
	Power Series Operations	16:40
	Lagrange Error	7:09
Section 6: Practice Test
	AP Calc BC Practice Test Section 1: Multi Choice Part 1	15:45
	AP Calc BC Practice Test Section 1: Multi Choice Part 2	21:38
	AP Calc BC Practice Test Section 1: Multi Choice Part 3	17:31
	AP Calc BC Practice Test Section 1: Multi Choice Part 4	14:17
	AP Calc BC Practice Test Section 1: Multi Choice Part 5	22:45
	AP Calc BC Practice Test Section 1: Free Response Part 1	9:55
	AP Calc BC Practice Test Section 1: Free Response Part 2	10:08
	AP Calc BC Practice Test Section 1: Free Response Part 3	13:58

Related Books

	Cracking the AP Calculus BC Exam, 2015 Edition Author: Princeton Review ISBN: 804124825 Publisher: Princeton Review Year: 2014 The book includes 3 full length practice tests with detailed explanations, a review of all the key concepts, and targeted strateeies to ace th exam for your highest score.
	Cliffs AP: Calculus AB & BC, 3rd Edition Authors: Dale W Johnson, Kerry J King ISBN: 764586831 Publisher: Cliffs Notes Year: 2001 The book has a topic-by topic breakdown and lots of problem approach suggestions for both free response and multiple choice calculus questions. The book also includes two full length tests.

Related Books

Name	Description	Link
BookRenter.com	BookRenter.com is simply the most reliable online textbook rental service.	Visit BookRenter.com
PhysicsForums.com Homework Help	Physics Forums is a scientific community for students looking for math & science help.	Visit PhysicsForums.com Homework Help

John Zhu

Integration By Parts

Slide Duration:

Table of Contents

Section 1: Prerequisites

Parametric Curves

10m 10s

Intro