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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????

Integration By Parts

  • Know by heart:
  • Choosing u and dv:
    • dv can 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 exdx
  • Let u = x2 and dv = exdx
  • du = 2x dx
  • dv = exdx
  • v = ex + C
  • Apply Integration by Parts
  • 3x2 exdx = 3∫ x2 exdx
  • = 3( exx2 − ∫ ex( 2x dx ) )
  • Apply Integration by Parts again on ∫ ex( 2x dx )
  • ex( 2x dx ) = ex(2x) + ∫ ex(2dx)
  • = 2xex + 2ex + C
Apply substitution
3( exx2 − ∫ ex( 2x dx ) ) = 3( exx2 − 2xex + 2ex + C )
0π (x − 1)sinxdx
  • Let u = x − 1 and dv = sinxdx
  • du = dx
  • dv = sinxdx
  • v = − cosx
  • Apply Integration by Parts
  • 0π (x − 1)sinxdx = − cosx(x − 1) − ∫0π − cosx(dx)
  • = (cosx − xcosx + sinx)0π
  • = cosπ− πcos π+ sinπ− (cos0 − (0)cos0 + sin0)
  • = − 1 + π+ 0 − (1)
= π− 2
010 x exdx
  • Let u = x and dv = exdx
  • du = dx
  • dv = exdx
  • v = ex
  • Apply Integration by Parts
  • 010 x exdx = [ ex(x) ]010 + ∫010 exdx
= e10(10) + e10 − (e0(0) + e0) = e10(10) + e10 − 1 = 11e10 − 1
0e [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
  • 0e [x/(x + 1)]dx = xln(x + 1) − ∫0e 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) − ∫0e ln(x + 1) dx = xln(x + 1) − (xln(x + 1) − x + ln(x + 1))
  • = [ xln(x + 1) − xln(x + 1) + x − ln(x + 1) ]0e
  • = 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

*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 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