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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of Calculus AB
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Lecture Comments (4)

1 answer

Last reply by: Angela Patrick
Sun Dec 15, 2013 7:26 AM

Post by sergey smith on October 3, 2013

on the second to last example isn't tan(3t) divided by the a term or 3? so wouldn't it equal tan(3t)/9 + C?

1 answer

Last reply by: Vicente Orellana
Wed Jul 17, 2013 8:08 PM

Post by Huseyin Kayahan on March 3, 2013

for example 1, isnt it supposed to be -cos2x/2x + C ?

Chain Rule

  • Tip: choose u so that dx cancels out rest of x terms
  • Don’t forget to substitute original expression back into u!

Chain Rule

Solve ∫sin(x − π) dx using u substitution
  • u = x − π
  • du = dx
  • ∫sin(x − π) dx = ∫sinu  du
  • ∫sin(x − π) dx = −cosu + c
  • ∫sin(x − π) dx = −cos(x − π) + c
  • −cos(x − π) = cosx
  • In reality, we could have simplified the original integral, but this works too
∫sin(x − π) dx = cosx + c
∫x sin(x2) dx
  • u = x2
  • du = 2x dx
  • ∫x sin(x2) dx = ∫x sinu [du/2x]
  • ∫x sin(x2) dx = [1/2] ∫sinu  du
  • ∫x sin(x2) dx = [1/2] (−cosu) + c
∫x sin(x2) dx = −[1/2] cosx2 + c
∫(3x − 5)4 dx
  • u = 3x − 5
  • du = 3  dx
  • ∫(3x − 5)4 dx = ∫u4 [du/3]
  • ∫(3x − 5)4 dx = [1/3] ∫u4 du
  • ∫(3x − 5)4 dx = [1/3] [(u5)/5] + c
∫(3x − 5)4 dx = [1/15] (3x − 5)5 + c
∫cos√{x2 + 2x + 1}  dx with x ≥ 0
  • There is a constraint that means the square root will never be negative
  • ∫cos√{x2 + 2x + 1}  dx = ∫cos√{(x + 1)2} dx
  • ∫cos√{x2 + 2x + 1}  dx = ∫cos(x + 1) dx
  • u = x + 1
  • du = dx
  • ∫cos√{x2 + 2x + 1}  dx = ∫cosu  du
  • ∫cos√{x2 + 2x + 1}  dx = sinu + c
∫cos√{x2 + 2x + 1}  dx = sin(x + 1) + c given x ≥ 0
∫(6x + 3)(x2 + x)7 dx
  • u = x2 + x
  • du = (2x + 1) dx
  • ∫(6x + 3)(x2 + x)7 dx = ∫(6x + 3) u7 [du/(2x + 1)]
  • ∫(6x + 3)(x2 + x)7 dx = ∫[(6x + 3)/(2x + 1)] u7 du
  • ∫(6x + 3)(x2 + x)7 dx = 3 ∫u7 du
  • ∫(6x + 3)(x2 + x)7 dx = 3 [(u8)/8] + c
∫(6x + 3)(x2 + x)7 dx = [3/8] (x2 + x)8 + c
∫√{7x + 9}  dx
  • u = 7x + 9
  • du = 7  dx
  • ∫√{7x + 9}  dx = ∫√u [du/7]
  • ∫√{7x + 9}  dx = [1/7] ∫√u  du
  • ∫√{7x + 9}  dx = [1/7] ∫u[1/2] du
  • ∫√{7x + 9}  dx = [1/7] [(u[3/2])/([3/2])] + c
  • ∫√{7x + 9}  dx = [2/21] u[3/2] + c
∫√{7x + 9}  dx = [2/21] (7x + 9)[3/2] + c
∫[(x2 + 1)/((x3 + 3x − 11)4)] dx
  • u = x3 + 3x − 11
  • du = (3x2 + 3) dx
  • ∫[(x2 + 1)/((x3 + 3x − 11)4)] dx = ∫[(x2 + 1)/(u4)] [du/(3x2 + 3)]
  • ∫[(x2 + 1)/((x3 + 3x − 11)4)] dx = ∫[(x2 + 1)/(3x2 + 3)] u−4 du
  • ∫[(x2 + 1)/((x3 + 3x − 11)4)] dx = [1/3] ∫u−4 du
  • ∫[(x2 + 1)/((x3 + 3x − 11)4)] dx = [1/3] [(u−3)/(−3)] + c
∫[(x2 + 1)/((x3 + 3x − 11)4)] dx = −[1/9] [1/((x3 + 3x − 11)3)] + c
∫cosx (sinx)6 dx
  • u = sinx
  • du = cosx  dx
  • ∫cosx (sinx)6 dx = ∫(sinx)6 cosx  dx
  • ∫cosx (sinx)6 dx = ∫u6 du
  • ∫cosx (sinx)6 dx = [(u7)/7] + c
∫cosx (sinx)6 dx = [1/7] (sinx)7 + c
∫[(sec2 √x)/(√x)] dx
  • u = √x
  • du = [1/2] [1/(√x)] dx
  • ∫[(sec2 √x)/(√x)] dx = 2 ∫sec2 √x [1/(2√x)] dx
  • ∫[(sec2 √x)/(√x)] dx = 2 ∫sec2 u  du
  • ∫[(sec2 √x)/(√x)] dx = 2 tanu + c
∫[(sec2 √x)/(√x)] dx = 2 tan√x + c
∫x2 (x3 + 4)5 dx
  • u = x3 + 4
  • du = 3x2 dx
  • ∫x2 (x3 + 4)5 dx = [1/3] ∫(x3 + 4)5 3x2 dx
  • ∫x2 (x3 + 4)5 dx = [1/3] ∫u5 du
  • ∫x2 (x3 + 4)5 dx = [1/3] [(u6)/6] + c
∫x2 (x3 + 4)5 dx = [1/18] (x3 + 4)6 + 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

Chain Rule

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
  • Chain Rule 0:07
    • Example 1
  • Example 2 3:17
  • Example 3 5:09
  • Example 4 7:53
  • Example 5 9:40
  • Example 6 11:39