INSTRUCTORS Raffi Hovasapian John Zhu

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For more information, please see full course syllabus of Calculus AB

• ## Related Books

Lecture Comments (4)
 1 answerLast reply by: Angela PatrickSun Dec 15, 2013 7:26 AMPost by sergey smith on October 3, 2013on 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 answerLast reply by: Vicente OrellanaWed Jul 17, 2013 8:08 PMPost by Huseyin Kayahan on March 3, 2013for 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