integration by parts

KG THWALA MATE1B1 2020
Assignment HW7 due 10/02/2020 at 11:59pm SAST



and we can find the second integral by using integration by parts
1. (1 point) again. Here, we let u = t and v0 = sin(t), so that u0 = 1 and
Antidifferentiate using a table of integrals. You may need to v = − cos(t), and
transform the integrand first. Z Z
5
Z
dz = 2t 2 cos(t) dt = 2t 2 sin(t) + 4t cos(t) − 4 cos(t) dt
4 + (z + 2)2
Solution:
SOLUTION = 2t 2 sin(t) + 4t cos(t) − 4 sin(t) +C.
Substituting w = z + 2, we get Answer(s) submitted:
Z
5 5 z+2 • 2((tˆ(2)-2)sin(t)+2t cos(t))+C
dz = arctan( ) +C.
4 + (z + 2)2 2 2 (correct)
Correct Answers:
Answer(s) submitted:
• 2*[tˆ2*sin(t)+2*t*cos(t)-2*sin(t)]+C
• [[5arctan((2z+4)/(4))]/2]+C
(correct) 4. (1 point)
Correct Answers:
Find the integral
• 2.5*atan((z+2)/2)+C (z + 1) e3z dz =
R

Solution:
2. (1 point) SOLUTION
Antidifferentiate using a table of integrals. You may need to We use integration by parts ( u v0 dx = u · v − u0 v dx) with
R R
transform the integrand first. u = z + 1 and v0 = e3z , so that u0 = 1 and v = 13 e3z . Then
dx
Z
√ =
z + 1 3z 1 3z z + 1 3z 1 3z
Z Z
36 − 25x2 (z + 1) e3z dz = e − e dz = e − e +C.
Solution: 3 3 3 9
SOLUTION
Answer(s) submitted:
We first factor out the 25 and then use the table to get
• [[[3z+2]eˆ(3z)]/9]+C
dx dx 1 dx
Z Z Z
√ = q = q = (correct)
36 − 25x2 25( 36 2 5 ( 56 )2 − x2
25 − x ) Correct Answers:
!   • [(1+z)/3-0.111111]*eˆ(3*z)+C
1 x 1 5
arcsin 6
+C = arcsin x +C.
5 5
5 6 5. (1 point)
Answer(s) submitted:
Find
R 6
the integral
x ln(x) dx =
• [[sinˆ(-1)((5x)/(6))]/5]+C
Solution:
(correct) SOLUTION
Correct Answers: We use integration by parts ( u v0 dx = u · v − u0 v dx) with
R R

• 0.2*asin(5*x/6)+C u = ln(x) and v0 = x6 . Then u0 = 1x and v = 71 x7 , so

1 1 6 1 1
Z Z
3. (1 point) x6 ln(x) dx = x7 ln(x) − x dx = x7 ln(x) − x7 +C.
Find the integral 7 7 7 49
R 2
2t cos(t) dt = Answer(s) submitted:
Solution:
• [[xˆ(7)(7ln(x)-1)]/49]+C
SOLUTION
We use integration by parts ( u v0 dx = u · v − u0 v dx) with (correct)
R R
2 0 0
u = t and v = cos(t), so that u = 2t and v = sin(t). Then Correct Answers:
Z Z • 0.142857*xˆ7*ln(x)-0.0204082*xˆ7+C
2t 2 cos(t) dt = 2t 2 sin(t) − 4t sin(t) dt,
1

, • w*asin(2*w)+0.5*sqrt(1-4*wˆ2)+C
6. (1 point)
Find the integral 9. (1 point)
R √
y y + 8 dy = Find
R 5
the integral
Solution: x sin(x3 ) dx =
SOLUTION Solution:
We use integration by parts ( u v0 dx = u · v − u0 v dx) with
R R
SOLUTION
u = y and v0 = (y + 8)1/2 . Then u0 = 1 and v = 23 (y + 8)3/2 . Thus We use integration by parts with v0 containing the sin(x3 )
term; to be able to integrate this, we must also include a factor
2 2
Z Z
1/2
y(y + 8) dy = ( ) · y · (y + 8)3/2 − (y + 8)3/2 dy = of x2 , and so let u = x3 and v0 = x2 sin(x3 ), so that u0 = 3x2 and
3 3 v = − 13 cos(x3 ). Thus
2 4
( ) · y · (y + 8)3/2 − · (y + 8)5/2 +C. 1
Z Z
3 15 x5 sin(x3 ) dx = − x3 cos(x3 ) + x2 cos(x3 ) dx =
Answer(s) submitted: 3
• [[(y+8)ˆ(3/2)(6y-32)]/15]+C 1 3 1
− x cos(x3 ) + sin(x3 ) +C.
(correct) 3 3
Correct Answers: Answer(s) submitted:
• 0.666667*y*(y+8)ˆ1.5-0.266667*(y+8)ˆ2.5+C • [[sin(xˆ(3))-xˆ(3)cos(xˆ(3))]/3]+C
(correct)
7. (1 point) Correct Answers:
Find
R 3 ln x
the integral • 0.333333*[-xˆ3*cos(xˆ3)+sin(xˆ3)]+C
x 5 dx =
Solution: 10. (1 point)
SOLUTION For each of the following integrals, indicate whether integra-
We use integration by parts ( u v0 dx = u · v − u0 v dx) with tion by substitution or integration by parts is more appropriate,
R R

u = ln(x) and v0 = x−5 . Thus u0 = 1x and v = − 4x14 , so that or if neither
R
method is appropriate. Do not evaluate the integrals.
  1. x sin x dx
3 ln x 1 1
Z Z
dx = 3 − 4 · ln(x) + dx = • A. neither
x5 4x 4x5
  • B. integration by parts
1 1 • C. substitution
3 − 4 · ln(x) − +C. R x2
4x 16x4 2. 1+x3 dx
Answer(s) submitted:
• A. substitution
• -[[12ln(x)+3]/(16xˆ(4))]+C • B. neither
(correct) • C. integration by parts
Correct Answers: R 2 x3
3. x e dx
• 3*(-[1/(4*xˆ4)]*ln(x)-1/(16*xˆ4))+C
• A. integration by parts
8. (1 point) • B. neither
Find
R
the integral • C. substitution
arcsin(2w) dw = 4. x2 cos(x3 ) dx
R
Solution:
• A. neither
SOLUTION
• B. substitution
We use integration by parts ( u v0 dx = u · v − u0 v dx) with
R R
• C. integration by parts
u = arcsin(2w) and v0 = 1. Then u0 = √ 2
2
and v = w, so that R 1
1−4w 5. √4x+1 dx
2w
Z Z
arcsin(2w) dw = w · arcsin(2w) − √ dw = • A. neither
1 − 4w2 • B. substitution
1p • C. integration by parts
w · arcsin(2w) + 1 − 4w2 +C.
2 (Note that because this is multiple choice, you will not be
The second integral is found using substitution with a substitu- able to see which parts of the problem you got correct.)
tion variable W = 1 − 4w2 . Solution:
Answer(s) submitted: SOLUTION
• [[2wsinˆ(-1)(2w)+sqrt(1-4wˆ(2))]/2]+C For each of these, we’re looking to see if there is a good sub-
(correct) stitution (we can take w to be the argument of a function, etc.,
Correct Answers: such that its derivative, dw = w0 dx, appears in the integrand; or,
2