Math 131 Lab 8
Z Do * problems first.
Then do the others. Brief answers are on the back. Reduction formulas on back
1. Higher Powers of Trig Functions Use the rules we developed in class to determine
Z
Z
Z
2
3
5
a) ∗ sin 2x cos 2x dx
b) ∗ cos 2x dx
c)
sec5 (2x) dx
Z
Z
Z
d) ∗ sin3 (5x) cos−7 (5x) dx
e)
cos2 (10x) sin2 (10x) dx
f)
tan2 (πx) sec2 (πx) dx Think first!
2. Low Powers of Trig Functions which you should already know.
Z
Z
Z
a)
cos 9x dx
b) ∗ tan2 (πx) dx
c) ∗ sin2 (−7x) dx
3. Before working these out, classify each by the technique that applies: mentally adjust, substitution, parts, parts twice,
or ordinary methods. Check your answers by differentiating.
Z
Z 1
Z π
2
a) ∗ xex dx
b) ∗
(x2 − x + 1)ex dx
c) ∗
(x − π) sin x dx
0
0
Z
Z
Z
d)
ex sin(ex ) dx
e)
e3x sin x dx
f ) ∗ sin(5x) cos(x) dx
R
4. a) *Assume that n 6= −1. Determine a formula for xn ln x dx.
R
b) Use your answer to quickly determine (without further integration) lnx6x dx.
R
5. a) *Determine (ln x)2 dx.
b) *Let R be the region enclosed by y = ln x, the x-axis, and x = e in the first quadrant. Rotate R about the x-axis
and find the volume. Use your answer to part (a).
√
6. Area Review: Find the area of the region enclosed by y = arcsin x, the x-axis, and x = 2/2 in the first quadrant
by integrating along the x-axis.
7. a) *Work Review: A tank is formed from rotating the region enclosed by y = arcsin(x2 ), the y-axis, y = π/2, and
the x-axis. The tank is full of whale oil (60 lbs/ft3 ). Find the work “lost” (the work will be negative) by draining
the tank through a hole in the bottom. Hint: Remember H represents the height where the liquid ends up. What
is H in this problem? (Set up the integral first and decide on a technique. Solve later.)
b) Just set up the integral for the work is lost if only the top half is drained.
8. *Volume Review: Let R be the region in the first quadrant enclosed by y = cos x, y = sin x, and the y-axis.
a) Rotate R about the x-axis and find the volume using the disk method. (Do set-up now, check answer later.)
b) Rotate R about the y-axis and find the volume by using the shell method. (Do set-up now, check answer later.)
9. WeBWorK setDay21. Let R be the region enclosed by y = ln x, the y-axis, the x-axis, and y = 1 in the first quadrant.
Rotate R about the y-axis to form a tank. If liquid has density 60 lbs/cu. ft., how much work is required to fill it
through a hole in the bottom of the tank? Hint: During the integration you can use Problem 4.
10. [Harder] Let R be the region enclosed by y = ex , the y-axis, and y = e in the first quadrant. Rotate R about the y-axis
to form a tank. If it is full of a liquid whose density is 60 lbs/cu. ft., how much work is lost if it leaks out the bottom
and drops to ground level?
Z
ln x
dx by u-substitution. Show that you can do it by parts—but as a
11. a) In the past we have done the problem
x
problem that cycles back on itself (a technique that we had not seen until Day 20–21).
b) Check your answer by doing the same problem via substitution.
R
12. a) WeBWorK setExtraCredit2. Assume that n is a positive integer. Create a reduction formula for xn ex dx by
using integration by parts once. Hint: Let u = xn .
R
b) Extra Credit: Use your reduction formula (repeatedly) to find x3 ex dx.
1
Math 131 Lab 8 Brief Answers. Caution: Be careful of typos. Extra Credit if you are the first to find one.
1. a)
b)
c)
d)
e)
sin3 2x sin5 2x
−
+c
6
10
cos4 2x sin 2x 2 cos2 2x sin 2x 4 sin 2x
+
+
+c
10
15
15
sec3 (2x) tan(2x) 3 sec(2x) tan(2x) 3 ln | sec(2x) + tan(2x)|
+
+
+c
8
16
16
cos−6 (5x) cos−4 (5x)
−
+c
30
20
1
3
3π tan (πx) + c
2. Use half-angle for (c).
a)
1
9
sin 9x + c
b)
1
π
tan(πx) − x + c
c)
x
1
+
sin(−14x) + c
2 28
3. Sketches:
a)
1 x2
2e
0
0
d) − cos ex + c
4. a)
π
c) −(x − π) cos x + sin x = −π
1
b) (x2 − 3x + 4)ex = 2e − 4
+c
1
n+1
n+1 x
ln x −
e)
1
3x
10 (3e
1
n+1
(n+1)2 x
1
f ) − [sin(5x) sin x + 5 cos(5x) cos x] + c
24
sin x − e3x cos x) + c
+ c; (b) − 15 x−5 ln(x) −
1 −5
25 x
+c
5. a) Use parts twice: x(ln x)2 − 2x ln x + 2x + c.
b) Use (a): π(e − 2).
6. By parts: x arcsin x +
7. a) W = 60
8. a) V = π
R π/2
0
R π/4
0
√
1−
√2/2
x2 0
√
=
2
8 π
π(0 − y)(sin y) dy = −60π
+
√
R π/2
0
2/2 − 1.
y sin y dy. Use parts: −60π ft-lbs. (b) −60π
cos2 x − sin2 x dx = 12 π. (b) V = 2π
R π/4
0
√
x cos x − x sin x dx =
9. −15π(e2 + 1) ft-lbs.
10. −15π(e2 − 1) ft-lbs.
11.
(ln x)2
2
+ c.
12. a) xn ex − n
R
xn−1 ex dx. (b) x3 ex − 3x2 ex + 6xex − 6ex + c.
Reduction Formulas for Large Powers.
Z
R
1)
cosn x dx = n1 cosn−1 x sin x + n−1
cosn−2 x dx
n
Z
R
2)
sinn x dx = − n1 sinn−1 x cos x + n−1
sinn−2 x dx
n
Z
R
1
3)
tann x dx = n−1
tann−1 x − tann−2 x dx
Z
R
1
4)
secn x dx = n−1
secn−2 x tan x + n−2
secn−2 x dx
n−1
2
2 2
2 π
− 2π.
R π/2
π/4
y sin y dy.
Math 131 Lab 8 Answers
1. a) Since the power of the cosine function is odd, we use Guideline #2: Split off a power of cos 2x, then let u = sin 2x,
so 12 du = cos 2x dx.
Z
sin2 2x cos3 2x dx =
Z
sin2 2x cos2 2x · cos 2x dx =
Z
sin2 2x(1 − sin2 2x) · cos 2x dx
1 u4
u5
sin3 2x sin5 2x
−
+c=
−
+c
2 3
5
6
10
Z
Z
1
cos4 3x dx =
cos4 u du
b) Use reduction formulas. First let u = 3x, so 13 du = dx. So
3
Z
Z
1 cos3 u sin u 3
1 cos3 u sin u 3 cos u sin u 1
2
=
+
cos u du =
+
+
1 du
3
4
4
3
4
4
2
2
=
=
1
3
1
2
Z
cos3 u sin u 3 cos u sin u u
cos3 3x sin 3x cos 3x sin 3x 3x
+
+
+c=
+
+
+c
4
4
2
2
12
8
8
u2 (1 − u2 ) du =
1
2
Z
u2 − u4 du =
c) Use reduction formula and substitution: u = 2x ⇒ 12 du = dx. So
Z
sec5 (2x) dx =
1
2
Z
sec5 u du =
Z
Z
1 sec3 u tan u 3
1 sec3 u tan u 3 sec u tan u 1
+
sec3 u du =
+
+
sec du
2
4
4
2
4
4
2
2
1 sec3 u tan u 3 sec u tan u 3 ln | sec u + tan u|
sec3 (2x) tan(2x) 3 sec(2x) tan(2x) 3 ln | sec(2x) + tan(2x)|
+
+
+
+
+c
=
+c =
2
4
8
8
8
16
16
d) Since the power of the sine function is odd, we use Guideline #1: Split off a power of sin 5x, then let u = cos 5x, so
Use u = cos(5x) so − 15 du = sin(5x)dx.
Z
sin3 (5x) cos−7 (5x) dx =
1
=−
5
Z
2
(1 − u )u
−7
Z
sin2 (5x) cos−7 (5x) sin(5x) dx =
1
du = −
5
Z
u−7 − u−5 ) du =
Z
[1 − cos2 (5x)] cos−7 (5x) sin(5x) dx
u−6
u−4
cos−6 (5x) cos−4 (5x)
−
+c=
−
+c
30
20
30
20
e) Use Half-angle forumlas:
Z
2
2
Z cos (10x) sin (10x) dx =
Z
=
f ) u-substitution.
Z
1 1
1 1
1 1
+ cos(20x)
− cos(20x) dx =
− cos2 (20x) dx
2 2
2 2
4 4
Z
1 1
x
1
1 1 1 1
−
+ cos(40x) dx =
− cos(40x) dx = −
sin(40x) + c
4 4 2 2
8 8
8 320
Substitution:
du = π sec2 πx dx
u = tan πx
1
π du
=
1
π
R
u2 du =
1 3
3π u
+c=
1
3π
tan3 (πx) + c
= x dx
R
R
2. a) Mental adjustment (u = 9x). cos 9x dx = 19 cos u du = 91 sin u + c =
R
R
b) Trig id: tan2 (πx) dx = sec2 (πx) − 1 dx = π1 tan(πx) − x + c.
1
9
sin 9x + c.
c) Use half-angle formula and followed by a mental adjustment:
Z
Z
1
1
1 1
sin2 (7x) dx =
− cos(−14x) dx = x +
sin(−14x) + c
2 2
2
28
3
u = x2 − x + 1
du = (2x − 1)dx
b) Parts twice:
u = 2x − 1
du = 2dx
1
2
2
eu du = 12 eu + c = 21 ex + c.
1 R 1
dv = ex dx = (x2 − x + 1)ex 0 − 0 (2x − 1)ex dx
v = ex
1
1 R 1
dv = ex dx = (x2 − x + 1)ex 0 − [(2x − 1)ex 0 − 0 2ex dx]
1
v = ex
= (x2 − x + 1)ex − (2x − 1)ex + 2ex 0 = 2e − 4
3. a) Substitution (u = x2 ⇒ du = 2x dx). ⇒
R
π R π
= −(x − π) cos x0 + 0 cos x dx
c) Parts:
π
du = dx
v = − cos x = −(x − π) cos x + sin x0 = (0 + 0) − (π) = −π
R
d) Substitution (u = ex ⇒ du = ex dx). ⇒ sin u du = − cos u + c = − cos(ex ) + c.
e) Parts twice, “circle around”. Note signs and constants:
u=x−π
dv = sin xdx
u = e3x
du = 3e3x dx
dv = sin xdx
v = − cos x
R
u = 3e3x
dv = cos xdx
R
du = 9e3x dx
v = sin x
e3x sin x dx = −e3x cos x +
R
3e3x cos x dx
R
e3x sin x dx = −e3x cos x + 3e3x sin x − 9 e3x sin x dx
R
So, 10 e3x sin x dx = e3x (3 sin x − cos x) + c
R
1 3x
Thus, e3x sin x dx = 10
e (3 sin x − cos x) + c
f ) Parts twice, “circle around”. Note signs and constants:
u = sin(5x)
du = 5 cos(5x) dx
dv = cos x dx
v = sin x
R
u = 5 cos(5x)
dv = sin x dx
R
du = −25 sin(5x) dx
v = − cos x
4. a) Parts:
b)
du = x−1 dx
− 15 x−5 (ln(x)
5. a) Parts twice:
+
1
5)
v=
1
n+1
n+1 x
R
5 cos(5x) sin x dx
R
sin(5x) cos x dx = sin(5x) sin x + 5 cos(5x) cos x + 25 sin(5x) cos x dx
R
So, −24 sin(5x) cos x dx = sin(5x) sin x + 5 cos(5x) cos x + c
R
1
Thus, sin(5x) cos x dx = − 24
[sin(5x) sin x + 5 cos(5x) cos x] + c
R
xn ln x dx =
R
xn ln x dx =
1
n+1
n+1 x
1
n+1
n+1 x
ln x −
ln x −
1
n
n+1 x dx
1
n+1
+
(n+1)2 x
R
c
+ c.
u = (ln x)2
du =
2 ln x
x dx
dv = dx
v=x
= x(ln x)2 − 2
R
ln x dx. Use parts again in problem #4 with n = 0.
2
= x(ln x) − 2x ln x + 2x + c
...
y = ln .x
....... ..
..... ...
1
b) 0
6. A =
dv = xn dx
u = ln x
sin(5x) cos x dx = sin(5x) sin x −
..
......
......
.....
.....
.
.
.
....
....
.....
...
...
.
.
...
...
...
..
...
...
...
...
...
..
1
e
R b=√2/2
0
arcsin x dx
From (a):
.
......
.... .
.... ..
.... ....
.
.
.
.
....
....
.....
...
....
....
...
...
.
.
...
.
...
.
...
.
..
.
.
.
.
...
.
....
0
Re
1
e
π(ln x)2 dx = π x(ln x)2 − 2x ln x + 2x = π(e − 2).
1
Parts:
b
u = arcsin x
1
du = √1−x
dx
2
dv = dx
v=x
Substitution:
u = 1 − x2
du = −2x dx
− 12 du = x dx
√2/2 R √2/2
= x arcsin x0
− 0
√ x
1−x2
dx
√2/2 1 R −1/2
√2/2
= x arcsin x0
+2 u
du
=
x
arcsin
x
+ u1/2
0
√
√
√2/2
√
= x arcsin x + 1 − x2 0
= 82 π + 22 − 1
4
7. a) Save work! Since the radius of a cross-section of the tank is x, we need to solve for x2 (not x): y = arcsin(x2 ), so
Z π/2
Z π/2
x2 = sin y. W = 60
π(0 − y)(sin y) dy = −60π
y sin y dy. Use parts:
0
0
h
i
π/2 R π/2
u=y
dv = sin y dy = −60π −y cos y 0 + 0 cos y dy
h
π/2 i
du = dy
v = − cos y
= −60π (0 − 0) + sin y 0
= (0 + 0) + (1 − 0)) = −60π lbf
Z π/2
π(0 − y)(sin y) dy
b) Only the lower limit changes to π/4 since the upper part of the tank drains out: W = 60
π/4
8. a) V = π
R π/4
b) V = 2π
u=x
du = dx
So π
R e2
1
0
cos2 x − sin2 x dx = π
R π/4
0
R π/4
0
x cos x − x sin x dx = 2π
dv = (cos x − sin x)dx
v = sin x + cos x
( 12 +
R π/4
0
2π
R π/4
2π
R π/4
0
0
1
2
cos 2x) − ( 21 −
1
2
cos 2x) dx = π
R π/4
0
π/4
cos 2x) dx = 21 π sin(2x)0 = 12 π.
x(cos x − sin x) dx. Use parts
π/4
R
x(cos x − sin x) dx = 2π x(sin x + cos x) − (sin x + cos x) dx 0
π/4
x cos x − x sin x dx = 2π [x(sin x + cos x) + cos x − sin x dx]0
√
=
2 2
2 π
− 2π
e2
(ln x)2 dx = π[x(ln x)2 − 2x ln x + 2x]1 = π[(4e2 − 4e2 + 2e2 ) − (0 − 0 + 1) = π(2e2 − 1).
9. Since y = ln x, then x = ey . So W =
u=y
du = dy
R1
0
60π(ey )2 (y − 0) dy = 60π
R1
0
ye2y dy. By parts
1 R 1
= 60π[ 12 ye2y 0 − 0 21 e2y dy]
1
=60π[ 12 ye2y − 41 e2y ]0
dv = e2y dy
v = 12 e2y
−60π[( 12 e2 − 14 e2 ) − (0 − 41 )] = −60π[ 14 e2 + 14 ] = −15π(e2 + 1)
10. Since y = ex , then x = ln y. So W =
using #4 with n = 1:
u = (ln y)2
y
du = 2 ln
y dy
Re
1
60π(ln y)2 (0 − y) dy = −60π
dv = y dy
v = 12 y 2
Re
1
y(ln y)2 dy. By parts twice, the second time
e R e
= −60π[y(ln y)2 1 − 1 y ln y dy]
e
= −60π[ 21 y 2 (ln y)2 − 12 y 2 ln y + 14 y 2 ]1
Use #4 with n = 1:
=−60π[( 21 e2 − 21 e2 + 14 e2 ) − (0 + 0 + 14 )] = −15π(e2 − 1)
1
x
ln x
x
R
dx = (ln x)2 − lnxx dx Cycle!
11. a) Parts:
R
R
2
du = x1 dx
So 2 lnxx dx = (ln x)2 or lnxx dx = (ln2x) + c
Z
Z
ln x
u2
(ln x)2
b) EZ u-substitution: u = ln x, du = x1 dx. So
dx = u du =
+c=
+c
x
2
2
u = ln x
12. a) Parts:
Z
b)
u = xn
dv =
dx
R
dv = ex dx
R
xn ex dx = xn ex − n
R
xn−1 ex dx
du = nxn−1 dx
v = ex
Z
Z
x3 ex dx = x3 ex − 3 x2 ex dx = x3 ex − 3 x2 ex − 2 xex dx
Z
= x3 ex − 3 x2 ex − 2(xex − ex dx) = x3 ex − 3x2 ex + 6xex − 6ex + c.
5