HW 5 Solution
1
Z
(7.1 Problem 20)
2
2
(x2 − 2x)(x3 − 3x2 + 3) 3 dx
Solution: Let u = x3 −3x2 +3. Then du = (3x2 −6x)dx and du
= (x2 −2x)dx.
3
R
R 2
5
5
2 du
2
So (x − 2x)(x3 − 3x2 + 3) 3 dx = u 3 3 = 13 · 53 u 3 + C = 15 u 3 + C =
5
1
(x3 − 3x2 + 3) 3 + C.
5
x2 − 1
dx
x3 − 3x + 1
Z
(7.1 Problem 22)
3
Solution: Let u = x3 − 3x + 1. Then du = (3x2 − 3)dx and du
= (x2 − 1)dx.
3
R x2 −1
R
R
R
1
So x3 −3x+1 dx = x3 −3x+1
· (x2 − 1)dx = u1 · du
= 13 u1 du = 31 ln |u| + C =
3
1
ln |x3 − 3x + 1| + C.
3
Z
sec2 (x)etan(x) dx
(7.1 Problem 28)
Solution:
Then du = sec2 (x)dx. So
R tan(x) 2Let u = Rtan(x).
e
sec (x)dx = eu du = eu + C = etan(x) + C.
4
5
R
sec2 (x)etan(x) dx =
Z √
ln x
(7.1 Problem 36)
1 + ln x
dx
x
R√
Solution: Let u = 1+ln x. Then du = x1 dx and ln x = u−1 So
1 + ln x lnxx dx =
R√
R
R
R
√
3
1
1
1 + ln x · ln x · x1 dx =
u · (u − 1)du = u 2 · (u − 1)du = (u 2 − u 2 )du =
3
5
3
2 52
u − 32 u 2 + C = 25 (1 + ln x) 2 − 23 (1 + ln x) 2 + C.
5
Z
2
(7.1 Problem 44)
√
x5 x3 + 2dx
1
Solution: Let u = x3 + 2. Then du = 3x2 dx, du
= x2 dx and x3 = u − 2.
3
3
Also
x = 1 gives uR= 1 √+ 2 = 3 and x R= 2 gives u = 23 + 2R= 8 + 2 = 10. So
R 2 5√
√
1
10
2
10
x x3 + 2dx = 1 x3 x3 + 2x2 dx = 3 (u − 2) u du
= 13 3 (u − 2)u 2 du =
3
1
10
10
R
5
3
1
1 2 25
2 32 2 52
4 32 2
1 10
2 − 2u 2 )du =
(u
(
u
−
2
·
u
)
=
(
u
−
u
)
· 10 2 − 49 ·
= ( 15
3 3
3 5
3
15
9
3
3
3
5
3
5
3
5
3
2
2
2
10 2 ) − 15
· 3 2 − 49 · 3 2 = 15
· 10 2 − 49 · 10 2 − 15
· 3 2 + 49 · 3 2 .
6
Z
(7.1 Problem 52)
0
π
3
sin x
dx
cos2 (x)
Solution: Let u = cos(x). Then du = − sin(x)dx and sin(x)dx = −du.
Also x = 0 gives u = cos(0) = 1 and x = π3 gives u = cos( π3 ) = 12 . So
MATH 1760 : page 1 of 2
HW 5 Solution
Rπ
3
0
7
1
1
sin x
dx
cos2 (x)
MATH 1760 page 2 of 2
=
Rπ
3
0
1
cos2 (x)
· sin(x)dx =
R1
1
(−1)du
1 u2
2
=−
R1
1
du
1 u2
2
1
2
= u1 =
1
1
1
2
−
= 2 − 1 = 1.
Z
(7.2 Problem 10)
2x2 e−x dx
R −x
Solution: RLet u = 2x2 and dv = e−x dx.
Then
du
=
4xdx
and
v
=
dx =
R
R e −x
−x
2 −x
2
−x
−x
2 −x
−e . So 2x e R dx = 2x (−e ) − (−e ) · 4xdx = −2x e + 4 xe dx.
Next we find xe−x dx using integration
R by parts again. RLet u = x
and dv = eR−x dx. Then du = dx Rand v = e−x dx = −e−x . So xe−x dx =
x(−e−x ) − (−eR−x )dx = −xe−x + e−x dx R= −xe−x − e−xR+ C.
Combining 2xR2 e−x dx = −2x2 e−x + 4 xe−x dx and xe−x dx = −xe−x −
e−x + C, we have 2x2 e−x dx = −2x2 e−x − 4xe−x − 4e−x + C.
8
Z
(7.2 Problem 12)
x2 ln xdx
R
Solution: Let u = ln(x) and dv = x2 dx. Then du = x1 dx and v = x2 dx =
R
R
R 3
R 2
3
3
x3
. So x2 ln xdx = ln(x) · x2 dx = ln(x) · x3 − x3 · x1 dx = x ln(x)
− x3 dx =
3
3
3
x3 ln(x)
− x9 + C.
3
9
Z π
4
(7.2 Problem 18)
2x cos(x)dx
0
R
Let Ru = 2x and dv = cos(x)dx. Then
du
=
2dx
and
v
=
cos(x)dx
R
R = sin(x).
So 2x cos(x)dxdx = 2x sin(x) − sin(x) · 2dx = 2x sin(x) − 2 sin(x)dx =
2x sin(x) + 2 cos(x) + C.
π
Rπ
Thus 04 2x cos(x)dx = 2x sin(x) + 2 cos(x)04 = 2 · π4 sin( π4 ) + 2 cos( π4 ) − (2 ·
√
π
0 sin(0) + 2 cos(0)) = π2 · √12 + 2 · √12 − 2 = 2√
2 − 2. We have used
+
2
π
π
1
sin( 4 ) = cos( 4 ) = √2 and cos(0) = 1.