Math 1152 Assignment 6 Solutions
DUE DATE: Mar.10,2017.
Page 342
40)
y=
Z
√
√
sin xdx
x
dy= 12 x−1/2 dx,
which implies, dy =
1
dx,
2y
hence dy2y = dx.
Z
Z
sin(y)dy(2y) = 2
ysin(y)dy
Integration By Parts
u=y
du = dy
Z
2[−ycos(y) −
dv = sin(y)dy
v = −cos(y)
√
√
√
−cos(y)dy] = −2ycos(y) + 2cos(y)dy = −2 xcos x + 2(sin x) + C
44)
Z
2
sinxcos3 xe1−sin x dx
Using sin2 x + cos2 x = 1, sin2 x = 1 − cos2 x
Z
Z
2
3
(1−(1−cos2 x))
sinxcos xe
dx = sinxcos3 xecos x dx
u=cosx, du=-sinxdx
Z
du
)=
u e (sinx)(
−sinx
3 u2
Z
3 u2
Z
−u e du = −
2
u3 eu du
v = u2 , dv=2udu
Z
=−
u3
e dv( ) = −
2u
v
y=v
dy = dv
1
−
2
Z
Z
u2
1
e dv( ) = −
2
2
v
Z
vev dv
dw = ev dv
v = ev
Z
−1 v 1 v
−1 v
ve dv =
ve − ev dv =
ve + e + C = ...
2
2
2
−1
1
2
2
=
(cos2 x)ecos x + ecos x + C
2
2
v
1
48)
1
Z
x3 ln(x2 + 1)dx
0
u = (lnx2 + 1)
du =
1
x4
(lnx2 + 1)0 −
4
=
1
Z
0
dv = x3 dx
2
1
(2x)(dx) = dx
2
x
x
2 x4
1
( )dx = −
x 4
4
Z
1
v=
x4
4
1 3
1 1
x dx = −
2
4 2
0
Z
1
x3 dx = ...
0
4 1 1 x 1 1
1 1
2 1
1
−
= − ( )( ) = − =
0
4 2 4
4
2 4
8 8
8
50)
Z
2
xe−2x dx
u=−2x2
du=-4xdx
−1
4
54)
Z
Z
−1 −2x2
e
+C
4
eu du =
Z
2
2x sinxdx = 2
x2 sinxdx
Integration By Parts
u = x2
dv = sinxdx
v = −cosx
du = 2xdx
2
2[−x cosx −
Z
2
2x(−cosx)dx] = −2x cosx + 4
Z
xcosxdx
Integration By Parts
u=x
du = dx
2
−2x cosx + 4[xsinx −
58)
Z
dv = cosxdx
v = sinx
sinxdx] = −2x2 cosx + 4xsinx + 4cosx + C
Z
x2
1
dx
+3
2
Note that, with constant "a",
Z
Z
Z
1
1
1
1
dx
=
dx
=
dx∗ = ...
∗
∗
a2 + x2∗
a2 (1 + (x2∗ /a2 ))
a2
1 + (x∗ /a)2
v= xa∗ , dv=(1/a)dx∗ , which implies (a)dv=dx∗
Z
Z
Z
Z
1
1
1
1
1
1
a
1
dx∗ = 2
(a)dv = 2
dv =
dv = ...
2
2
2
2
a
1 + (x∗ /a)
a
1+v
a
1+v
a
1 + v2
=
Hence,
1
x∗
1
arctan(v) + C = arctan( ) + C
a
a
a
Z
x2
1
1
x
dx = √ arctan( √ ) + C
+3
3
3
66)
2
Z
ln(x2 ex )dx
1
u = ln(x2 ex )
1
2+x
du = 2 x (2xex ) + ex x2 =
xe
x
dv = dx
v=x
(Note that, a(ln b)=(ln(ba )))
Z 2
Z 2
2
2
2
+
x
2 x
2
x
ln(x e )(x)1 −
)(x)dx = ln(x e )(x)1 −
(
(2 + x)dx = ...
x
1
1
Z 2
Z 2
2
2
2 x2 2
2 x
= ln(x e )(x) 1 −
2dx −
xdx = ln(x2 ex )(x)1 − 2x1 − 1 = ...
2
1
1
1
= 2(ln(4e2 )) − 4 − 2 − [ln(e) − 2 − ] = ...
2
1
−3
1
= ln(16) − 2 − [1 − 2 − ] = ln(16) − 2 − ( ) = ln(16) −
2
2
2
70)
Z
π/6
(1 + tan2 x)dx
0
2
2
Using 1 + tan x = sec x (From table of integrals,
Z
π/6
2
Z
(1 + tan x)dx =
0
0
π/6
R
sec2 xdx = tanx + C)
π/6
1
1
sec2 xdx = (tanx)0 = √ − 0 = √
3
3
Page 350
2)
x2 − 4x − 1
x−1
Degree of numerator is higher than the degree of the denominator.
f (x) = −
Using Long Division
3
−x+3
x − 1 − x2 + 4x + 1
x2 − x
3x + 1
− 3x + 3
4
Hence,
f (x) = −
x2 − 4x − 1
4
= −x + 3 +
x−1
x−1
4)
x3 − 3x2 − 15
x2 + x + 3
Degree of numerator is higher than the degree of the denominator.
f (x) =
Using Long Division
x −4
x2 + x + 3
x3 − 3x2
− 15
− x3 − x2 − 3x
− 4x2 − 3x − 15
4x2 + 4x + 12
x −3
Hence,
f (x) =
x−3
x3 − 3x2 − 15
=x−4+ 2
2
x +x+3
x +x+3
8)
f (x) =
16x − 6
(2x − 5)(3x + 1)
Using Partial Fractions
A
B
A(3x + 1) + B(2x − 5)
3Ax + A + 2Bx − 5B
+
=
=
2x − 5 3x + 1
(2x − 5)(3x + 1)
(2x − 5)(3x + 1)
Hence,
(3A + 2B)x = 16(x)
A − 5B = −6
(1)
(2)
From (2), A=5B-6,
Sub in A=5B-6 into (1),
3(5B-6)+2B=16
15B-18+2B=16
17B=34
Therefore, B=2
Sub B=-2 into (2)
A-5(2)=-6
4
A-10=-6
A=4
Therefore,
f (x) =
16x − 6
4
2
=
+
(2x − 5)(3x + 1)
2x − 5 3x + 1
10)
f (x) =
9x − 7
− 7x + 3
2x2
Factor the denominator,
2x2 − 7x + 3 = 2x2 − 6x − x + 3 = 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3)
f (x) =
9x − 7
(2x − 1)(x − 3)
Using Partial Fractions,
B
A(x − 3) + B(2x − 1)
Ax − 3A + 2Bx − B
A
+
=
=
2x − 1 x − 3
(2x − 1)(x − 3)
(2x − 1)(x − 3)
Hence,
(A+2B)x=9x
-3A-B=-7
(1)
(2)
From (2), B=7-3A
Sub in B=7-3A into (1)
A+2(7-3A)=9
A+14-6A=9
-5A=-5
A=1
Sub in A=1 into B=7-3A,
B=7-3(1)
B=4
Therefore,
f (x) =
14)
9x − 7
1
4
=
+
(2x − 1)(x − 3)
2x − 1 x − 3
Z
1
dx
x(2x + 1)
Using Partial Fractions,
B
A(2x + 1) + B(x)
2Ax + A + Bx
A
+
=
=
x 2x + 1
x(2x + 1)
x(2x + 1)
Therefore,
(2A+B)x=0(x) (1)
A=1
(2)
5
Sub A=1 into (1)
2(1)+B=0
B=-2
Hence,
A
B
1
−2
+
= +
x 2x + 1
x 2x + 1
Therefore,
Z
1
dx =
x(2x + 1)
Z
1
dx +
x
Z
−2
dx = ln|x| − 2
2x + 1
Z
1
dx
2x + 1
R 1
Note that, 2x+1
dx, u=2x+1, du=2dx, which implies dx = du
.
2
R 1 du
R 1
R 1
1
1
1
dx = u ( 2 ) = 2 u du = 2 ln|u| + C = 2 ln|2x + 1| + C
2x+1
Therefore,
Z
1
dx =
x(2x + 1)
Z
1
dx +
x
Z
1
−2
dx = ln|x| − 2( ln|2x + 1|) + C = ...
2x + 1
2
= ln|x| − ln|2x + 1| + C
18)
Z
4x2 − x − 1
dx
(x + 1)2 (x − 3)
Using Partial Fractions
A
B
C
+
+
= ...
2
x + 1 (x + 1)
x−3
=
A(x + 1)(x − 3) + B(x − 3) + C(x + 1)2
= ...
(x + 1)2 (x − 3)
A(x2 − 2x − 3) + B(x − 3) + C(x2 + 2x + 1)
=
(x + 1)2 (x − 3)
Hence,
(A + C)x2 = 4x2 , which implies A+C=4
(1)
(−2A + B + 2C)x = −x, which implies -2A+B+2C=-1 (2)
(-3A-3B+C)=-1
(3)
From (1), A=4-C
From (2), B=-1+2A-2C
From (3), -3B=-1+3A-C
Sub B=-1+2A-2C into -3B=-1+3A-C
-3B=-1+3A-C
-3(-1+2A-2C)=-1+3A-C
3-6A+6C=-1+3A-C
7C=-4+9A
6
Sub in A=4-C into 7C=-4+9A
Hence,
7C=-4+9(4-C)
7C=-4+36-9C
16C=32
C=2
Sub C=2 into (1)
A=4-C=4-2=2
Hence, A=2
Sub in A=2 and C=2 into (2)
B=-1+2(2)-2(2)=-1+4-4=-1
Therefore, A=2, B=-1, and C=2
Z
Z
Z
4x2 − x − 1
2
−1
2
dx =
dx +
dx = ...
dx +
2
2
(x + 1) (x − 3)
x+1
(x + 1)
x−3
Z
Z
Z
1
1
1
=2
dx −
dx
+
2
dx = ...
x+1
(x + 1)2
x−3
Z
1
= 2ln|x + 1| −
dx + 2ln|x − 3| + C
(x + 1)2
R 1
Note that, (x+1)
2 dx, u=x+1, du=dx
R 1
R 1
1
dx = u2 dx = − u1 + C = − x+1
+C
(x+1)2
Z
Therefore,
Z
Z
= 2ln|x + 1| −
4x2 − x − 1
dx = ...
(x + 1)2 (x − 3)
1
1
dx
+
2ln|x
−
3|
+
C
=
2ln|x
+
1|
+
+ 2ln|x − 3| + C
(x + 1)2
x+1
7