Web 7.1
In w1–w4, integrate by parts.
w1.
Z
x
3
e
x
2
dx
=
Z
[
x
2
]
e
x
2
x dx
=
x
2
1
2
e
x
2
-
Z
1
2
e
x
2
2
x dx
by parts
=
”
-
1
2
Z
e
x
2
2
x dx
=
”
-
1
2
e
x
2
+
C
by B3
w2.
Do 35 in Ex 7.1 in the book without first substituting.
Solution.
Z
√
π
√
π
2
θ
3
cos
θ
2
dθ
=
Z
√
π
√
π
2
[
θ
2
] (cos
θ
2
θ dθ
)
=
θ
2
1
2
sin
θ
2
-
Z
1
2
sin
θ
2
2
θ dθ
√
π
√
π
2
=
· · ·
w3.
Do 30 in Ex 7.1 in the book.
Solution.
Z
1
0
r
3
√
4 +
r
2
dr
=
Z
1
0
[
r
2
]
(4 +
r
2
)
-
1
2
r dr
=
r
2
(4 +
r
2
)
1
2
-
Z
(4 +
r
2
)
1
2
2
r dr
1
0
=
· · ·
w4.
Z
x
3
dx
5
√
9 +
x
2
=
Z
[
x
2
]
(9 +
x
2
)
-
1
5
x dx
=
x
2
1
2
5
4
(9 +
x
2
)
4
5
-
Z
5
8
(9 +
x
2
)
4
5
2
x dx
=
5
8
x
2
(9 +
x
2
)
4
5
-
25
72
(9 +
x
2
)
9
5
+
C
In w5 and w6, integrate by parts twice. Then transpose the desired unknown integral from
the right side to the left side of the equation.
Then solve the equation for this unknown
integral.
w5.
Z
[
e
ax
] (cos
bx dx
) =
e
ax
1
b
sin
bx
-
a
b
Z
[
e
ax
] (sin
bx dx
)
This
preview
has intentionally blurred sections.
Sign up to view the full version.
=
1
b
e
ax
sin
bx
-
a
b
e
ax
-
cos
bx
b
-
Z
-
cos
bx
b
e
ax
a dx
!
=
1
b
e
ax
sin
bx
+
a
b
2
e
ax
cos
bx
-
a
2
b
2
Z
e
ax
cos
bx dx
Now we write the integral at the beginning of this ”string” equal to the expression after
the last ”equal sign” and we get:
This is the end of the preview.
Sign up
to
access the rest of the document.
- Spring '08
- Bonner
- Calculus, sin bx dx, eax sin bx, eax cos, sin bx, eax cos bx
-
Click to edit the document details