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Week 2
Outline
Integration by parts
– The formula
– How to decide what should be
u
and what should be d
v
– Reduction formula
The Inverse Trig Substitution
– Integrals involving
√
a
2

x
2
– Integrals involving
√
a
2
+
x
2
or
1
x
2
+
a
2
– Integrals involving
√
x
2

a
2
Completing the square
Integrals involving
n
√
ax
+
b
Multiplying by a form of 1
The tan
θ
2
substitution
Integration by parts
If we reverse the product rule
(
u
(
x
)
v
(
x
))
0
=
u
0
(
x
)
v
(
x
) +
u
(
x
)
v
0
(
x
)
to do integration, we will get the formula of integration by parts
Z
u
(
x
)
v
0
(
x
)d
x
=
u
(
x
)
v
(
x
)

Z
u
0
(
x
)
v
(
x
)d
x.
By using the shortened forms
d
u
=
d
u
d
x
d
x
=
u
0
d
x
and
d
v
=
d
v
d
x
d
x
=
v
0
d
x,
we get the usual form for integration by parts
Z
u
d
v
=
uv

Z
v
d
u.
Remark
The trick in applying this powerful method is to decide which part of the original
integral should be
u
and which part should be d
v
.
1
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View Full DocumentExamples
1.
R
x
ln
x
d
x
2.
R
ln
x
d
x
3.
R
tan

1
x
d
x
4.
R
x
2
tan

1
x
d
x
5.
R
x
2
e
x
d
x
6.
R
e
x
sin
x
d
x
How to decide what should be
u
and what should be
d
v
Rule 1. The d
v
has to be something one can integrate. The
u
is what’s left over.
Rule 2. The resulting integral
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