math 54 lesson 1 - review of integration
TRANSCRIPT
Review of Integration-Substitution Rule
Review of Integration–Substitution RuleMathematics 54–Elementary Analysis 2
Institute of MathematicsUniversity of the Philippines-Diliman
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Review of Integration-Substitution Rule
Review of IntegrationIntegration Formulas
1
∫un du =
un+1
n+1+C, n 6= −1
ln |u|+C, n =−1
2
∫au du = au
lna+C, a > 0,a 6= 1
3
∫sinudu =−cosu+C
4
∫cosudu = sinu+C
5
∫sec2 udu = tanu+C
6
∫csc2 udu =−cotu+C
7
∫secu tanudu = secu+C
8
∫cscucotudu =−cscu+C
9
∫cotudu = ln |sinu|+C
10
∫tanudu = ln |secu|+C
11
∫secudu = ln |secu+ tanu|+C
12
∫cscudu = ln |cscu−cotu|+C
13
∫du
a2 +u2= 1
atan−1
( u
a
)+C,a 6= 0
14
∫sinhudu = coshu+C
15
∫coshudu = sinhu+C
16
∫sech2 udu = tanhu+C
17
∫csch2 udu =−cothu+C
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Review of Integration-Substitution Rule
Review of Integration
Example 1. Evaluate ∫x3
√x2 +1 dx.
Let u = x2 +1 =⇒ du = 2x dx∫x3
√x2 +1 dx = 1
2
∫x2 ·2x
√x2 +1 dx
= 1
2
∫(u−1)u
12 du
= 1
2
∫ (u
32 −u
12
)du
= 1
2
[u
52
52
− u32
32
]+C
= (x2 +1)52
5− (x2 +1)
32
3+C
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Review of Integration-Substitution Rule
Review of Integration
Example 2. Evaluate∫5xecsch5x
csch(5x)coth(5x) dx.
Let u = csch5x =⇒ du =−csch(5x
)coth
(5x
) (5x
)ln5 dx∫
5xecsch5xcsch(5x)coth(5x) dx = −1
ln5
∫eu du
= −1
ln5eu +C
= −1
ln5ecsch5x +C
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Review of Integration-Substitution Rule
Review of Integration
Example 3. Evaluate ∫dx
4x2 −12x+18.
∫dx
4x2 −12x+18=
∫dx
4x2 −12x+9+9
=∫
dx
(2x−3)2 +32
Let u = 2x−3 ⇒ du = 2 dx∫dx
(2x−3)2 +32 = 1
2
∫du
u2 +32
= 1
2· 1
3tan−1
(u
3
)+C
= 1
6tan−1
(2x−3
3
)+C
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Review of Integration-Substitution Rule
Review of Integration
Example 4. Evaluate ∫ 0
−12cosh[ln(1−2x)] dx.
∫ 0
−12cosh[ln(1−2x)] dx = 2
∫ 0
−1
eln(1−2x) +e− ln(1−2x)
2dx
=∫ 0
−1(1−2x)+ (1−2x)−1 dx
Let u = 1−2x =⇒ du =−2 dxNote: x = 0 ⇒ u = 1 & x =−1 ⇒ u = 3∫ 0
−1
((1−2x)+ (1−2x)−1) dx = −1
2
∫ 1
3
(u+u−1) du
= −1
2
(u2
2+ lnu
)]1
3=−1
2[−4− ln3]
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