Guide to Integration
Mathematics 101
Mark MacLean and Andrew Rechnitzer
Winter 2006/2007
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Guide to Integration
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1
Elementary Integrals Substitution Trigonometric integrals Integration by parts Trigonometric substitutions Partial Fractions 100 Integrals to do
2
3
4
5
6
7
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Table of integrals
Recognise these from a table of derivatives.
The very basics
1
1 dx = x + c 1 dx = log |x| + c x 1 x n+1 + c x n dx = n+1 1 e ax dx = e ax + c a
2
3
4
()
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Table of integrals
Recognise these from a table of derivatives.
The very basics
1
1 dx = x + c 1 dx = log |x| + c — don’t forget the |.|. x 1 x n+1 + c — if n = −1. x n dx = n+1 1 e ax dx = e ax + c a
2
3
4
()
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Table of integrals
Recognise these from a table of derivatives.
Trigonometry
1
sin(ax) dx =
2
3
4
−1 cos(ax) + c a 1 cos(ax) dx = sin(ax) + c a 1 sec2 (ax) dx = tan(ax) + c a 1 sec(ax) tan(ax) dx = sec(ax) + c a
()
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Table of integrals
Recognise these from a table of derivatives.
Trigonometry
1
sin(ax) dx =
2
3
4
−1 cos(ax) + c a 1 cos(ax) dx = sin(ax) + c a 1 sec2 (ax) dx = tan(ax) + c a 1 sec(ax) tan(ax) dx = sec(ax) + c a
Inverse trig
1
√
2
1 dx = sin−1 (x/a) + c − x2 1 1 dx = tan−1 (x/a) + c. 2 + x2 a a a2
() Guide to Integration Winter 2006/2007 3 / 24
Table of integrals
Recognise these from a table of derivatives.
Trigonometry
1
sin(ax) dx =
2
3
4
−1 cos(ax) + c a 1 cos(ax) dx = sin(ax) + c a 1 sec2 (ax) dx = tan(ax) + c a 1 sec(ax) tan(ax) dx = sec(ax) + c a
Inverse trig
1
√
2
1 dx = sin−1 (x/a) + c — need a > 0. − x2 1 1 dx = tan−1 (x/a) + c. 2 + x2 a a a2
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Substitution rule
From the chain rule we get
f (g (x))g (x) dx = f (u) du u = g (x)
()
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Substitution rule
From the chain rule we get
f (g (x))g (x) dx = f (u) du u = g (x)
= f (u) + c = f (g (x)) + c Look for a function and its derivative in the integrand.
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Substitution example
Example sin(3 log x) dx x
()
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Substitution example
Example sin(3 log x) dx x
1 x
Let u = log x so du =
dx.
We then completely transform all x’s into u’s.
()
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Substitution example
Example sin(3 log x) dx x
1 x
Let u = log x so du =
dx.
We then completely transform all x’s into u’s. sin(3 log x) dx = x = sin 3u du −1 cos(3u) + c 3
()
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Substitution example
Example sin(3 log x) dx x
1 x
Let u = log x so du =
dx.
We then completely transform all x’s into u’s. sin(3 log x) dx = x = We have to turn all the u’s back into x’s sin 3u du −1 cos(3u) + c 3
()
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Substitution example
Example sin(3 log x) dx x
1 x
Let u = log x so du =
dx.
We then completely transform all x’s into u’s. sin(3 log x) dx = x = We have to turn all the u’s back into x’s = −1 cos(3 log x) + c 3 sin 3u du −1 cos(3u) + c 3
()
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Substitution example
Example sin(3 log x) dx x
1 x
Let u = log x so du =
dx.
We then completely transform all x’s into u’s. sin(3 log x) dx = x = We have to turn all the u’s back into x’s = −1 cos(3 log x) + c 3 sin 3u du −1 cos(3u) + c 3
WARNING — you must turn all the x’s into the new variable.
() Guide to Integration Winter 2006/2007 5 / 24
Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Transform terminals
We make u = log x — so change the terminals too.
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Transform terminals
We make u = log x — so change the terminals too.
2 1
sin(3 log x) dx = x
log 2
sin 3u du
log 1
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Transform terminals
We make u = log x — so change the terminals too.
2 1
sin(3 log x) dx = x =
log 2
sin 3u du
log 1
−1 cos(3u) 3
log 2 log 1=0
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Transform terminals
We make u = log x — so change the terminals too.
2 1
sin(3 log x) dx = x =
log 2
sin 3u du
log 1
−1 cos(3u) 3
log 2 log 1=0
−1 = cos(3 log 2) + 3 −1 = cos(3 log 2) + 3
1 cos(0) 3 1 3
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Keep terminals, remember to change everything back to x
2 1
sin(3 log x) dx = x
x=2
sin 3u du
x=1
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Keep terminals, remember to change everything back to x
2 1
sin(3 log x) dx = x =
x=2
sin 3u du
x=1
−1 cos(3u) 3
x=2 x=1
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Keep terminals, remember to change everything back to x
2 1
sin(3 log x) dx = x =
x=2
sin 3u du
x=1
−1 cos(3u) 3
x=2 x=1 2 1
−1 cos(3 log x) = 3
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Keep terminals, remember to change everything back to x
2 1
sin(3 log x) dx = x =
x=2
sin 3u du
x=1
−1 cos(3u) 3
x=2 x=1 2
−1 cos(3 log x) = 3 1 −1 1 = cos(3 log 2) + 3 3
()
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Substitution with definite integrals
You have 2 choices of what to do with the integration terminals.
Keep terminals, remember to change everything back to x
2 1
sin(3 log x) dx = x =
x=2
sin 3u du
x=1
−1 cos(3u) 3
x=2 x=1 2
−1 cos(3 log x) = 3 1 −1 1 = cos(3 log 2) + 3 3 Of course the answers are the same.
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Trigonometric integrals
Trig integrals are really just special cases of substitution.
()
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Trigonometric integrals
Trig integrals are really just special cases of substitution. Usually we need trig identities like
()
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Trigonometric integrals
Trig integrals are really just special cases of substitution. Usually we need trig identities like
Useful trig identities
cos2 x + sin2 x = 1 1 + tan2 x = sec2 x 1 cos2 x = (1 + cos 2x) 2 1 sin2 x = (1 − cos 2x) 2
()
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Trigonometric integrals
Example
sina x cosb x dx
()
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Trigonometric integrals
Example
sina x cosb x dx
If a and b are both even then use cos2 x = 1 (1 + cos 2x) 2 1 sin2 x = (1 − cos 2x) 2
()
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Trigonometric integrals
Example
sina x cosb x dx
If a and b are both even then use cos2 x = 1 (1 + cos 2x) 2 1 sin2 x = (1 − cos 2x) 2
If a or b is odd then use cos2 x = 1 − sin2 x sin2 x = 1 − cos2 x
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Trigonometric integrals
Example sec x dx
()
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Trigonometric integrals
Example sec x dx
Now this is not at all obvious, but you should see it. . .
()
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Trigonometric integrals
Example sec x dx
Now this is not at all obvious, but you should see it. . . sec x dx = sec x sec x + tan x sec x + tan x dx
()
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Trigonometric integrals
Example sec x dx
Now this is not at all obvious, but you should see it. . . sec x dx = Now set u = sec x + tan x: du = sec x tan x + sec2 x = sec x (tan x + sec x) dx sec x sec x + tan x sec x + tan x dx
()
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Trigonometric integrals
Example sec x dx
Now this is not at all obvious, but you should see it. . . sec x dx = Now set u = sec x + tan x: du = sec x tan x + sec2 x = sec x (tan x + sec x) dx Hence we have sec x sec x + tan x sec x + tan x dx = 1 du · dx u dx sec x sec x + tan x sec x + tan x dx
()
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Trigonometric integrals
Example sec x dx
Now this is not at all obvious, but you should see it. . . sec x dx = Now set u = sec x + tan x: du = sec x tan x + sec2 x = sec x (tan x + sec x) dx Hence we have sec x sec x + tan x sec x + tan x dx = = 1 u 1 u · du dx dx sec x sec x + tan x sec x + tan x dx
du = log |u| + c
()
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Trigonometric integrals
Example sec x dx
Now this is not at all obvious, but you should see it. . . sec x dx = Now set u = sec x + tan x: du = sec x tan x + sec2 x = sec x (tan x + sec x) dx Hence we have sec x sec x + tan x sec x + tan x dx = 1 du · dx u dx 1 du = log |u| + c = u = log | sec x + tan x| + c
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sec x
sec x + tan x sec x + tan x
dx
()
Guide to Integration
Integration by parts
From the product rule we get
f (x)g (x) dx = f (x)g (x) − g (x)f (x) dx
Frequently used when you have the product of 2 different types of functions.
()
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Integration by parts
From the product rule we get
f (x)g (x) dx = f (x)g (x) − g (x)f (x) dx
Frequently used when you have the product of 2 different types of functions. You have to choose f (x) and g (x) — there are 2 options. Usually one will work and the other will not.
()
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Integration by parts
Example
xe x dx
()
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Integration by parts
Example
xe x dx
Choose f = x and g = e x
()
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Integration by parts
Example
xe x dx
Choose f = x and g = e x — so f = 1 and g = e x :
()
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Integration by parts
Example
xe x dx
Choose f = x and g = e x — so f = 1 and g = e x : f (x)g (x) dx = f (x)g (x) − g (x)f (x) dx
()
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Integration by parts
Example
xe x dx
Choose f = x and g = e x — so f = 1 and g = e x : f (x)g (x) dx = f (x)g (x) − xe x dx = xe x − g (x)f (x) dx
e x · 1 dx
= xe x − e x + c
()
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Integration by parts
Example
xe x dx
What if we choose f and g the other way around?
()
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Integration by parts
Example
xe x dx
What if we choose f and g the other way around? f = e x and g = x — so f = e x and g = x 2 /2
()
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Integration by parts
Example
xe x dx
What if we choose f and g the other way around? f = e x and g = x — so f = e x and g = x 2 /2 xe x dx = x2 x e − 2 x2 x e dx 2
()
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Integration by parts
Example
xe x dx
What if we choose f and g the other way around? f = e x and g = x — so f = e x and g = x 2 /2 xe x dx = x2 x e − 2 x2 x e dx 2
This is not getting easier, so stop!
()
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Integration by parts
Sometimes one of the parts is “1”.
Example
log x dx
()
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Integration by parts
Sometimes one of the parts is “1”.
Example
log x dx
Choose f = log x and g = 1
()
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Integration by parts
Sometimes one of the parts is “1”.
Example
log x dx
Choose f = log x and g = 1 — so f = 1/x and g = x:
()
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Integration by parts
Sometimes one of the parts is “1”.
Example
log x dx
Choose f = log x and g = 1 — so f = 1/x and g = x: f (x)g (x) dx = f (x)g (x) − g (x)f (x) dx
()
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Integration by parts
Sometimes one of the parts is “1”.
Example
log x dx
Choose f = log x and g = 1 — so f = 1/x and g = x: f (x)g (x) dx = f (x)g (x) − log x dx = x log x − g (x)f (x) dx x/x dx
= x log x − x + c
()
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Trigonometric substitutions
Based on
sin2 θ = 1 − cos2 θ tan2 θ + 1 = sec2 θ
()
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Trigonometric substitutions
Based on
sin2 θ = 1 − cos2 θ tan2 θ + 1 = sec2 θ
Things to associate
If the integrand contains a2 − x 2 −→ a2 + x 2 −→ sin2 θ = 1 − cos2 θ 1 + tan2 θ = sec2 θ
()
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Trigonometric substitutions
Compute
Contains
(5 − x 2 )−3/2 dx
√ a2 − x 2 so put x = √ 5 sin θ.
()
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Trigonometric substitutions
Compute
Contains Hence
dx dθ
(5 − x 2 )−3/2 dx
√ a2 − x 2 so put x = √ = 5 cos θ and √ 5 sin θ. √
(5 − x 2 )−3/2 dx =
5 cos θ dθ 53/2 (1 − sin2 θ)3/2
()
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Trigonometric substitutions
Compute
Contains Hence
dx dθ
(5 − x 2 )−3/2 dx
√ a2 − x 2 so put x = √ = 5 cos θ and √ 5 sin θ. √
(5 − x 2 )−3/2 dx = =
5 cos θ dθ 53/2 (1 − sin2 θ)3/2 cos θ dθ 5(cos2 θ)3/2
()
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Trigonometric substitutions
Compute
Contains Hence
dx dθ
(5 − x 2 )−3/2 dx
√ a2 − x 2 so put x = √ = 5 cos θ and √ 5 sin θ. √
(5 − x 2 )−3/2 dx = = =
5 cos θ dθ 53/2 (1 − sin2 θ)3/2 cos θ dθ 5(cos2 θ)3/2 1 1 dθ = sec2 θ dθ 2 5 cos 5
()
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Trigonometric substitutions
Compute
Contains Hence
dx dθ
(5 − x 2 )−3/2 dx
√ a2 − x 2 so put x = √ = 5 cos θ and √ 5 sin θ. √
(5 − x 2 )−3/2 dx = = = =
5 cos θ dθ 53/2 (1 − sin2 θ)3/2 cos θ dθ 5(cos2 θ)3/2 1 1 dθ = sec2 θ dθ 2 5 cos 5 1 tan θ + c 5
()
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Trigonometric substitutions
Compute
Contains Hence
dx dθ
(5 − x 2 )−3/2 dx
√ a2 − x 2 so put x = √ = 5 cos θ and √ 5 sin θ. √
(5 − x 2 )−3/2 dx = = = =
5 cos θ dθ 53/2 (1 − sin2 θ)3/2 cos θ dθ 5(cos2 θ)3/2 1 1 dθ = sec2 θ dθ 2 5 cos 5 1 tan θ + c 5
We aren’t done yet — we have to change back to the x variable.
() Guide to Integration Winter 2006/2007 14 / 24
Trigonometric substitutions
Compute (5 − x 2 )−3/2 dx
√ 5 sin θ and got
1 5
We substituted x =
tan θ + c
()
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Trigonometric substitutions
Compute (5 − x 2 )−3/2 dx
√ 1 We substituted x = 5 sin θ and got 5 tan θ + c We can express tan θ in terms of sin θ tan θ = = sin θ = cos θ √ x/ 5 1− sin θ 1 − sin2 θ =√ x 5 1− x 2 /5 =√ x 5 − x2
x 2 /5
()
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Trigonometric substitutions
Compute (5 − x 2 )−3/2 dx
√ 1 We substituted x = 5 sin θ and got 5 tan θ + c We can express tan θ in terms of sin θ tan θ = = Hence the integral is x (5 − x 2 )−3/2 dx = √ +c 5 5 − x2 sin θ = cos θ √ x/ 5 1− sin θ 1 − sin2 θ =√ x 5 1− x 2 /5 =√ x 5 − x2
x 2 /5
()
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Trigonometric substitutions
1 dx 4 + x2
Compute
√
()
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Trigonometric substitutions
1 dx 4 + x2
Compute
√
Contains a2 + x 2 , so sub x = 2 tan θ.
()
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Trigonometric substitutions
1 dx 4 + x2
Compute
dx dθ
√
Contains a2 + x 2 , so sub x = 2 tan θ. Hence = 2 sec2 θ and √ 1 dx = 4 + x2 √ 2 sec2 θ 4 + 4 tan2 θ dθ
()
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Trigonometric substitutions
1 dx 4 + x2
Compute
dx dθ
√
Contains a2 + x 2 , so sub x = 2 tan θ. Hence = 2 sec2 θ and √ 1 dx = 4 + x2 = dθ 4 + 4 tan2 θ 2 sec2 θ √ dθ = 2 1 + tan2 θ √ 2 sec2 θ
sec2 θ √ dθ sec2 θ
()
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Trigonometric substitutions
1 dx 4 + x2
Compute
dx dθ
√
Contains a2 + x 2 , so sub x = 2 tan θ. Hence = 2 sec2 θ and √ 1 dx = 4 + x2 = = dθ 4 + 4 tan2 θ 2 sec2 θ √ dθ = 2 1 + tan2 θ sec θ dθ √ 2 sec2 θ
sec2 θ √ dθ sec2 θ
()
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Trigonometric substitutions
1 dx 4 + x2
Compute
dx dθ
√
Contains a2 + x 2 , so sub x = 2 tan θ. Hence = 2 sec2 θ and √ 1 dx = 4 + x2 = = dθ 4 + 4 tan2 θ 2 sec2 θ √ dθ = 2 1 + tan2 θ sec θ dθ √ 2 sec2 θ
sec2 θ √ dθ sec2 θ
We have assumed sec θ > 0. We did similarly in the previous example.
()
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Trigonometric substitutions
Compute √ 1 dx 4 + x2
1 dx = 4 + x2
We substituted x = 2 tan θ and got √ secθ dθ
()
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Trigonometric substitutions
Compute √ 1 dx 4 + x2
1 dx = secθ dθ 4 + x2 = log | sec θ + tan θ| + c
We substituted x = 2 tan θ and got √
previous work
()
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Trigonometric substitutions
Compute √ 1 dx 4 + x2
1 dx = secθ dθ 4 + x2 = log | sec θ + tan θ| + c
We substituted x = 2 tan θ and got √
previous work
So now we need to rewrite in terms of x.
()
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Trigonometric substitutions
Compute √ 1 dx 4 + x2
1 dx = secθ dθ 4 + x2 = log | sec θ + tan θ| + c
We substituted x = 2 tan θ and got √
previous work
So now we need to rewrite in terms of x. The tan θ = x/2 is easy. But sec θ is harder: sec2 θ = 1 + tan2 θ sec θ = 1 + tan2 θ = 1 + x 2 /4
()
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Trigonometric substitutions
Compute √ 1 dx 4 + x2
1 dx = secθ dθ 4 + x2 = log | sec θ + tan θ| + c
We substituted x = 2 tan θ and got √
previous work
So now we need to rewrite in terms of x. The tan θ = x/2 is easy. But sec θ is harder: sec2 θ = 1 + tan2 θ sec θ = Hence: √ 1 + tan2 θ = 1 + x 2 /4
1 dx = log | 1 + x 2 /4 + x/2| + c. 4 + x2
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()
Trigonometric substitutions
Sometimes you need to complete the square in order to get started.
Try
dx 4x 2 + 12x + 13
()
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Partial fractions
Based on partial fraction decomposition of rational functions
There are some very general rules for this technique. It is one of the few very formulaic techniques of integration.
()
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Partial fractions
Based on partial fraction decomposition of rational functions
There are some very general rules for this technique. It is one of the few very formulaic techniques of integration. Any polynomial with real coefficients can be factored into linear and quadratic factors with real coefficients Q(x) = k(x − a1 )m1 (x − a2 )m2 · · · (x − aj )mj × (x 2 + b1 x + c1 )n1 (x 2 + b2 x + c2 )n2 · · · (x 2 + bl x + cl )nl
()
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Partial fractions
Suppose f (x) = P(x)/Q(x) with deg(P) < deg(Q). You might have to do division to arrive at this.
()
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Partial fractions
Suppose f (x) = P(x)/Q(x) with deg(P) < deg(Q). You might have to do division to arrive at this. Factorise Q(x) as on the previous slide.
()
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Partial fractions
Suppose f (x) = P(x)/Q(x) with deg(P) < deg(Q). You might have to do division to arrive at this. Factorise Q(x) as on the previous slide. Rewrite f (x) as f (x) = A12 A1m1 A11 + + ··· + 1 2 (x − a1 ) (x − a1 ) (x − a1 )m1 + similar terms for each linear factor B12 x + C12 B1n x + C1n1 B11 x + C11 + 2 + ··· 2 1 + 2 1 2 (x + b1 x + c1 ) (x + b1 x + c1 ) (x + b1 x + c1 )n1 + similar terms for each quadratic factor
()
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Partial fractions
Suppose f (x) = P(x)/Q(x) with deg(P) < deg(Q). You might have to do division to arrive at this. Factorise Q(x) as on the previous slide. Rewrite f (x) as f (x) = A12 A1m1 A11 + + ··· + 1 2 (x − a1 ) (x − a1 ) (x − a1 )m1 + similar terms for each linear factor B12 x + C12 B1n x + C1n1 B11 x + C11 + 2 + ··· 2 1 + 2 1 2 (x + b1 x + c1 ) (x + b1 x + c1 ) (x + b1 x + c1 )n1 + similar terms for each quadratic factor
Once in this form, we can integrate term-by-term.
()
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Partial fractions
dx x(x − 1)
Write in partial fraction form 1 A B = + x(x − 1) x x −1 A(x − 1) + Bx = x(x − 1) Now find A and B. Compare numerators
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Partial fractions
dx x(x − 1)
Write in partial fraction form 1 A B = + x(x − 1) x x −1 A(x − 1) + Bx = x(x − 1) Now find A and B. Compare numerators
Compare coefficients of x in the numerators to get equations for A and B. x(A + B) + (−A) = 0x + 1
()
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Partial fractions
dx x(x − 1)
Write in partial fraction form 1 A B = + x(x − 1) x x −1 A(x − 1) + Bx = x(x − 1) Now find A and B. Compare numerators
Compare coefficients of x in the numerators to get equations for A and B. x(A + B) + (−A) = 0x + 1 Hence we have 2 equations A+B −A
()
=0 =1
⇒ A = −1, B = 1
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Partial fractions
dx x(x − 1)
Hence in partial fraction form we have 1 −1 1 = + x(x − 1) x x −1 always check this!
()
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Partial fractions
dx x(x − 1)
Hence in partial fraction form we have 1 −1 1 = + x(x − 1) x x −1 Now integrate term-by-term 1 1 1 dx = − dx + dx x(x − 1) x x −1 = − log |x| + log |x − 1| + c = log x −1 +c x always check this!
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Guide to Integration
Winter 2006/2007
22 / 24
Partial fractions
dx x(x − 1)
Hence in partial fraction form we have 1 −1 1 = + x(x − 1) x x −1 Now integrate term-by-term 1 1 1 dx = − dx + dx x(x − 1) x x −1 = − log |x| + log |x − 1| + c = log Try 1 dx. x 2 − a2
() Guide to Integration Winter 2006/2007 22 / 24
always check this!
x −1 +c x
Partial fractions
1 dx x(x − 1)2
Start by writing in partial fraction form:
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Guide to Integration
Winter 2006/2007
23 / 24
Partial fractions
1 dx x(x − 1)2
Start by writing in partial fraction form: A B C 1 = + + 2 x(x − 1) x x − 1 (x − 1)2 A(x − 1)2 + Bx(x − 1) + Cx = x(x − 1)2
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Guide to Integration
Winter 2006/2007
23 / 24
Partial fractions
1 dx x(x − 1)2
Start by writing in partial fraction form: A B C 1 = + + 2 x(x − 1) x x − 1 (x − 1)2 A(x − 1)2 + Bx(x − 1) + Cx = x(x − 1)2 Comparing numerators gives A + B + 0C = 0 −2A − B + C = 0 A + 0B + 0C = 1
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Guide to Integration
Winter 2006/2007
23 / 24
Partial fractions
1 dx x(x − 1)2
Start by writing in partial fraction form: A B C 1 = + + 2 x(x − 1) x x − 1 (x − 1)2 A(x − 1)2 + Bx(x − 1) + Cx = x(x − 1)2 Comparing numerators gives A + B + 0C = 0 −2A − B + C = 0 A + 0B + 0C = 1
Solve these equations to get A = 1, B = −1, C = 1.
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Guide to Integration
Winter 2006/2007
23 / 24
Partial fractions
1 dx x(x − 1)2
Start by writing in partial fraction form: A B C 1 = + + 2 x(x − 1) x x − 1 (x − 1)2 A(x − 1)2 + Bx(x − 1) + Cx = x(x − 1)2 Comparing numerators gives A + B + 0C = 0 −2A − B + C = 0 A + 0B + 0C = 1
Solve these equations to get A = 1, B = −1, C = 1. Integrate term-by-term 1 dx = x(x − 1)2
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1 dx + x
Guide to Integration
−1 dx + x −1
1 dx (x − 1)2
Winter 2006/2007 23 / 24
Partial fractions
1 dx x(x − 1)2
Integrate term-by-term 1 dx = x(x − 1)2 1 dx + x −1 dx + x −1 1 dx (x − 1)2
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Guide to Integration
Winter 2006/2007
24 / 24
Partial fractions
1 dx x(x − 1)2
Integrate term-by-term 1 dx = x(x − 1)2 −1 1 dx dx + x −1 (x − 1)2 1 = log |x| − log |x − 1| − +c x −1 x = log +c x −1 1 dx + x
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Guide to Integration
Winter 2006/2007
24 / 24
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Guide to Integration
Winter 2006/2007
24 / 24