Download Strategy for Integration

Standard Table
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Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
Faced with an integral, we must use a problem solving approach to finding the
right method or combination of methods to apply.
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
Faced with an integral, we must use a problem solving approach to finding the
right method or combination of methods to apply.
I
It may be possible to Simplify the integral e.g.
Z
Z
cos x
cot xdx =
dx.
sin x
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
Faced with an integral, we must use a problem solving approach to finding the
right method or combination of methods to apply.
I
It may be possible to Simplify the integral e.g.
Z
Z
cos x
cot xdx =
dx.
sin x
I
It may be possible to simplify or solve the integral with a substitution
e.g.
Z
1
dx
x(ln x)10
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
Faced with an integral, we must use a problem solving approach to finding the
right method or combination of methods to apply.
I
It may be possible to Simplify the integral e.g.
Z
Z
cos x
cot xdx =
dx.
sin x
I
It may be possible to simplify or solve the integral with a substitution
e.g.
Z
1
dx
x(ln x)10
I
if it is of the form
Z
sinn x cosm xdx,
Z
tann x secm xdx
Z
sin(nx) cos(mx)dx
we can deal with it using the standard methods for trigonometric
functions we have studied.
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
Faced with an integral, we must use a problem solving approach to finding the
right method or combination of methods to apply.
I
It may be possible to Simplify the integral e.g.
Z
Z
cos x
cot xdx =
dx.
sin x
I
It may be possible to simplify or solve the integral with a substitution
e.g.
Z
1
dx
x(ln x)10
I
if it is of the form
Z
sinn x cosm xdx,
Z
tann x secm xdx
Z
sin(nx) cos(mx)dx
we can deal with it using the standard methods for trigonometric
functions we have studied.
I
If we are trying to integrate a rational function, we apply the techniques
of the previous section.
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
Faced with an integral, we must use a problem solving approach to finding the
right method or combination of methods to apply.
I
It may be possible to Simplify the integral e.g.
Z
Z
cos x
cot xdx =
dx.
sin x
I
It may be possible to simplify or solve the integral with a substitution
e.g.
Z
1
dx
x(ln x)10
I
if it is of the form
Z
sinn x cosm xdx,
Z
tann x secm xdx
Z
sin(nx) cos(mx)dx
we can deal with it using the standard methods for trigonometric
functions we have studied.
I
If we are trying to integrate a rational function, we apply the techniques
of the previous section.
I
We should check if integration by parts will work.
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
I
√
If the integral contains an expression of the form ±x 2 ± a2 we can use a
trigonometric
substitution. If the integral contains an expression of the
√
form√n ax + b, the function may become a rational function when we use
u = n ax + b, a rationalizing substitution.
This may also
p
p work for
integrals with expressions of the form n g (x) with u = n g (x)
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
I
I
√
If the integral contains an expression of the form ±x 2 ± a2 we can use a
trigonometric
substitution. If the integral contains an expression of the
√
form√n ax + b, the function may become a rational function when we use
u = n ax + b, a rationalizing substitution.
This may also
p
p work for
integrals with expressions of the form n g (x) with u = n g (x)
You may be able to manipulate the integrand to change its form. e.g.
Z
Z
sec x(sec x + tan x)
sec xdx =
sec x + tan x
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
I
√
If the integral contains an expression of the form ±x 2 ± a2 we can use a
trigonometric
substitution. If the integral contains an expression of the
√
form√n ax + b, the function may become a rational function when we use
u = n ax + b, a rationalizing substitution.
This may also
p
p work for
integrals with expressions of the form n g (x) with u = n g (x)
I
You may be able to manipulate the integrand to change its form. e.g.
Z
Z
sec x(sec x + tan x)
sec xdx =
sec x + tan x
I
The integral may resemble something you have already seen and you
may see that a change of format or substitution will convert the integral
to some basic integral that you have already worked through e.g.
Z
sin x cos xe sin x dx
Annette Pilkington
Strategy for Integration
Standard Table
Problem Solving approach
I
√
If the integral contains an expression of the form ±x 2 ± a2 we can use a
trigonometric
substitution. If the integral contains an expression of the
√
form√n ax + b, the function may become a rational function when we use
u = n ax + b, a rationalizing substitution.
This may also
p
p work for
integrals with expressions of the form n g (x) with u = n g (x)
I
You may be able to manipulate the integrand to change its form. e.g.
Z
Z
sec x(sec x + tan x)
sec xdx =
sec x + tan x
I
The integral may resemble something you have already seen and you
may see that a change of format or substitution will convert the integral
to some basic integral that you have already worked through e.g.
Z
sin x cos xe sin x dx
I
Your solution may involve several steps.
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤
Annette Pilkington
π
,
2
R
√
Strategy for Integration
1
25−x 2
dx = sin−1
“ ”
x
5
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
I
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤
R
sec x dx
Annette Pilkington
π
,
2
R
√
Strategy for Integration
1
25−x 2
dx = sin−1
“ ”
x
5
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
I
I
“ ”
R
−1 x
√ 1
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 ,
dx
=
sin
5
25−x 2
R
sec x dx
R
R
x(sec x+tan x)
sec xdx = sec sec
Let u = sec x + tan x.
x+tan x
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
I
I
I
“ ”
R
−1 x
√ 1
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 ,
dx
=
sin
5
25−x 2
R
sec x dx
R
R
x(sec x+tan x)
sec xdx = sec sec
Let u = sec x + tan x.
x+tan x
R √x
e dx
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
I
I
I
I
“ ”
R
−1 x
√ 1
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 ,
dx
=
sin
5
25−x 2
R
sec x dx
R
R
x(sec x+tan x)
sec xdx = sec sec
Let u = sec x + tan x.
x+tan x
R √x
e dx
√
Try the substitution w = x
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
I
I
I
I
I
“ ”
R
−1 x
√ 1
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 ,
dx
=
sin
5
25−x 2
R
sec x dx
R
R
x(sec x+tan x)
sec xdx = sec sec
Let u = sec x + tan x.
x+tan x
R √x
e dx
√
Try the substitution w = x
1
dw = 2√1 x dx = 2w
dx, dx = 2w dw
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
I
I
I
I
I
I
“ ”
R
−1 x
√ 1
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 ,
dx
=
sin
5
25−x 2
R
sec x dx
R
R
x(sec x+tan x)
sec xdx = sec sec
Let u = sec x + tan x.
x+tan x
R √x
e dx
√
Try the substitution w = x
1
dw = 2√1 x dx = 2w
dx, dx = 2w dw
R √x
R w
R
e dx = e 2w dw = 2 we w dw
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
ln x dx
I Integration by parts ; let u = ln x, dv = dx ......
R
I
tan x dx
R
R sin x
I write as
dx, let u = cos x......
tan x dx = cos
x
R
I
sin3 x cos x dx
I Substitution, let u = sin x......
R
I
√ 1
dx
2
25−x
I
I
I
I
I
I
I
I
“ ”
R
−1 x
√ 1
Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 ,
dx
=
sin
5
25−x 2
R
sec x dx
R
R
x(sec x+tan x)
sec xdx = sec sec
Let u = sec x + tan x.
x+tan x
R √x
e dx
√
Try the substitution w = x
1
dw = 2√1 x dx = 2w
dx, dx = 2w dw
R √x
R w
R
e dx = e 2w dw = 2 we w dw
Use integration by parts now with u = w , dv = e w dw .
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
ˆ
˜
I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x
2
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
ˆ
˜
I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x
2
R
I
cos2 x dx
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
ˆ
˜
I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x
2
R
I
cos2 x dx
I
Use the half angle formula: cos2 x = 21 (1 + cos 2x).
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
ˆ
˜
I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x
2
R
I
cos2 x dx
I
I
Use the half angle formula: cos2 x = 21 (1 + cos 2x).
R 1
dx
x 2 −9
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
ˆ
˜
I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x
2
R
I
cos2 x dx
I
I
I
Use the half angle formula: cos2 x = 21 (1 + cos 2x).
R 1
dx
x 2 −9
R
A
B
Partial Fractions x 21−9 dx = x−3
+ x+3
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
ˆ
˜
I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x
2
R
I
cos2 x dx
I
I
I
I
Use the half angle formula: cos2 x = 21 (1 + cos 2x).
R 1
dx
x 2 −9
R
A
B
Partial Fractions x 21−9 dx = x−3
+ x+3
R x
dx
x+3
Annette Pilkington
Strategy for Integration
Standard Table
Review
Outline How you would approach the following integrals:
R
I
sin(7x) cos(4x) dx
ˆ
˜
I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x
2
R
I
cos2 x dx
I
I
I
I
I
Use the half angle formula: cos2 x = 21 (1 + cos 2x).
R 1
dx
x 2 −9
R
A
B
Partial Fractions x 21−9 dx = x−3
+ x+3
R x
dx
x+3
R x
R
dx = u−3
du where u = x + 3.
x+3
u
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
Substitute u = x 3 , du = 3x 2 dx
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
I
Use tan−1 formula or substitute u = tan θ.
I
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
Use tan−1 formula or substitute u = tan θ.
R
1
dx
x 2 +27x+26
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
I
Use tan−1 formula or substitute u = tan θ.
R
1
dx
x 2 +27x+26
I
Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1).
I
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
I
I
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
Use tan−1 formula or substitute u = tan θ.
R
1
dx
x 2 +27x+26
Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1).
R x arctan x
dx
(1+x 2 )2
Annette Pilkington
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
Use tan−1 formula or substitute u = tan θ.
R
1
dx
x 2 +27x+26
I
Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1).
R x arctan x
dx
(1+x 2 )2
I
integration by parts with u = arctanx and dv =
I
Annette Pilkington
x
.
(1+x 2 )2
Strategy for Integration
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
Use tan−1 formula or substitute u = tan θ.
R
1
dx
x 2 +27x+26
I
Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1).
R x arctan x
dx
(1+x 2 )2
I
integration by parts with u = arctanx and dv =
I
1
and
du = 1+x
R 2x
R
v = (1+x 2 )2 dx = (w = 1 + x 2 ) =
I
Annette Pilkington
1
2
·
1
w2
=
x
.
(1+x 2 )2
−1
2w
=
Strategy for Integration
−1
2(1+x 2 )
.
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
Use tan−1 formula or substitute u = tan θ.
R
1
dx
x 2 +27x+26
I
Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1).
R x arctan x
dx
(1+x 2 )2
I
integration by parts with u = arctanx and dv =
I
1
and
du = 1+x
R 2x
R
v = (1+x 2 )2 dx = (w = 1 + x 2 ) = 12 ·
R x arctan x
R
1
dx = −(arctanx)
+ 2(1+x
2 )2 dx
(1+x 2 )2
2(1+x 2 )
I
I
Annette Pilkington
1
w2
=
x
.
(1+x 2 )2
−1
2w
=
Strategy for Integration
−1
2(1+x 2 )
.
Standard Table
More Challenging Integrals
The following integrals may require multiple steps:
R x2
I
dx
9+x 6
I
I
I
I
Substitute u = x 3 , du = 3x 2 dx
R x2
R 1
dx = 31 9+u
2 du
9+x 6
Use tan−1 formula or substitute u = tan θ.
R
1
dx
x 2 +27x+26
I
Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1).
R x arctan x
dx
(1+x 2 )2
I
integration by parts with u = arctanx and dv =
I
1
and
du = 1+x
R 2x
R
v = (1+x 2 )2 dx = (w = 1 + x 2 ) = 12 ·
R x arctan x
R
1
dx = −(arctanx)
+ 2(1+x
2 )2 dx
(1+x 2 )2
2(1+x 2 )
I
I
I
1
w2
=
x
.
(1+x 2 )2
−1
2w
=
−1
2(1+x 2 )
.
latter integral
use trig substitution x = tan θ, we get
RFor the
R
sec −2 x dx = cos2 x dx, we can use the half angle formula.
Annette Pilkington
Strategy for Integration
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
Annette Pilkington
Strategy for Integration
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
I
1+(ln x)
u = ln x followed by w = 1 + u 2
Annette Pilkington
Strategy for Integration
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
I
u = ln x followed by w = 1 + u 2
I
R
1+sin x
dx.
1−sin x
(Requires True Grit
Annette Pilkington
)
Strategy for Integration
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
I
u = ln x followed by w = 1 + u 2
I
R
I
1+sin x
dx.
1−sin x
multiply by
(Requires True Grit
)
1+sin x
.
1+sin x
Annette Pilkington
Strategy for Integration
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
I
u = ln x followed by w = 1 + u 2
I
R
I
I
1+sin x
dx.
1−sin x
(Requires True Grit
)
1+sin x
.
1+sin x
multiply by
R 1+sin x 1+sin x
R
· 1+sin x dx =
1−sin x
(1+sin x)2
1−sin2 x
Annette Pilkington
dx =
R
(1+sin x)2
cos2 x
dx.
Strategy for Integration
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
I
u = ln x followed by w = 1 + u 2
I
R
I
I
I
1+sin x
dx.
1−sin x
(Requires True Grit
)
1+sin x
.
1+sin x
multiply by
R 1+sin x 1+sin x
R
R
x)2
x)2
· 1+sin x dx = (1+sin
dx = (1+sin
dx.
1−sin x
1−sin2 x
cos2 x
R (1+sin x)2
R 1+2 sin x+sin2 x
R 1
R
R
dx =
dx = cos2 x dx + 2cossin2 xx dx +
cos2 x
cos2 x
Annette Pilkington
Strategy for Integration
sin2 x
cos2 x
dx.
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
I
u = ln x followed by w = 1 + u 2
I
R
I
I
I
I
1+sin x
dx.
1−sin x
(Requires True Grit
)
1+sin x
.
1+sin x
multiply by
R 1+sin x 1+sin x
R
R
x)2
x)2
· 1+sin x dx = (1+sin
dx = (1+sin
dx.
1−sin x
1−sin2 x
cos2 x
R (1+sin x)2
R 1+2 sin x+sin2 x
R 1
R
R
dx =
dx = cos2 x dx + 2cossin2 xx dx +
2
2
R cos x2
R 2 sin xcos x R
= sec x dx + cos2 x dx + tan2 x dx =
R
R
tan x − u22 du + sec2 x − 1 dx, where u = cos x.
Annette Pilkington
Strategy for Integration
sin2 x
cos2 x
dx.
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
I
u = ln x followed by w = 1 + u 2
I
R
I
I
I
I
I
1+sin x
dx.
1−sin x
(Requires True Grit
)
1+sin x
.
1+sin x
multiply by
R 1+sin x 1+sin x
R
R
x)2
x)2
· 1+sin x dx = (1+sin
dx = (1+sin
dx.
1−sin x
1−sin2 x
cos2 x
R (1+sin x)2
R 1+2 sin x+sin2 x
R 1
R
R
dx =
dx = cos2 x dx + 2cossin2 xx dx +
2
2
R cos x2
R 2 sin xcos x R
= sec x dx + cos2 x dx + tan2 x dx =
R
R
tan x − u22 du + sec2 x − 1 dx, where u = cos x.
= 2 tan x + 2 sec x − x + C
Annette Pilkington
Strategy for Integration
sin2 x
cos2 x
dx.
Standard Table
More More Challenging Integrals
The following integrals may require multiple steps:
R
I
√ ln x
dx
2
x
1+(ln x)
I
u = ln x followed by w = 1 + u 2
I
R
I
I
I
I
I
I
1+sin x
dx.
1−sin x
(Requires True Grit
)
1+sin x
.
1+sin x
multiply by
R 1+sin x 1+sin x
R
R
x)2
x)2
· 1+sin x dx = (1+sin
dx = (1+sin
dx.
1−sin x
1−sin2 x
cos2 x
R (1+sin x)2
R 1+2 sin x+sin2 x
R 1
R
R sin2 x
dx =
dx = cos2 x dx + 2cossin2 xx dx + cos
2
2
2 x dx.
R cos x2
R 2 sin xcos x R
2
= sec x dx + cos2 x dx + tan x dx =
R
R
tan x − u22 du + sec2 x − 1 dx, where u = cos x.
= 2 tan x + 2 sec x − x + C
Note if you integrate this in Mathematica you get a different looking
answer, but both differ by a constant
In[10]:=
Out[10]=
x
x
SimplifyB - x CosB F + H4 + xL SinB F ì
2
2
x
x
CosB F - SinB F
2
2
- H2 Tan@xD + 2 Sec@xD - xLF
-2
Annette Pilkington
Strategy for Integration
Standard Table
Even More More Challenging Integrals
The following integrals may require multiple steps:
R ln x
I
√ dx
x
Annette Pilkington
Strategy for Integration
Standard Table
Even More More Challenging Integrals
The following integrals may require multiple steps:
R ln x
I
√ dx
x
√
1
1
I Let u = x, du = √
dx = 2u
dx
2 x
Annette Pilkington
Strategy for Integration
Standard Table
Even More More Challenging Integrals
The following integrals may require multiple steps:
R ln x
I
√ dx
x
√
1
1
I Let u = x, du = √
dx = 2u
dx
2 x
R ln x
R
2
I Let
√ dx = 2
ln u du
x
Annette Pilkington
Strategy for Integration
Standard Table
Even More More Challenging Integrals
The following integrals may require multiple steps:
R ln x
I
√ dx
x
√
1
1
I Let u = x, du = √
dx = 2u
dx
2 x
R ln x
R
2
I Let
√ dx = 2
ln u du
x
R
R
I = 4 ln u du, we use integration by parts on
ln u du.
Annette Pilkington
Strategy for Integration