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The Method of Trigonometric Substitution
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Main Idea
The method helps dealing with integrals, where the integrand contains
one of the following expressions: (where a and b are constants)
bax
bxa
bxa
2
2
2
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A simpler forms of the former expressions are the following ones:
1
1
1
2
2
2
x
x
x
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To get rid of the root, we substitute sinθ, tanθ or secθ respectively
The radicalSubstitutionThe radical becomes
dx becomes
x = sinθ cosθcosθ dθ
x = tanθsecθsec2θ dθ
x = secθtanθsecθ tanθ dθ
21 x
21 x
12 x
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To apply that to the general cases, we transfer the radical to a form similar to the respective simple form
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2
222
)(1 xa
xba
ab
2
222
)(1 xa
xba
ab
1)( 2
222
xa
axb
ab
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The First Case
We let:sinθ = (b/a)x→ x = asinθ/b→ dx = (a/b) cosθ dθ
The radical becomes a cosθ
2
222
)(1 xa
xba
ab
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The Second Case
We let:
tanθ = (b/a)x→ x = (a /b) tanθ→
dx = (a/b) sec2θ dθ
The radical becomes a sec θ
2
222
)(1 xa
xba
ab
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The Third Case
We let:
secθ = (b/a) x
→ x = (a /b) secθ→
dx = (a /b) secθ tanθ dθ
The radical becomes a tanθ
1)( 2
222
xa
axb
ab
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Examples
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Example 1
The First case
222 xba
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dI
So
x
ddxx
x
Let
xx
dxxI
coscos4
cos4916
cossin
sin
)1(16916
916
34
2
34
34
43
21692
2
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c
c
d
d
d
2sin
)2sin(
)2cos1(
cos
34
38
21
38
38
22cos1
316
2316
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We must write the answer in terms of x
We have: sinθ = 3x/4Hence
θ = arcsin(3x/4)cosθ = √[1 - (3x/4)2 ] = (1/4) √[16 - 9x2] And so:sin2θ = 2 sin θ cosθ= 2 (3x/4) . (1/4) . √[16 - 9x2]
= (3/8)x . √[16 - 9x2]
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A faster way to find cosθ is by using the triangle method, starting from the fact
that sin θ = 3x / 4
x3 4
2916 x
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cxx
cxx
cxx
cI
x
x
x
221
43
38
234
932
43
38
243
43
38
43
38
34
38
916arcsin
)(1arcsin
])(1[arcsin
)cossin2
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Example 2
The second case
222 xba
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dxx
dxx
xI
x 23
])(1[
.)925(
253
3
1251
32
3
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3
2353
27125
332
235
35
53
22592
sec125
sectan
sec125)925(
sectan
tan:
)1(25925
dI
x
ddxx
xLet
xx
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Let’s simplify the intrgrand (The expression inside the integral sign)
32815
cossin
815
3
2353
27125
sincos
]cos[
sec125
sectan
3
3
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c
c
dd
d
d
dI
]cos[sec
]cos[
]sin)sin(cos[
sin)1(cos
sin)cos1(cos
sincos
815
)1(cos
815
2815
2815
22815
32815
1
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We must write the answer in terms of x
We have: tanθ = 3x/5Hence,
secθ = √[ 1 + (3x/5)2 ] = √[ 1 + (9x2/25)]
= 5 √[25 + 9x2]
& cosθ = 1/ 5 √[25 + 9x2]
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We could have also found secθ and cosθ using the triangle method,
starting from ths fact that tan θ = 3x / 5
x3
5
2925 x
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2
2
2
2
259
259
815
815
925
1
81
25925
81
1
]925
5
5
925[
81
5
]1
11[
]cos[sec
2
2
xx
cx
x
c
ccI
x
x
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If we wish, we can simplify the answer
cx
x
cx
x
cIx
x
281252
811
2
251
815
259
259
815
925
1925
]925
15925[
]1
11[
2
2
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Example 3
The Third case
222 axb
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23
]1)[(
)254(
252125
1
32
x
dx
x
dxI
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325
332
25
25
52
22542
tan125
tansec
tan125)254(
tansecsec
sec:
)1(25254
dI
x
ddxx
xLet
xx
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c
c
d
d
d
dI
sec
cossin
tan
sectan125
tansec
501
)1(sin
501
2501
sin
coscos1
501
2501
325
1
2
2
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We must write the answer in terms of x
We have: secθ = 2x/5
Hence,
cscθ = 1 / √[1 –(5/2x)2]
= 2x / √[4x2 –25]2
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Here is the triangle method to find cscθ, starting from the fact that
sec θ = 2x / 5
x2
5
254 2 x
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cx
x
cx
x
cI
Thus
254
254
2
sec
,
2251
2501
501
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Integrals Involving an Expression of the Form √(ax2+bx+c)
We simply complete the square and rewrite the expression in one of the previous three forms
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Example 1
X2 + 10x + 16
= (x + 5)2 – 25 + 16
= (x + 5)2 - 9
= 9 [ ( ( x + 5) / 3 )2 - 1 ]
Let (x + 5 ) / 3 = secθ→ x = 3 secθ – 5→ dx = 3 secθ
tanθ dθ
32 )1610( xx
dx
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Example 2
11 - x2 - 10x
= 11 - [ x2 + 10x ]
= 11 - [ ( x + 5 )2 – 25 ]
= 11 - ( x + 5 )2 + 25
= 36 - ( x + 5 )2
= 36 [ 1 - ( (x+5) / 6)2 ]
Let ( x + 5 ) / 6 = sinθ→ x = 6sinθ – 5→ dx = 6 cosθ dθ
32 )1011( xx
dx
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Example 3
50x - 25x2 - 16= - 25 (x2 - 2x ) - 16= - 25 [ (x - 1)2 - 1] - 16= 9 – 25 ( x – 1 )2
= 9[ 1 - ( 5(x -1) / 3)2 ]Let
(5/3) (x-1) = sinθ→ x = (3 / 5) sinθ
+ 1→ dx = (3 / 5) cosθ
32 )162550(
)15(
xx
dxx