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Trigonometric Substitutions
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2.
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
For a2 – x2 , if x = a*sin( )
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
For a2 – x2 , if x = a*sin( ) a2 – x2 = a*cos( )
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
For a2 – x2 , if x = a*sin( ) a2 – x2 = a*cos( )-π/2 π/2 (If < < )
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
For a2 – x2 , if x = a*sin( ) a2 – x2 = a*cos( )-π/2 π/2 (If < < )
For a2 + x2 , if x = a*tan( )
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
For a2 – x2 , if x = a*sin( ) a2 – x2 = a*cos( )-π/2 π/2 (If < < )
For a2 + x2 , if x = a*tan( ) a2 + x2 = a*sec( ) -π/2 π/2 (If < < )
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
For a2 – x2 , if x = a*sin( ) a2 – x2 = a*cos( )-π/2 π/2 (If < < )
For a2 + x2 , if x = a*tan( ) a2 + x2 = a*sec( ) -π/2 π/2 (If < < )
For x2 – a2 , if x = a*sec( )
In this section we evaluate integrals with integrands containing expressions of the form
Trigonometric Substitutions
a2 – x2 , a2 + x2 , and x2 – a2. If we set x to be a well chosen trig-function, the expressions may be rooted and simplified.
For a2 – x2 , if x = a*sin( ) a2 – x2 = a*cos( )-π/2 π/2 (If < < )
For a2 + x2 , if x = a*tan( ) a2 + x2 = a*sec( ) -π/2 π/2 (If < < )
For x2 – a2 , if x = a*sec( ) x2 – a2 = a*tan( )-π/2 π/2 (If < < )
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = xa
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = x
a xa
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = x
a xa
a2 – x2
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = x
a
or = sin-1( )xa
xa
a2 – x2
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = x
a
or = sin-1( )xa
xa
a2 – x2
For x = a*tan( )
we've tan( ) = xa
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = x
a
or = sin-1( )xa
xa
a2 – x2
For x = a*tan( )
we've tan( ) = xa x
a
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = x
a
or = sin-1( )xa
xa
a2 – x2
For x = a*tan( )
we've tan( ) = xa x
a
a2 + x2
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sin( )
we've sin( ) = x
a
or = sin-1( )xa
xa
a2 – x2
For x = a*tan( )
we've tan( ) = xa
or = tan-1( )xa
x
a
a2 + x2
The following are the algebraic relations between a, x and for each substitutions.
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa x
a
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa x
a
x2 – a2
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( )
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( ) = 2c( )
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( ) = 2c( )
= 2*c( )dxd
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( ) = 2c( )
= 2*c( )dxd
dx = 2c( ) d
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( ) = 2c( )
= 2*c( )dxd
dx = 2c( ) d
∫ dx = x2
4 – x2 So,
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( ) = 2c( )
= 2*c( )dxd
dx = 2c( ) d
∫ dx = x2
4 – x2 ∫ 2c( )So,
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( ) = 2c( )
= 2*c( )dxd
dx = 2c( ) d
∫ dx = x2
4 – x2 ∫ 4s2( )2c( )So,
Trigonometric Substitutions
For x = a*sec( )
we've sec( ) = xa
or = sec-1( )xa
x
a
x2 – a2
Example: Find ∫ dx. x2
4 – x2
set x = 2*s( )a = 2,
4 – x2 = So, 4 – 4s2( ) = 2c( )
= 2*c( )dxd
dx = 2c( ) d
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )dSo,
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
We need the diagram to put the answer in x:
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
x2
We need the diagram to put the answer in x:
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
x2
4 – x2 We need the diagram to put the answer in x:
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
x2
4 – x2 We need the diagram to put the answer in x:
= sin-1( ) 2x
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
x2
4 – x2 We need the diagram to put the answer in x:
= 2
= sin-1( ) 2x
sin-1( ) 2x
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
x2
4 – x2 We need the diagram to put the answer in x:
= 2
= sin-1( ) 2x
sin-1( ) 2x
– 2* 2x
24 – x2
+ k
Trigonometric Substitutions
∫ dx = x2
4 – x2 ∫ 4s2( )2c( ) 2c( )d
= ∫ 4s2( )d
= 4( 2
2s() c() ) + k–
x2
4 – x2 We need the diagram to put the answer in x:
= 2
= sin-1( ) 2x
sin-1( ) 2x
– 2* 2x
24 – x2
+ k
= 2 sin-1( ) 2x – x4 – x2
+ k 2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9 x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3, x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9
x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = So,
x2
dxx2 – 9 x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2So,
x2
dxx2 – 9 x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3tSo,
x2
dxx2 – 9 x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3t3et d So,
x2
dxx2 – 9 x2
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3t3et d So,
x2
dxx2 – 9 x2 = ∫
9ed
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3t3et d So,
x2
dxx2 – 9 x2 = ∫
9ed
91
= ∫c
d
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3t3et d So,
x2
dxx2 – 9 x2 = ∫
9ed
91
= ∫c
d 91= s( ) + k
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3t3et d So,
x2
dxx2 – 9 x2 = ∫
9ed
91
= ∫c
d 91= s( ) + k
x
3
We've the diagram:
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3t3et d So,
x2
dxx2 – 9 x2 = ∫
9ed
91
= ∫c
d 91= s( ) + k
x
3
x2 – 9
We've the diagram:
Trigonometric Substitutions
Example: Find ∫
dxx2 – 9
set x = 3*e( )a = 3,
x2 – 9 = So, 9e2( ) – 9 = 3t( )
= 3*etdxd
dx = 3et d
∫ = ∫ 9e2*3t3et d So,
x2
dxx2 – 9 x2 = ∫
9ed
91
= ∫c
d 91= s( ) + k
x
3
x2 – 9
We've the diagram:
=
x2 – 9 9x + k
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( )
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
= 16e4
2e2
∫
d
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
= 16e4
2e2
∫
d
= ∫
d 8e2
1
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
= 16e4
2e2
∫
d
= ∫
d 8e2
1= ∫
d 8 1
c2
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
= 16e4
2e2
∫
d
= ∫
d 8e2
1= ∫
d 8 1
c2
= [ ] + k
8 1
2 + c( )s( )
2
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
= 16e4
2e2
∫
d
= ∫
d
x
2
x2 + 4
We've the diagram:
8e2
1= ∫
d 8 1
c2
= [ ] + k
8 1
2 + c( )s( )
2
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
= 16e4
2e2
∫
d
= ∫
d
x
2
x2 + 4
We've the diagram:
8e2
1= ∫
d 8 1
c2
= [ ] + k
8 1
2 + c( )s( )
2
= 16 1 [ sin-1( ) x
x2 + 4
Trigonometric Substitutions
x4 + 8x2 + 16 Example: Find
dx∫
x4 + 8x2 + 16 dx
∫
= (x2 + 4)2 dx
∫
set x = 2t( ) so x2 + 4 = 4t2( ) + 4 = 4e2( )
dx d = 2e2( ) or dx = 2e2( )d
= 16e4
2e2
∫
d
= ∫
d
x
2
x2 + 4
We've the diagram:
8e2
1= ∫
d 8 1
c2
= [ ] + k
8 1
2 + c( )s( )
2
= 16 1 [ sin-1( ) + x
x2 + 4 2x
x2 + 4 + k ]
Trigonometric Substitutions
One type of integrals that we will encounter later are the integrals of the form
cx + dax2 + bx + c
where the denominator is irreducible.
Trigonometric Substitutions
One type of integrals that we will encounter later are the integrals of the form
xx2 + 6x + 10
Example: Find ∫
dx
cx + dax2 + bx + c
where the denominator is irreducible.
Trigonometric Substitutions
One type of integrals that we will encounter later are the integrals of the form
xx2 + 6x + 10
Example: Find ∫
dx
cx + dax2 + bx + c
where the denominator is irreducible.
x2 + 6x + 10. Complete the square on
Trigonometric Substitutions
One type of integrals that we will encounter later are the integrals of the form
xx2 + 6x + 10
Example: Find ∫
dx
cx + dax2 + bx + c
where the denominator is irreducible.
x2 + 6x + 10. Complete the square on
x2 + 6x + 10 = (x2 + 6x ) + 10
Trigonometric Substitutions
One type of integrals that we will encounter later are the integrals of the form
xx2 + 6x + 10
Example: Find ∫
dx
cx + dax2 + bx + c
where the denominator is irreducible.
x2 + 6x + 10. Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
Trigonometric Substitutions
One type of integrals that we will encounter later are the integrals of the form
xx2 + 6x + 10
Example: Find ∫
dx
cx + dax2 + bx + c
where the denominator is irreducible.
x2 + 6x + 10. Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dx
= ∫
x(x + 3)2 + 1
dx
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3
= ∫
x(x + 3)2 + 1
dx
substitution
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
= ∫
u u2 + 1 du – 3 ∫
1 u2 + 1 du
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
= ∫
u u2 + 1 du – 3 ∫
1 u2 + 1 du
Set w = u2 + 1
substitution
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
= ∫
u u2 + 1 du – 3 ∫
1 u2 + 1 du
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
= ∫
u u2 + 1 du – 3 ∫
1 u2 + 1 du
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
= ∫
u u2 + 1 du – 3 ∫
1 u2 + 1 du
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
= ∫
u u2 + 1 du – 3 ∫
1 u2 + 1 du
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
Trigonometric Substitutionsx
x2 + 6x + 10
Hence ∫
dxSet u = x + 3 x = u – 3 dx = du
= ∫
x(x + 3)2 + 1
dx
substitution
= ∫
u – 3 u2 + 1
du
= ∫
u u2 + 1 du – 3 ∫
1 u2 + 1 du
Set w = u2 + 1
= 2u
substitution
dwdudu =
dw2u
= ∫
u w
– 3 ∫
1 u2 + 1 du
dw2u
= ½ ∫
1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
