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