The document discusses the substitution method of integration. It explains that while the derivative of an elementary function is another elementary function, the antiderivative may not be. There are two main integration methods: substitution and integration by parts. Substitution reverses the chain rule by letting u be a function of x with derivative u', then substituting u for x and replacing dx with du/u' in the integral.

1. The Substitution Method

2. The Substitution Method
The derivative of an elementary function is another
elementary function, but the antiderivative of an
elementary function might not be elementary.

3. The Substitution Method
The derivative of an elementary function is another
elementary function, but the antiderivative of an
elementary function might not be elementary.
For example,
∫ sin ( ) x 1
dx is not an elementary function.

4. The Substitution Method
The derivative of an elementary function is another
elementary function, but the antiderivative of an
elementary function might not be elementary.
For example,
∫ sin ( ) x 1
dx is not an elementary function.
That is to say, there is no formula that we may build,
using the algebraic functions, the trig–functions, or
the exponential and log–functions
.

5. The Substitution Method
The derivative of an elementary function is another
elementary function, but the antiderivative of an
elementary function might not be elementary.
For example,
∫ sin ( ) x 1
dx is not an elementary function.
That is to say, there is no formula that we may build,
using the algebraic functions, the trig–functions, or
the exponential and log–functions in finitely many
steps

6. The Substitution Method
The derivative of an elementary function is another
elementary function, but the antiderivative of an
elementary function might not be elementary.
For example,
∫ sin ( ) x 1
dx is not an elementary function.
That is to say, there is no formula that we may build,
using the algebraic functions, the trig–functions, or
the exponential and log–functions in finitely many
steps, whose derivative is sin(1/x).

7. The Substitution Method
The derivative of an elementary function is another
elementary function, but the antiderivative of an
elementary function might not be elementary.
For example,
∫ sin ( ) x 1
dx is not an elementary function.
That is to say, there is no formula that we may build,
using the algebraic functions, the trig–functions, or
the exponential and log–functions in finitely many
steps, whose derivative is sin(1/x). Because there is
no Product Rule, no Quotient Rule, and no Chain
Rule for integrations, it means that the algebra of
calculating antiderivatives is a lot more complex than
derivatives.

8. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.

9. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule

10. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.

11. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.
The Substitution Method–Chain Rule Reversed

12. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.
The Substitution Method–Chain Rule Reversed
Let f = f(u) be a function of u, and u = u(x) be a
function x.

13. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.
The Substitution Method–Chain Rule Reversed
Let f = f(u) be a function of u, and u = u(x) be a
function x. By the Chain Rule,
the derivative of f with respect to x is
(f ○ u)'(x) = [f(u(x))]'

14. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.
The Substitution Method–Chain Rule Reversed
Let f = f(u) be a function of u, and u = u(x) be a
function x. By the Chain Rule,
the derivative of f with respect to x is
(f ○ u)'(x) = [f(u(x))]' = f'(u(x))u'(x),

15. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.
The Substitution Method–Chain Rule Reversed
Let f = f(u) be a function of u, and u = u(x) be a
function x. By the Chain Rule,
the derivative of f with respect to x is
(f ○ u)'(x) = [f(u(x))]' = f'(u(x))u'(x), or that
∫ f'(u(x))u'(x) dx f(u(x)) + k

16. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.
The Substitution Method–Chain Rule Reversed
Let f = f(u) be a function of u, and u = u(x) be a
function x. By the Chain Rule,
the derivative of f with respect to x is
(f ○ u)'(x) = [f(u(x))]' = f'(u(x))u'(x), or that
∫ f'(u(x))u'(x) dx
f(u(x)) + k
This integration rule is more useful when stated with
the f’s as specific basic functions.

17. The Substitution Method
There are two main methods of integration,
the Substitution Method and Integration by Parts.
The Substitution Method reverses the Chain Rule
and Integration by Parts unwinds the Product Rule.
The Substitution Method–Chain Rule Reversed
Let f = f(u) be a function of u, and u = u(x) be a
function x. By the Chain Rule,
the derivative of f with respect to x is
(f ○ u)'(x) = [f(u(x))]' = f'(u(x))u'(x), or that
∫ f'(u(x))u'(x) dx
f(u(x)) + k
This integration rule is more useful when stated with
the f’s as specific basic functions. We list the integral
formulas from last section in their substitution–form .

18. Antiderivatives
In the following formulas, u = u(x) is a function of x
and u' = du/dx is the derivative with respect to x.

19. Antiderivatives
In the following formulas, u = u(x) is a function of x
and u' = du/dx is the derivative with respect to x.
∫ uP+1
The Power Functions up u'dx =
(P ≠ –1)
P + 1
+ k

20. In the following formulas, u = u(x) is a function of x
and u' = du/dx is the derivative with respect to x.
up u'dx =
sin(u) u'dx
Antiderivatives
∫
∫ uP+1
P + 1
(P ≠ –1)
∫ cos(u) u'dx= sin(u) + k
The Power Functions
The Trig–Functions
= –cos(u) + k
+ k

21. Antiderivatives
In the following formulas, u = u(x) is a function of x
and u' = du/dx is the derivative with respect to x.
up u'dx =
sin(u) u'dx
sec(u)tan(u) u'dx
∫
∫ uP+1
P + 1
(P ≠ –1)
∫ cos(u) u'dx= sin(u) + k
∫sec2(u) u'dx = tan(u) + k ∫ csc2(u) u'dx
= –cot(u) + k
∫
= sec(u) + k
∫csc(u)cot(u) u'dx
= –csc(u) + k
The Power Functions
The Trig–Functions
= –cos(u) + k
+ k

22. In the following formulas, u = u(x) is a function of x
and u' = du/dx is the derivative with respect to x.
up u'dx =
sin(u) u'dx
= –cos(u) + k
sec(u)tan(u) u'dx
eu u'dx = eu + k
Antiderivatives
∫
∫
∫ uP+1
P + 1
(P ≠ –1)
∫ cos(u) u'dx= sin(u) + k
∫sec2(u) u'dx = tan(u) + k ∫ csc2(u) u'dx
= –cot(u) + k
∫
= sec(u) + k
∫csc(u)cot(u) u'dx
= –csc(u) + k
The Power Functions
The Trig–Functions
The Log and Exponential Functions
+ k

23. Antiderivatives
In the following formulas, u = u(x) is a function of x
and u' = du/dx is the derivative with respect to x.
up u'dx =
sin(u) u'dx
sec(u)tan(u) u'dx
eu u'dx = eu + k
csc2(u) u'dx
u'
u dx = ln(u) + k
∫
∫
∫
∫ uP+1
P + 1
(P ≠ –1)
∫ cos(u) u'dx= sin(u) + k
∫sec2(u) u'dx = tan(u) + k ∫ = –cot(u) + k
∫
= sec(u) + k
∫csc(u)cot(u) u'dx
= –csc(u) + k
The Power Functions
The Trig–Functions
The Log and Exponential Functions
or
∫ u–1u'dx = ln(u) + k
= –cos(u) + k
+ k

24. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand.

25. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.

26. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx

27. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.

28. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.

29. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x)

30. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x) 2. Set du/dx= u' so
dx = du/ u'.

31. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.

32. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.

33. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
∫esin(x)cos(x) dx
Sub. with the u
Replace the dx

34. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
∫esin(x)cos(x) dx
= ∫ eu cos(x) du
Sub. with the u
Replace the dx
cos(x)

35. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
∫esin(x)cos(x) dx
= ∫ eu cos(x) du
Sub. with the u
Replace the dx
cos(x)

36. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
∫esin(x)cos(x) dx
= ∫ eu cos(x) du
cos(x)
= ∫ eu du
Sub. with the u
Replace the dx

37. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
∫esin(x)cos(x) dx
= ∫ eu cos(x) du
cos(x)
= ∫ eu du = eu+ k
Sub. with the u
Replace the dx

38. The Substitution Method
Note the presence of a “u(x)” and its derivative u' as a
factor in each integrand. Below is one way for tracking
the integrals using “substitution” after the respective
u(x) and u' are identified in the integrand.
Example A. Find the following antiderivative.
a. ∫esin(x)cos(x) dx
1. Identify u(x) whose
derivative u' is a factor of
Let u(x) = sin(x) where the integrand.
u' = cos(x), 2. Set du/dx= u' so
so dx = du/cos(x). dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
∫esin(x)cos(x) dx
= ∫ eu cos(x) du
Sub. with the u
Replace the dx
cos(x)
= ∫ eu du = eu+ k = esin(x) + k 4. Write the answer in x.

39. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.

40. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.
1
= ∫ x cos(ln(x)) dx

41. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
1
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x,

42. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
1
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x

43. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x
so xdu = dx
1

44. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x
so xdu = dx
1
substitute with the u,
replace the dx,

45. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x
so xdu = dx
1
substitute with the u,
replace the dx,
∫ 1
cos(ln(x)) dx
x

46. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x
so xdu = dx
1
substitute with the u,
replace the dx,
∫ 1
cos(ln(x)) dx
x
=
1
∫ cos(u) x du
x

47. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x
so xdu = dx
1
substitute with the u,
replace the dx,
∫ 1
cos(ln(x)) dx
x
=
1
∫ cos(u) x du
x
= ∫ cos(u) du

48. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x
so xdu = dx
1
substitute with the u,
replace the dx,
∫ 1
cos(ln(x)) dx
x
=
1
∫ cos(u) x du
x
= ∫ cos(u) du = sin(u) + k

49. The Substitution Method
b. ∫ cos(xln(x))dx
1. Identify u(x) whose
derivative u' is a factor of
the integrand.
2. Set du/dx= u' so
dx = du/ u'.
3. Substitute the u into the
integrand, and replace
dx by du/u' then integrate
the new integral in u.
4. Write the answer in x.
= ∫ x cos(ln(x)) dx
Let u(x) = ln(x) where
u' = 1/x, or
du
dx = 1x
so xdu = dx
1
substitute with the u,
replace the dx,
∫ 1
cos(ln(x)) dx
x
=
1
∫ cos(u) x du
x
= ∫ cos(u) du = sin(u) + k = sin(ln(x)) + k

50. The Substitution Method
By the Chain Rule
e3x + 2]' e3x + 2(1
[= 3), so [ (e3x + 2)]' = e3x + 2
3

51. The Substitution Method
By the Chain Rule
[e3x + 2]' = e3x + 2(3), so [ (e3x + 2)]' = e3x + 2
Hence ∫ e3x + 2 dx =
1
e3x + 2
3
+ k
3

52. The Substitution Method
By the Chain Rule
1
[e3x + 2]' = e3x + 2(3), so [ (e3x + 2)]' = e3x + 2
Hence ∫ e3x + 2 dx =
3
e3x + 2
3
+ k
[sin(3x + 2)]' = cos(3x + 2)(3),

53. The Substitution Method
By the Chain Rule
1
[e3x + 2]' = e3x + 2(3), so [ (e3x + 2)]' = e3x + 2
Hence ∫ e3x + 2 dx =
3
e3x + 2
3
+ k
[sin(3x + 2)]' = cos(3x + 2)(3),
1
so [ sin(3x + 2)]' = cos(3x + 2)
3

54. The Substitution Method
By the Chain Rule
1
[e3x + 2]' = e3x + 2(3), so [ (e3x + 2)]' = e3x + 2
Hence ∫ e3x + 2 dx =
3
e3x + 2
3
+ k
[sin(3x + 2)]' = cos(3x + 2)(3),
1
so [ sin(3x + 2)]' = cos(3x + 2)
3
Hence ∫ cos(3x + 2) dx =
sin(3x + 2)
3
+ k

55. The Substitution Method
By the Chain Rule
1
[e3x + 2]' = e3x + 2(3), so [ (e3x + 2)]' = e3x + 2
Hence ∫ e3x + 2 dx =
3
e3x + 2
3
+ k
[sin(3x + 2)]' = cos(3x + 2)(3),
1
so [ sin(3x + 2)]' = cos(3x + 2)
3
Hence ∫ cos(3x + 2) dx =
sin(3x + 2)
3
+ k
In fact if F'(x) = f(x), then F'(ax + b) = a*f(ax + b) where
a and b are constants.

56. The Substitution Method
By the Chain Rule
1
[e3x + 2]' = e3x + 2(3), so [ (e3x + 2)]' = e3x + 2
Hence ∫ e3x + 2 dx =
3
e3x + 2
3
+ k
[sin(3x + 2)]' = cos(3x + 2)(3),
1
so [ sin(3x + 2)]' = cos(3x + 2)
3
Hence ∫ cos(3x + 2) dx =
sin(3x + 2)
3
+ k
In fact if F'(x) = f(x), then F'(ax + b) = a*f(ax + b) where
a and b are constants.
(Linear Substitution Integrations) Let F'(x) = f(x),
then 1
∫f(u(x)) dx = a F(u(x)) + k where u(x) = ax + b.

57. The Substitution Method
Therefore
∫ e – ½ x + 5 dx =
∫ csc( √ 2 x + 2) cot( x + 2) dx
π
√2
π
=
1
∫ dx =
i.
ii.
iii.
√–5x + 3)

58. The Substitution Method
Therefore
∫ e – ½ x + 5 dx = –2e – ½ x + 5
∫ csc( √ 2 x + 2) cot( x + 2) dx
π
√2
π
+ k
1
∫ dx =
i.
ii.
iii.
√–5x + 3)

59. The Substitution Method
Therefore
∫ e – ½ x + 5 dx = –2e – ½ x + 5
+ k
∫ csc( √ 2 x + 2) cot( x + 2) dx
π
√2
π
= –csc( x + 2)* √2
π
π √2
+ k
1
∫ dx =
i.
ii.
iii.
√–5x + 3)

60. The Substitution Method
Therefore
∫ e – ½ x + 5 dx = –2e – ½ x + 5
+ k
∫ csc( √ 2 x + 2) cot( x + 2) dx
π
√2
π
= –csc( x + 2)* √2
π
π √2
+ k
1
∫ dx =
i.
ii.
iii. ∫(–5x + 3) –½ dx
√–5x + 3)

61. The Substitution Method
Therefore
∫ e – ½ x + 5 dx = –2e – ½ x + 5
+ k
∫ csc( √ 2 x + 2) cot( x + 2) dx
π
√2
π
= –csc( x + 2)* √2
π
π √2
+ k
∫ dx =
–5
1
i.
ii.
iii. ∫(–5x + 3) –½ dx
√–5x + 3)
= 2 * (–5x + 3) ½ *
1 + k

62. The Substitution Method
Therefore
∫ e – ½ x + 5 dx = –2e – ½ x + 5
+ k
∫ csc( √ 2 x + 2) cot( x + 2) dx
π
√2
π
= –csc( x + 2)* √2
π
π √2
+ k
1
∫ dx =
i.
ii.
iii. ∫(–5x + 3) –½ dx
√–5x + 3)
= 2 * (–5x + 3) –5 ½ *
= + k 2(–5x + 3) ½
–5
1 + k