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Integral Calculus
Dr. M. Alghamdi
Department of Mathematics
January 2, 2019
Dr. M. Alghamdi MATH 106 January 2, 2019 1 / 47
Chapter 5: Techniques of Integration
Main Contents
1 Integration by parts.
2 Trigonometric functions.
3 Trigonometric substitutions.
4 Integrals of rational functions.
5 Integrals of quadratic forms.
6 Miscellaneous substitutions.
Dr. M. Alghamdi MATH 106 January 2, 2019 2 / 47
Integration by PartsIntegration by parts is a method to transfer the original integral to an easier one thatcan be evaluated. Practically, the integration by parts divides the original integral intotwo parts u and dv , then we find the du by deriving u and v by integrating dv .
Theorem
If u = f (x) and v = g(x) such that f ā² and g ā² are continuous, thenā«u dv = uv ā
ā«v du.
Proof: We know that ddx
(f (x)g(x)
)= f (x)g ā²(x) + f ā²(x)g(x). Thus,
f (x)g ā²(x) = ddx
(f (x)g(x)
)ā f ā²(x)g(x).
By integrating both sides, we obtainā«f (x)g ā²(x) dx =
ā«d
dx
(f (x)g(x)
)dx ā
ā«f ā²(x)g(x) dx
= f (x)g(x)āā«
f ā²(x)g(x) dx .
Since u = f (x) and v = g(x), then du = f ā²(x) dx and dv = g ā²(x) dx . Therefore,ā«u dv = uv ā
ā«v du. ļæ½
Dr. M. Alghamdi MATH 106 January 2, 2019 3 / 47
Integration by PartsIntegration by parts is a method to transfer the original integral to an easier one thatcan be evaluated. Practically, the integration by parts divides the original integral intotwo parts u and dv , then we find the du by deriving u and v by integrating dv .
Theorem
If u = f (x) and v = g(x) such that f ā² and g ā² are continuous, thenā«u dv = uv ā
ā«v du.
Proof: We know that ddx
(f (x)g(x)
)= f (x)g ā²(x) + f ā²(x)g(x). Thus,
f (x)g ā²(x) = ddx
(f (x)g(x)
)ā f ā²(x)g(x).
By integrating both sides, we obtainā«f (x)g ā²(x) dx =
ā«d
dx
(f (x)g(x)
)dx ā
ā«f ā²(x)g(x) dx
= f (x)g(x)āā«
f ā²(x)g(x) dx .
Since u = f (x) and v = g(x), then du = f ā²(x) dx and dv = g ā²(x) dx . Therefore,ā«u dv = uv ā
ā«v du. ļæ½
Dr. M. Alghamdi MATH 106 January 2, 2019 3 / 47
Example
Evaluate the integral
ā«x cos x dx .
Solution:
Let I =
ā«x cos x dx . Let u = x and dv = cos x dx . Hence,
u = x ā du = dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
From the theorem, we have
I = x sin x āā«
sin x dx = x sin x + cos x + c.
Try to chooseu = cos x and dv = x dx
Do you have the same result?
Dr. M. Alghamdi MATH 106 January 2, 2019 4 / 47
Example
Evaluate the integral
ā«x cos x dx .
Solution:
Let I =
ā«x cos x dx . Let u = x and dv = cos x dx . Hence,
u = x ā du = dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
From the theorem, we have
I = x sin x āā«
sin x dx = x sin x + cos x + c.
Try to chooseu = cos x and dv = x dx
Do you have the same result?
Dr. M. Alghamdi MATH 106 January 2, 2019 4 / 47
Example
Evaluate the integral
ā«x cos x dx .
Solution:
Let I =
ā«x cos x dx . Let u = x and dv = cos x dx . Hence,
u = x ā du = dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
From the theorem, we have
I = x sin x āā«
sin x dx = x sin x + cos x + c.
Try to chooseu = cos x and dv = x dx
Do you have the same result?
Dr. M. Alghamdi MATH 106 January 2, 2019 4 / 47
Example
Evaluate the integral
ā«x cos x dx .
Solution:
Let I =
ā«x cos x dx . Let u = x and dv = cos x dx . Hence,
u = x ā du = dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
From the theorem, we have
I = x sin x āā«
sin x dx = x sin x + cos x + c.
Try to chooseu = cos x and dv = x dx
Do you have the same result?
Dr. M. Alghamdi MATH 106 January 2, 2019 4 / 47
Example
Evaluate the integral
ā«x cos x dx .
Solution:
Let I =
ā«x cos x dx . Let u = x and dv = cos x dx . Hence,
u = x ā du = dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
From the theorem, we have
I = x sin x āā«
sin x dx = x sin x + cos x + c.
Try to chooseu = cos x and dv = x dx
Do you have the same result?
Dr. M. Alghamdi MATH 106 January 2, 2019 4 / 47
Example
Evaluate the integral
ā«x ex dx .
Solution:
Let I =
ā«x ex dx . Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = ex dx ā v =
ā«ex dx = ex .
From Theorem .1, we have
I = x ex āā«
ex dx = x ex ā ex + c.
Try to choose
u = ex and dv = x dx
We will obtain
I =x2
2ex ā
ā«x2
2ex dx .
However, the integral
ā«x2
2ex dx is
more difficult than the original oneā«xex dx .
Dr. M. Alghamdi MATH 106 January 2, 2019 5 / 47
Example
Evaluate the integral
ā«x ex dx .
Solution:
Let I =
ā«x ex dx . Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = ex dx ā v =
ā«ex dx = ex .
From Theorem .1, we have
I = x ex āā«
ex dx = x ex ā ex + c.
Try to choose
u = ex and dv = x dx
We will obtain
I =x2
2ex ā
ā«x2
2ex dx .
However, the integral
ā«x2
2ex dx is
more difficult than the original oneā«xex dx .
Dr. M. Alghamdi MATH 106 January 2, 2019 5 / 47
Example
Evaluate the integral
ā«x ex dx .
Solution:
Let I =
ā«x ex dx . Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = ex dx ā v =
ā«ex dx = ex .
From Theorem .1, we have
I = x ex āā«
ex dx = x ex ā ex + c.
Try to choose
u = ex and dv = x dx
We will obtain
I =x2
2ex ā
ā«x2
2ex dx .
However, the integral
ā«x2
2ex dx is
more difficult than the original oneā«xex dx .
Dr. M. Alghamdi MATH 106 January 2, 2019 5 / 47
Example
Evaluate the integral
ā«x ex dx .
Solution:
Let I =
ā«x ex dx . Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = ex dx ā v =
ā«ex dx = ex .
From Theorem .1, we have
I = x ex āā«
ex dx = x ex ā ex + c.
Try to choose
u = ex and dv = x dx
We will obtain
I =x2
2ex ā
ā«x2
2ex dx .
However, the integral
ā«x2
2ex dx is
more difficult than the original oneā«xex dx .
Dr. M. Alghamdi MATH 106 January 2, 2019 5 / 47
Example
Evaluate the integral
ā«x ex dx .
Solution:
Let I =
ā«x ex dx . Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = ex dx ā v =
ā«ex dx = ex .
From Theorem .1, we have
I = x ex āā«
ex dx = x ex ā ex + c.
Try to choose
u = ex and dv = x dx
We will obtain
I =x2
2ex ā
ā«x2
2ex dx .
However, the integral
ā«x2
2ex dx is
more difficult than the original oneā«xex dx .
Dr. M. Alghamdi MATH 106 January 2, 2019 5 / 47
Remark
1 Remember that when we consider the integration by parts, we want to obtain an easierintegral. As we saw in the previous example, if we choose u = ex and dv = x dx , we haveā«
x2
2ex dx which is more difficult than the original one.
2 When considering the integration by parts, we have to choose dv a function that can beintegrated.
3 Sometimes we need to use the integration by parts twice as in the upcoming examples.
Example
Evaluate the integral
ā«ln x dx .
Solution: Let I =
ā«ln x dx . Let u = ln x and dv = dx . Hence,
u = ln x ā du =1
xdx ,
dv = dx ā v =
ā«1 dx = x .
From Theorem .1, we obtain I = x ln x āā«
x1
xdx = x ln x ā
ā«1 dx = x ln x ā x + c .
Dr. M. Alghamdi MATH 106 January 2, 2019 6 / 47
Remark
1 Remember that when we consider the integration by parts, we want to obtain an easierintegral. As we saw in the previous example, if we choose u = ex and dv = x dx , we haveā«
x2
2ex dx which is more difficult than the original one.
2 When considering the integration by parts, we have to choose dv a function that can beintegrated.
3 Sometimes we need to use the integration by parts twice as in the upcoming examples.
Example
Evaluate the integral
ā«ln x dx .
Solution: Let I =
ā«ln x dx . Let u = ln x and dv = dx . Hence,
u = ln x ā du =1
xdx ,
dv = dx ā v =
ā«1 dx = x .
From Theorem .1, we obtain I = x ln x āā«
x1
xdx = x ln x ā
ā«1 dx = x ln x ā x + c .
Dr. M. Alghamdi MATH 106 January 2, 2019 6 / 47
Remark
1 Remember that when we consider the integration by parts, we want to obtain an easierintegral. As we saw in the previous example, if we choose u = ex and dv = x dx , we haveā«
x2
2ex dx which is more difficult than the original one.
2 When considering the integration by parts, we have to choose dv a function that can beintegrated.
3 Sometimes we need to use the integration by parts twice as in the upcoming examples.
Example
Evaluate the integral
ā«ln x dx .
Solution: Let I =
ā«ln x dx . Let u = ln x and dv = dx . Hence,
u = ln x ā du =1
xdx ,
dv = dx ā v =
ā«1 dx = x .
From Theorem .1, we obtain I = x ln x āā«
x1
xdx = x ln x ā
ā«1 dx = x ln x ā x + c .
Dr. M. Alghamdi MATH 106 January 2, 2019 6 / 47
Remark
1 Remember that when we consider the integration by parts, we want to obtain an easierintegral. As we saw in the previous example, if we choose u = ex and dv = x dx , we haveā«
x2
2ex dx which is more difficult than the original one.
2 When considering the integration by parts, we have to choose dv a function that can beintegrated.
3 Sometimes we need to use the integration by parts twice as in the upcoming examples.
Example
Evaluate the integral
ā«ln x dx .
Solution: Let I =
ā«ln x dx . Let u = ln x and dv = dx . Hence,
u = ln x ā du =1
xdx ,
dv = dx ā v =
ā«1 dx = x .
From Theorem .1, we obtain I = x ln x āā«
x1
xdx = x ln x ā
ā«1 dx = x ln x ā x + c .
Dr. M. Alghamdi MATH 106 January 2, 2019 6 / 47
Example
Evaluate the integral
ā«ex cos x dx .
Solution: Let I =
ā«ex cos x dx . Let u = ex and dv = cos x dx .
u = ex ā du = ex dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
Hence, I = ex sin x āā«
ex sin x dx .
The integral
ā«ex sin x dx cannot be evaluated. Therefore, we use the integration by parts
again where we assume J =
ā«ex sin x dx . Let u = ex and dv = sin x dx . Hence,
u = ex ā du = ex dx ,
dv = sin x dx ā v =
ā«sin x dx = ā cos x .
Hence, J = āex cos x +
ā«ex cos x dx . By substituting the result of J into I , we have
I = ex sin x ā J = ex sin x + ex cos x āā«
ex cos x dx ā I = ex sin x + ex cos x ā I .
This implies 2I = ex sin x + ex cos x ā I = 12
(ex sin x + ex cos x) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 7 / 47
Example
Evaluate the integral
ā«ex cos x dx .
Solution: Let I =
ā«ex cos x dx . Let u = ex and dv = cos x dx .
u = ex ā du = ex dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
Hence, I = ex sin x āā«
ex sin x dx .
The integral
ā«ex sin x dx cannot be evaluated. Therefore, we use the integration by parts
again where we assume J =
ā«ex sin x dx . Let u = ex and dv = sin x dx . Hence,
u = ex ā du = ex dx ,
dv = sin x dx ā v =
ā«sin x dx = ā cos x .
Hence, J = āex cos x +
ā«ex cos x dx . By substituting the result of J into I , we have
I = ex sin x ā J = ex sin x + ex cos x āā«
ex cos x dx ā I = ex sin x + ex cos x ā I .
This implies 2I = ex sin x + ex cos x ā I = 12
(ex sin x + ex cos x) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 7 / 47
Example
Evaluate the integral
ā«ex cos x dx .
Solution: Let I =
ā«ex cos x dx . Let u = ex and dv = cos x dx .
u = ex ā du = ex dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
Hence, I = ex sin x āā«
ex sin x dx .
The integral
ā«ex sin x dx cannot be evaluated. Therefore, we use the integration by parts
again where we assume J =
ā«ex sin x dx . Let u = ex and dv = sin x dx . Hence,
u = ex ā du = ex dx ,
dv = sin x dx ā v =
ā«sin x dx = ā cos x .
Hence, J = āex cos x +
ā«ex cos x dx . By substituting the result of J into I , we have
I = ex sin x ā J = ex sin x + ex cos x āā«
ex cos x dx ā I = ex sin x + ex cos x ā I .
This implies 2I = ex sin x + ex cos x ā I = 12
(ex sin x + ex cos x) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 7 / 47
Example
Evaluate the integral
ā«ex cos x dx .
Solution: Let I =
ā«ex cos x dx . Let u = ex and dv = cos x dx .
u = ex ā du = ex dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
Hence, I = ex sin x āā«
ex sin x dx .
The integral
ā«ex sin x dx cannot be evaluated. Therefore, we use the integration by parts
again where we assume J =
ā«ex sin x dx . Let u = ex and dv = sin x dx . Hence,
u = ex ā du = ex dx ,
dv = sin x dx ā v =
ā«sin x dx = ā cos x .
Hence, J = āex cos x +
ā«ex cos x dx . By substituting the result of J into I , we have
I = ex sin x ā J = ex sin x + ex cos x āā«
ex cos x dx ā I = ex sin x + ex cos x ā I .
This implies 2I = ex sin x + ex cos x ā I = 12
(ex sin x + ex cos x) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 7 / 47
Example
Evaluate the integral
ā«ex cos x dx .
Solution: Let I =
ā«ex cos x dx . Let u = ex and dv = cos x dx .
u = ex ā du = ex dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
Hence, I = ex sin x āā«
ex sin x dx .
The integral
ā«ex sin x dx cannot be evaluated. Therefore, we use the integration by parts
again where we assume J =
ā«ex sin x dx . Let u = ex and dv = sin x dx . Hence,
u = ex ā du = ex dx ,
dv = sin x dx ā v =
ā«sin x dx = ā cos x .
Hence, J = āex cos x +
ā«ex cos x dx . By substituting the result of J into I , we have
I = ex sin x ā J = ex sin x + ex cos x āā«
ex cos x dx ā I = ex sin x + ex cos x ā I .
This implies 2I = ex sin x + ex cos x ā I = 12
(ex sin x + ex cos x) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 7 / 47
Example
Evaluate the integral
ā«ex cos x dx .
Solution: Let I =
ā«ex cos x dx . Let u = ex and dv = cos x dx .
u = ex ā du = ex dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
Hence, I = ex sin x āā«
ex sin x dx .
The integral
ā«ex sin x dx cannot be evaluated. Therefore, we use the integration by parts
again where we assume J =
ā«ex sin x dx . Let u = ex and dv = sin x dx . Hence,
u = ex ā du = ex dx ,
dv = sin x dx ā v =
ā«sin x dx = ā cos x .
Hence, J = āex cos x +
ā«ex cos x dx . By substituting the result of J into I , we have
I = ex sin x ā J = ex sin x + ex cos x āā«
ex cos x dx ā I = ex sin x + ex cos x ā I .
This implies 2I = ex sin x + ex cos x ā I = 12
(ex sin x + ex cos x) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 7 / 47
Example
Evaluate the integral
ā«ex cos x dx .
Solution: Let I =
ā«ex cos x dx . Let u = ex and dv = cos x dx .
u = ex ā du = ex dx ,
dv = cos x dx ā v =
ā«cos x dx = sin x .
Hence, I = ex sin x āā«
ex sin x dx .
The integral
ā«ex sin x dx cannot be evaluated. Therefore, we use the integration by parts
again where we assume J =
ā«ex sin x dx . Let u = ex and dv = sin x dx . Hence,
u = ex ā du = ex dx ,
dv = sin x dx ā v =
ā«sin x dx = ā cos x .
Hence, J = āex cos x +
ā«ex cos x dx . By substituting the result of J into I , we have
I = ex sin x ā J = ex sin x + ex cos x āā«
ex cos x dx ā I = ex sin x + ex cos x ā I .
This implies 2I = ex sin x + ex cos x ā I = 12
(ex sin x + ex cos x) + c.Dr. M. Alghamdi MATH 106 January 2, 2019 7 / 47
Example
Evaluate the integral
ā«x2ex dx .
Solution: Let I =
ā«x2ex dx . Let u = x2 and dv = ex dx . Hence,
u = x2 ā du = 2x dx ,
dv = exdx ā v =
ā«ex dx = ex .
This implies, I = x2ex ā 2
ā«xex dx .
We use the integration by parts again for the integral
ā«xex dx . Let J =
ā«xex dx .
Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = exdx ā v =
ā«ex dx = ex .
Therefore, J = xex āā«
ex dx = xex ā ex + c. By substituting the result into I , we have
I = x2ex ā 2(xex ā ex ) + c = ex (x2 ā 2x + 2) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 8 / 47
Example
Evaluate the integral
ā«x2ex dx .
Solution: Let I =
ā«x2ex dx . Let u = x2 and dv = ex dx . Hence,
u = x2 ā du = 2x dx ,
dv = exdx ā v =
ā«ex dx = ex .
This implies, I = x2ex ā 2
ā«xex dx .
We use the integration by parts again for the integral
ā«xex dx . Let J =
ā«xex dx .
Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = exdx ā v =
ā«ex dx = ex .
Therefore, J = xex āā«
ex dx = xex ā ex + c. By substituting the result into I , we have
I = x2ex ā 2(xex ā ex ) + c = ex (x2 ā 2x + 2) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 8 / 47
Example
Evaluate the integral
ā«x2ex dx .
Solution: Let I =
ā«x2ex dx . Let u = x2 and dv = ex dx . Hence,
u = x2 ā du = 2x dx ,
dv = exdx ā v =
ā«ex dx = ex .
This implies, I = x2ex ā 2
ā«xex dx .
We use the integration by parts again for the integral
ā«xex dx . Let J =
ā«xex dx .
Let u = x and dv = ex dx . Hence,
u = x ā du = dx ,
dv = exdx ā v =
ā«ex dx = ex .
Therefore, J = xex āā«
ex dx = xex ā ex + c. By substituting the result into I , we have
I = x2ex ā 2(xex ā ex ) + c = ex (x2 ā 2x + 2) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 8 / 47
Example
Evaluate the integral
ā« 1
0
tanā1 x dx .
Solution:
Let I =
ā«tanā1 x dx . Let u = tanā1 x and dv = dx . Hence,
u = tanā1 x ā du =1
x2 + 1dx ,
dv = dx ā v =
ā«1 dx = x .
By applying the theorem, we obtain
I = x tanā1 x āā«
x
x2 + 1dx = x tanā1 x ā 1
2ln(x2 + 1) + c.
Therefore,ā« 1
0
tanā1 x dx =[x tanā1 xā1
2ln(x2+1)
]1
0= (tanā1(1)ā1
2ln 2)ā(0ā1
2ln 1) =
Ļ
4ālnā
2.
Dr. M. Alghamdi MATH 106 January 2, 2019 9 / 47
Example
Evaluate the integral
ā« 1
0
tanā1 x dx .
Solution:
Let I =
ā«tanā1 x dx . Let u = tanā1 x and dv = dx . Hence,
u = tanā1 x ā du =1
x2 + 1dx ,
dv = dx ā v =
ā«1 dx = x .
By applying the theorem, we obtain
I = x tanā1 x āā«
x
x2 + 1dx = x tanā1 x ā 1
2ln(x2 + 1) + c.
Therefore,ā« 1
0
tanā1 x dx =[x tanā1 xā1
2ln(x2+1)
]1
0= (tanā1(1)ā1
2ln 2)ā(0ā1
2ln 1) =
Ļ
4ālnā
2.
Dr. M. Alghamdi MATH 106 January 2, 2019 9 / 47
Trigonometric Functions(A) Integration of Powers of Trigonometric Functions
In this section, we evaluate integrals of forms
ā«sinn x cosm x dx ,ā«
tann x secm x dx and
ā«cotn x cscm x dx .
Form 1:
ā«sinn x cosm x dx .
This form is treated as follows:
1 If n is an odd integer, write
sinn x cosm x = sinnā1 x cosm x sin x
Then, use the identity sin2 x = 1ā cos2 x and the substitution u = cos x .
2 If m is an odd integer, write
sinn x cosm x = sinn x cosmā1 x cos x
Then, use the identity cos2 x = 1ā sin2 x and the substitution u = sin x .
3 If m and n are even, use the identities cos2 x = 1+cos 2x2
andsin2 x = 1ācos 2x
2.
Dr. M. Alghamdi MATH 106 January 2, 2019 10 / 47
Trigonometric Functions(A) Integration of Powers of Trigonometric Functions
In this section, we evaluate integrals of forms
ā«sinn x cosm x dx ,ā«
tann x secm x dx and
ā«cotn x cscm x dx .
Form 1:
ā«sinn x cosm x dx .
This form is treated as follows:
1 If n is an odd integer, write
sinn x cosm x = sinnā1 x cosm x sin x
Then, use the identity sin2 x = 1ā cos2 x and the substitution u = cos x .
2 If m is an odd integer, write
sinn x cosm x = sinn x cosmā1 x cos x
Then, use the identity cos2 x = 1ā sin2 x and the substitution u = sin x .
3 If m and n are even, use the identities cos2 x = 1+cos 2x2
andsin2 x = 1ācos 2x
2.
Dr. M. Alghamdi MATH 106 January 2, 2019 10 / 47
ExampleEvaluate the integral.
1
ā«sin3 x dx
2
ā«cos4 x dx
3
ā«sin5 x cos4 x dx
4
ā«sin2 x cos2 x dx
Solution:1) Write sin3 x = sin2 x sin x = (1ā cos2 x) sin x . Hence,ā«
sin3 x dx =
ā«(1ā cos2 x) sin x dx .
Let u = cos x , then du = ā sin x dx . By substitution, we have
āā«
(1ā u2) du = āu +u3
3+ c.
This implies ā«sin3 x dx = ā cos x +
1
3cos3 x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 11 / 47
ExampleEvaluate the integral.
1
ā«sin3 x dx
2
ā«cos4 x dx
3
ā«sin5 x cos4 x dx
4
ā«sin2 x cos2 x dx
Solution:1) Write sin3 x = sin2 x sin x = (1ā cos2 x) sin x . Hence,ā«
sin3 x dx =
ā«(1ā cos2 x) sin x dx .
Let u = cos x , then du = ā sin x dx . By substitution, we have
āā«
(1ā u2) du = āu +u3
3+ c.
This implies ā«sin3 x dx = ā cos x +
1
3cos3 x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 11 / 47
2) Write cos4 x = (cos2 x)2 = ( 1+cos 2x2
)2. Hence,ā«cos4 x dx =
ā« (1 + cos 2x
2
)2
dx
=1
4
ā«(1 + 2 cos 2x + cos2 2x) dx
=1
4
(ā«1 dx +
ā«2 cos 2x dx +
ā«cos2 2x dx
)=
1
4
(x + sin 2x +
1
2
ā«(1 + cos 4x) dx
)=
1
4
(x + sin 2x +
1
2
(x +
sin 4x
4
))+ c.
3) Write sin5 x cos4 x = sin4 x cos4 x sin x = (1ā cos2 x)2 cos4 x sin x .Let u = cos x , then du = ā sin x dx . Thus, the integral becomes
āā«
(1ā u2)2u4 du = āā«
(u4 ā 2u6 + u8) du = ā(u5
5ā 2u7
7+
u9
9
)+ c.
This implies
ā«sin5 x cos4 x dx = ā cos5 x
5+ 2 cos7 x
7ā cos9 x
9+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 12 / 47
2) Write cos4 x = (cos2 x)2 = ( 1+cos 2x2
)2. Hence,ā«cos4 x dx =
ā« (1 + cos 2x
2
)2
dx
=1
4
ā«(1 + 2 cos 2x + cos2 2x) dx
=1
4
(ā«1 dx +
ā«2 cos 2x dx +
ā«cos2 2x dx
)=
1
4
(x + sin 2x +
1
2
ā«(1 + cos 4x) dx
)=
1
4
(x + sin 2x +
1
2
(x +
sin 4x
4
))+ c.
3) Write sin5 x cos4 x = sin4 x cos4 x sin x = (1ā cos2 x)2 cos4 x sin x .Let u = cos x , then du = ā sin x dx . Thus, the integral becomes
āā«
(1ā u2)2u4 du = āā«
(u4 ā 2u6 + u8) du = ā(u5
5ā 2u7
7+
u9
9
)+ c.
This implies
ā«sin5 x cos4 x dx = ā cos5 x
5+ 2 cos7 x
7ā cos9 x
9+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 12 / 47
4) The integrand
sin2 x cos2 x = ( 1ācos 2x2
)( 1+cos 2x2
) = 1ācos2 2x4
= sin2 2x4
= 14( 1ācos 4x
2). Hence,ā«
sin2 x cos2 x dx =1
8
ā«(1ā cos 4x) dx =
1
8
(x ā sin 4x
4
)+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 13 / 47
Form 2:
ā«tann x secm x dx .
This form is treated as follows:
1 If n = 0, write secm x = secmā2 x sec2 x .
1 If m > 1 is odd, use the integration by parts.2 If m is even, use the identity sec2 x = 1 + tan2 x and the
substitution u = tan x .
2 If m = 0 and n is odd or even, write tann x = tannā2 x tan2 x . Then, use theidentity tan2 x = sec2 x ā 1 and the substitution u = tan x .
3 If n is even and m is odd, use the identity tan2 x = sec2 x ā 1 to reduce thepower m and then use the integration by parts.
4 If m ā„ 2 is even, write tann x secm x = tann x secmā2 x sec2 x . Then, usethe identity sec2 x = 1 + tan2 x and the substitution u = tan x . Alternatively,write
tann x secm x = tannā1 x secmā1 x tan x sec x
Then, use the identity tan2 x = sec2 x ā 1 and the substitution u = sec x .
5 If n is odd and m ā„ 1, write tann x secm x = tannā1 x secmā1 x tan x sec x .Then, use the identity tan2 x = sec2 x ā 1 and the substitution u = sec x .
Dr. M. Alghamdi MATH 106 January 2, 2019 14 / 47
Example
Evaluate the integral.
1
ā«tan5 x dx
2
ā«tan6 x dx
3
ā«sec3 x dx
4
ā«tan5 x sec4 x dx
5
ā«tan4 x sec4 x dx
Solution:1) Write tan5 x = tan3 x tan2 x = tan3 x (sec2 x ā 1). Thus,ā«
tan5 x dx =
ā«tan3 x (sec2 x ā 1) dx
=
ā«tan3 x sec2 x dx ā
ā«tan3 x dx
=tan4 x
4āā«
tan x (sec2 x ā 1) dx
=tan4 x
4āā«
tan x sec2 x dx +
ā«tan x dx
=tan4 x
4ā tan2 x
2+ ln | sec x | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 15 / 47
Example
Evaluate the integral.
1
ā«tan5 x dx
2
ā«tan6 x dx
3
ā«sec3 x dx
4
ā«tan5 x sec4 x dx
5
ā«tan4 x sec4 x dx
Solution:1) Write tan5 x = tan3 x tan2 x = tan3 x (sec2 x ā 1). Thus,ā«
tan5 x dx =
ā«tan3 x (sec2 x ā 1) dx
=
ā«tan3 x sec2 x dx ā
ā«tan3 x dx
=tan4 x
4āā«
tan x (sec2 x ā 1) dx
=tan4 x
4āā«
tan x sec2 x dx +
ā«tan x dx
=tan4 x
4ā tan2 x
2+ ln | sec x | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 15 / 47
2) Write tan6 x = tan4 x tan2 x = tan4 x (sec2 x ā 1). The integral becomesā«tan6 x dx =
ā«tan4 x (sec2 x ā 1) dx
=
ā«tan4 x sec2 x dx ā
ā«tan4 x dx
=tan5 x
5āā«
tan2 x (sec2 x ā 1) dx
=tan5 x
5āā«
tan2 x sec2 x dx +
ā«tan2 x dx
=tan5 x
5ā tan3 x
3+
ā«(sec2 x ā 1) dx
=tan5 x
5ā tan3 x
3+ tan x ā x + c.
3) Write sec3 x = sec x sec2 x and let I =
ā«sec x sec2 x dx .
We use the integration by parts to evaluate the integral as follows:
u = sec x ā du = sec x tan x dx ,
dv = sec2 x dx ā v =
ā«sec2 x dx = tan x .
Dr. M. Alghamdi MATH 106 January 2, 2019 16 / 47
2) Write tan6 x = tan4 x tan2 x = tan4 x (sec2 x ā 1). The integral becomesā«tan6 x dx =
ā«tan4 x (sec2 x ā 1) dx
=
ā«tan4 x sec2 x dx ā
ā«tan4 x dx
=tan5 x
5āā«
tan2 x (sec2 x ā 1) dx
=tan5 x
5āā«
tan2 x sec2 x dx +
ā«tan2 x dx
=tan5 x
5ā tan3 x
3+
ā«(sec2 x ā 1) dx
=tan5 x
5ā tan3 x
3+ tan x ā x + c.
3) Write sec3 x = sec x sec2 x and let I =
ā«sec x sec2 x dx .
We use the integration by parts to evaluate the integral as follows:
u = sec x ā du = sec x tan x dx ,
dv = sec2 x dx ā v =
ā«sec2 x dx = tan x .
Dr. M. Alghamdi MATH 106 January 2, 2019 16 / 47
Hence,
I = sec x tan x āā«
sec x tan2 x dx
= sec x tan x āā«
(sec3 x ā sec x) dx
= sec x tan x ā I + ln | sec x + tan x |
I =1
2(sec x tan x + ln | sec x + tan x |) + c.
4) Express the integrand tan5 x sec4 x as follows
tan5 x sec4 x = tan5 x sec2 x sec2 x = tan5 x (tan2 x + 1) sec2 x .
This implies ā«tan5 x sec4 x dx =
ā«tan5 x (tan2 x + 1) sec2 x dx
=
ā«(tan7 x + tan5 x) sec2 x dx
=tan8 x
8+
tan6 x
6+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 17 / 47
Hence,
I = sec x tan x āā«
sec x tan2 x dx
= sec x tan x āā«
(sec3 x ā sec x) dx
= sec x tan x ā I + ln | sec x + tan x |
I =1
2(sec x tan x + ln | sec x + tan x |) + c.
4) Express the integrand tan5 x sec4 x as follows
tan5 x sec4 x = tan5 x sec2 x sec2 x = tan5 x (tan2 x + 1) sec2 x .
This implies ā«tan5 x sec4 x dx =
ā«tan5 x (tan2 x + 1) sec2 x dx
=
ā«(tan7 x + tan5 x) sec2 x dx
=tan8 x
8+
tan6 x
6+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 17 / 47
5) Write tan4 x sec4 x = tan4 x (tan2 x + 1) sec2 x . The integral becomesā«tan4 x sec4 x dx =
ā«tan4 x (tan2 x + 1) sec2 x dx
=
ā«(tan6 x + tan4 x) sec2 x dx
=tan7 x
7+
tan5 x
5+ c.
Form 3:
ā«cotn x cscm x dx .
The treatment of this form is similar to the integral
ā«tann x secm x dx , except we use
the identitycot2 x + 1 = csc2 x .
ExampleEvaluate the integral.
1
ā«cot3 x dx
2
ā«cot4 x dx
3
ā«cot5 x csc4 x dx
Dr. M. Alghamdi MATH 106 January 2, 2019 18 / 47
Solution:1) Write cot3 x = cot x (csc2 x ā 1). Then,ā«
cot3 x dx =
ā«cot x (csc2 x ā 1) dx
=
ā«(cot x csc2 x ā cot x) dx
=
ā«cot x csc2 x dx ā
ā«cot x dx
= ā1
2cot2 x ā ln | sin x | +c.
2) The integrand can be expressed as cot4 x = cot2 x (csc2 x ā 1). Thus,ā«cot4 x dx =
ā«cot2 x (csc2 x ā 1) dx
=
ā«cot2 x csc2 x dx ā
ā«cot2 x dx
= ācot3 x
3āā«
(csc2 x ā 1) dx
= ācot3 x
3+ cot x + x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 19 / 47
3) Write cot5 x csc4 x = csc3 x cot4 x csc x cot x . This impliesā«cot5 x csc4 x dx =
ā«csc3 x cot4 x csc x cot x dx
=
ā«csc3 x (csc2 x ā 1)2 csc x cot x dx
=
ā«(csc7 x ā 2 csc5 x + csc3 x) csc x cot x dx
= ācsc8 x
8+
csc6 x
3ā csc4 x
4+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 20 / 47
(B) Integration of Forms sin ux cos vx, sin ux sin vx and cos ux cos vxWe deal with these integrals by using the following formulas:
sin ux cos vx =1
2
(sin (u ā v) x + sin (u + v) x
)sin ux sin vx =
1
2
(cos (u ā v) x ā cos (u + v) x
)cos ux cos vx =
1
2
(cos (u ā v) x + cos (u + v) x
)
ExampleEvaluate the integral.
1
ā«sin 5x sin 3x dx
2
ā«sin 7x cos 2x dx
3
ā«cos 5x sin 2x dx
4
ā«cos 6x cos 4x dx
Dr. M. Alghamdi MATH 106 January 2, 2019 21 / 47
(B) Integration of Forms sin ux cos vx, sin ux sin vx and cos ux cos vxWe deal with these integrals by using the following formulas:
sin ux cos vx =1
2
(sin (u ā v) x + sin (u + v) x
)sin ux sin vx =
1
2
(cos (u ā v) x ā cos (u + v) x
)cos ux cos vx =
1
2
(cos (u ā v) x + cos (u + v) x
)
ExampleEvaluate the integral.
1
ā«sin 5x sin 3x dx
2
ā«sin 7x cos 2x dx
3
ā«cos 5x sin 2x dx
4
ā«cos 6x cos 4x dx
Dr. M. Alghamdi MATH 106 January 2, 2019 21 / 47
Solution:1) From the previous formulas, we have sin 5x sin 3x = 1
2
(cos 2x ā cos 8x
). Hence,ā«
sin 5x sin 3x dx =1
2
ā«(cos 2x ā cos 8x) dx
=1
4sin 2x ā 1
16sin 8x + c.
2) Since sin 7x cos 2x = 12
(sin 5x + sin 9x
), thenā«
sin 7x cos 2x dx =1
2
ā«(sin 5x + sin 9x) dx
= ā 1
10cos 5x ā 1
18cos 9x + c.
3) Since cos 5x sin 2x = 12
(sin 3x + sin 7x
), thenā«
cos 5x sin 2x dx =1
2
ā«(sin 3x + sin 7x) dx
= ā1
6cos 3x ā 1
14cos 7x + c.
4) Since cos 6x cos 4x = 12
(cos 2x + cos 10x
), thenā«
cos 6x cos 4x dx =1
2
ā«(cos 2x + cos 10x) dx
=1
4sin 2x +
1
20sin 10x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 22 / 47
Trigonometric SubstitutionsIn this section, we are going to study integrals containing the following expressionsāa2 ā x2,
āa2 + x2 and
āx2 ā a2 where a > 0.
āa2 ā x2 = a cos Īø if x =
a sin Īø.If x = a sin Īø where Īø ā[āĻ/2, Ļ/2], thenā
a2 ā x2 =ā
a2 ā a2 sin2 Īø
=ā
a2(1ā sin2 Īø)
=āa2 cos2 Īø
= a cos Īø.
āa2 ā x2
xa
Īø
If the expressionāa2 ā x2 is in a denominator, then we assume āĻ
2< Īø < Ļ
2.
Dr. M. Alghamdi MATH 106 January 2, 2019 23 / 47
āa2 + x2 = a sec Īø if x =
a tan Īø.If x = a tan Īø where Īø ā(āĻ/2, Ļ/2), thenā
a2 + x2 =ā
a2 + a2 tan2 Īø
=ā
a2(1 + tan2 Īø)
=āa2 sec2 Īø
= a sec Īø.
a
x
āa2 + x2
Īø
Dr. M. Alghamdi MATH 106 January 2, 2019 24 / 47
āx2 ā a2 = a tan Īø if x =
a sec Īø.If x = a sec Īø where Īø ā [0, Ļ/2)āŖ[Ļ, 3Ļ/2), thenā
x2 ā a2 =ā
a2 sec2 Īø ā a2
=ā
a2(sec2 Īø ā 1)
=āa2 tan2 Īø
= a tan Īø.
a
āx2 ā a2
x
Īø
Dr. M. Alghamdi MATH 106 January 2, 2019 25 / 47
ExampleEvaluate the integral.
1
ā«x2
ā1ā x2
dx 2
ā« 6
5
āx2 ā 25
x4dx
3
ā« āx2 + 9 dx
Solution:1) Let x = sin Īø where Īø ā (āĻ/2, Ļ/2), thus dx = cos Īø dĪø. By substitution, we haveā«
x2
ā1ā x2
dx =
ā«sin2 Īø dĪø
=1
2
ā«(1ā cos 2Īø) dĪø
=1
2
(Īø ā
1
2sin 2Īø
)+ c
=1
2
(Īø ā sin Īø cos Īø
)+ c.
ā1ā x2
x1
Īø
Now, we must return to the original variable x :ā«x2
ā1ā x2
dx =1
2(sinā1 x ā x
ā1ā x2) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 26 / 47
ExampleEvaluate the integral.
1
ā«x2
ā1ā x2
dx 2
ā« 6
5
āx2 ā 25
x4dx
3
ā« āx2 + 9 dx
Solution:1) Let x = sin Īø where Īø ā (āĻ/2, Ļ/2), thus dx = cos Īø dĪø. By substitution, we haveā«
x2
ā1ā x2
dx =
ā«sin2 Īø dĪø
=1
2
ā«(1ā cos 2Īø) dĪø
=1
2
(Īø ā
1
2sin 2Īø
)+ c
=1
2
(Īø ā sin Īø cos Īø
)+ c.
ā1ā x2
x1
Īø
Now, we must return to the original variable x :ā«x2
ā1ā x2
dx =1
2(sinā1 x ā x
ā1ā x2) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 26 / 47
2) Let x = 5 sec Īø where Īø ā [0, Ļ/2) āŖ [Ļ, 3Ļ/2), thus dx = 5 sec Īø tan Īø dĪø. Aftersubstitution, the integral becomes
ā« ā25 sec2 Īø ā 25
625 sec4 Īø5 sec Īø tan Īø dĪø =
1
25
ā«tan2 Īø
sec3 ĪødĪø
=1
25
ā«sin2 Īø cos Īø dĪø
=1
75sin3 Īø + c.
5
āx2 ā 25
x
Īø
We must return to the original variable x :ā« āx2 ā 25
x4dx =
(x2 ā 25)3/2
75x3
Hence, ā« 6
5
āx2 ā 25
x4dx =
1
75
[ (x2 ā 25)3/2
x3
]6
5=
1
600.
Dr. M. Alghamdi MATH 106 January 2, 2019 27 / 47
3) Let x = 3 tan Īø where Īø ā (āĻ/2, Ļ/2). This implies dx = 3 sec2 Īø dĪø. By substitution, wehave
ā« āx2 + 9 dx =
ā« ā9 tan2 Īø + 9 (3 sec2 Īø) dĪø
= 9
ā«sec3 Īø dĪø
=9
2
(sec Īø tan Īø + ln
ā£ā£ sec Īø + tan Īøā£ā£).
This impliesā« āx2 + 9 dx =
9
2
( xāx2 + 9
9+ ln
ā£ā£ā£āx2 + 9 + x
3
ā£ā£ā£)+ c.
3
x
āx2 + 9
Īø
Dr. M. Alghamdi MATH 106 January 2, 2019 28 / 47
Integrals of Rational FunctionsIn this section, we study rational functions of form q(x) = f (x)
g(x)where f (x) and g(x) are
polynomials.Step 1: If the degree of g(x) is less than the degree of f (x), we do polynomiallong-division; otherwise we move to step 2.
From the long division shown on theright side, we have
q(x) =f (x)
g(x)= h(x) +
r(x)
g(x),
where h(x) is the quotient and r(x)is the remainder.
h(x)g(x)
)f(x)
- ......
r(x)
Dr. M. Alghamdi MATH 106 January 2, 2019 29 / 47
Step 2: Factor the denominator g(x) into irreducible polynomials where the factors areeither linear or irreducible quadratic polynomials.Step 3: Find the partial fraction decomposition. This step depends on the result of step2 where the fraction f (x)
g(x)or r(x)
g(x)can be written as a sum of partial fractions:
q(x) = P1(x) + P2(x) + P3(x) + ...+ Pn(x) ,
each Pk(x) = Ak(ax+b)n
, n ā N or Pk(x) = Ak x+Bk(ax2+bx+c)n
if b2 ā 4ac < 0. The constants Ak
and Bk are real numbers and computed later.Step 4: Integrate the result of step 3.
Dr. M. Alghamdi MATH 106 January 2, 2019 30 / 47
Example
Evaluate the integral
ā«x + 1
x2 ā 2x ā 8dx .
Solution:Step 1: This step can be skipped since the degree of f (x) = x + 1 is less than thedegree of g(x) = x2 ā 2x ā 8.Step 2: Factor the denominator g(x) into irreducible polynomials
g(x) = x2 ā 2x ā 8 = (x + 2)(x ā 4).
Step 3: Find the partial fraction decomposition.
x + 1
x2 ā 2x ā 8=
A
x + 2+
B
x ā 4=
Ax ā 4A + Bx + 2B
(x + 2)(x ā 4).
We need to find the constants A and B.
Coefficients of the numerators:
A + B = 1ā 1
ā4A + 2B = 1ā 2
By doing some calculation, we obtain A = 16
and B = 56.
Illustration
Multiply equation 1 by 4and add theresult to equation 2
4A + 4B = 4
ā4A + 2B = 1
āāāāāāāāā6B = 5
Dr. M. Alghamdi MATH 106 January 2, 2019 31 / 47
Example
Evaluate the integral
ā«x + 1
x2 ā 2x ā 8dx .
Solution:Step 1: This step can be skipped since the degree of f (x) = x + 1 is less than thedegree of g(x) = x2 ā 2x ā 8.
Step 2: Factor the denominator g(x) into irreducible polynomials
g(x) = x2 ā 2x ā 8 = (x + 2)(x ā 4).
Step 3: Find the partial fraction decomposition.
x + 1
x2 ā 2x ā 8=
A
x + 2+
B
x ā 4=
Ax ā 4A + Bx + 2B
(x + 2)(x ā 4).
We need to find the constants A and B.
Coefficients of the numerators:
A + B = 1ā 1
ā4A + 2B = 1ā 2
By doing some calculation, we obtain A = 16
and B = 56.
Illustration
Multiply equation 1 by 4and add theresult to equation 2
4A + 4B = 4
ā4A + 2B = 1
āāāāāāāāā6B = 5
Dr. M. Alghamdi MATH 106 January 2, 2019 31 / 47
Example
Evaluate the integral
ā«x + 1
x2 ā 2x ā 8dx .
Solution:Step 1: This step can be skipped since the degree of f (x) = x + 1 is less than thedegree of g(x) = x2 ā 2x ā 8.Step 2: Factor the denominator g(x) into irreducible polynomials
g(x) = x2 ā 2x ā 8 = (x + 2)(x ā 4).
Step 3: Find the partial fraction decomposition.
x + 1
x2 ā 2x ā 8=
A
x + 2+
B
x ā 4=
Ax ā 4A + Bx + 2B
(x + 2)(x ā 4).
We need to find the constants A and B.
Coefficients of the numerators:
A + B = 1ā 1
ā4A + 2B = 1ā 2
By doing some calculation, we obtain A = 16
and B = 56.
Illustration
Multiply equation 1 by 4and add theresult to equation 2
4A + 4B = 4
ā4A + 2B = 1
āāāāāāāāā6B = 5
Dr. M. Alghamdi MATH 106 January 2, 2019 31 / 47
Example
Evaluate the integral
ā«x + 1
x2 ā 2x ā 8dx .
Solution:Step 1: This step can be skipped since the degree of f (x) = x + 1 is less than thedegree of g(x) = x2 ā 2x ā 8.Step 2: Factor the denominator g(x) into irreducible polynomials
g(x) = x2 ā 2x ā 8 = (x + 2)(x ā 4).
Step 3: Find the partial fraction decomposition.
x + 1
x2 ā 2x ā 8=
A
x + 2+
B
x ā 4=
Ax ā 4A + Bx + 2B
(x + 2)(x ā 4).
We need to find the constants A and B.
Coefficients of the numerators:
A + B = 1ā 1
ā4A + 2B = 1ā 2
By doing some calculation, we obtain A = 16
and B = 56.
Illustration
Multiply equation 1 by 4and add theresult to equation 2
4A + 4B = 4
ā4A + 2B = 1
āāāāāāāāā6B = 5Dr. M. Alghamdi MATH 106 January 2, 2019 31 / 47
Step 4: Integrate the result of step 3.ā«x + 1
x2 ā 2x ā 8dx =
ā«1/6
x + 2dx +
ā«5/6
x ā 4dx =
1
6ln | x + 2 | +
5
6ln | x ā 4 | +c.
Example
Evaluate the integral
ā«2x3 ā 4x2 ā 15x + 5
x2 + 3x + 2dx .
Solution:Step 1: Do the polynomial long-division.Since the degree of the denominator g(x) is less than the degree of the numerator f (x),we do the polynomial long-division given on the right side. Then, we have
q(x) = (2x ā 10) +11x + 25
x2 + 3x + 2.
2x ā10x2 + 3x + 2
)2x3 ā4x2 ā15x +5ā(2x3 +6x2 +4x)
ā10x2 ā19x +5ā(ā10x2 ā30x ā20)
11x +25
Dr. M. Alghamdi MATH 106 January 2, 2019 32 / 47
Step 4: Integrate the result of step 3.ā«x + 1
x2 ā 2x ā 8dx =
ā«1/6
x + 2dx +
ā«5/6
x ā 4dx =
1
6ln | x + 2 | +
5
6ln | x ā 4 | +c.
Example
Evaluate the integral
ā«2x3 ā 4x2 ā 15x + 5
x2 + 3x + 2dx .
Solution:Step 1: Do the polynomial long-division.Since the degree of the denominator g(x) is less than the degree of the numerator f (x),we do the polynomial long-division given on the right side. Then, we have
q(x) = (2x ā 10) +11x + 25
x2 + 3x + 2.
2x ā10x2 + 3x + 2
)2x3 ā4x2 ā15x +5ā(2x3 +6x2 +4x)
ā10x2 ā19x +5ā(ā10x2 ā30x ā20)
11x +25
Dr. M. Alghamdi MATH 106 January 2, 2019 32 / 47
Step 4: Integrate the result of step 3.ā«x + 1
x2 ā 2x ā 8dx =
ā«1/6
x + 2dx +
ā«5/6
x ā 4dx =
1
6ln | x + 2 | +
5
6ln | x ā 4 | +c.
Example
Evaluate the integral
ā«2x3 ā 4x2 ā 15x + 5
x2 + 3x + 2dx .
Solution:Step 1: Do the polynomial long-division.Since the degree of the denominator g(x) is less than the degree of the numerator f (x),we do the polynomial long-division given on the right side. Then, we have
q(x) = (2x ā 10) +11x + 25
x2 + 3x + 2.
2x ā10x2 + 3x + 2
)2x3 ā4x2 ā15x +5ā(2x3 +6x2 +4x)
ā10x2 ā19x +5ā(ā10x2 ā30x ā20)
11x +25
Dr. M. Alghamdi MATH 106 January 2, 2019 32 / 47
Step 2: Factor the denominator g(x) into irreducible polynomials
g(x) = x2 + 3x + 2 = (x + 1)(x + 2).
Step 3: Find the partial fraction decomposition.
q(x) = (2xā10)+11x + 25
x2 + 3x + 2= (2xā10)+
A
x + 1+
B
x + 2= (2xā10)+
Ax + 2A + Bx + B
(x + 1)(x + 2).
We need to find the constants A and B.Coefficients of the numerators:
A + B = 11ā 1
2A + B = 25ā 2
By doing some calculation, we have A = 14 andB = ā3.
Illustration
ā2Ć 1 + 2
ā2Aā 2B = ā22
2A + B = 25
āāāāāāāāāāB = 3
Step 4: Integrate the result of step 3.ā«q(x) dx =
ā«(2x ā 10) dx +
ā«14
x + 1dx +
ā«ā3
x + 2dx
= x2 ā 10x + 14 ln | x + 1 | ā3 ln | x + 2 | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 33 / 47
Step 2: Factor the denominator g(x) into irreducible polynomials
g(x) = x2 + 3x + 2 = (x + 1)(x + 2).
Step 3: Find the partial fraction decomposition.
q(x) = (2xā10)+11x + 25
x2 + 3x + 2= (2xā10)+
A
x + 1+
B
x + 2= (2xā10)+
Ax + 2A + Bx + B
(x + 1)(x + 2).
We need to find the constants A and B.Coefficients of the numerators:
A + B = 11ā 1
2A + B = 25ā 2
By doing some calculation, we have A = 14 andB = ā3.
Illustration
ā2Ć 1 + 2
ā2Aā 2B = ā22
2A + B = 25
āāāāāāāāāāB = 3
Step 4: Integrate the result of step 3.ā«q(x) dx =
ā«(2x ā 10) dx +
ā«14
x + 1dx +
ā«ā3
x + 2dx
= x2 ā 10x + 14 ln | x + 1 | ā3 ln | x + 2 | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 33 / 47
Step 2: Factor the denominator g(x) into irreducible polynomials
g(x) = x2 + 3x + 2 = (x + 1)(x + 2).
Step 3: Find the partial fraction decomposition.
q(x) = (2xā10)+11x + 25
x2 + 3x + 2= (2xā10)+
A
x + 1+
B
x + 2= (2xā10)+
Ax + 2A + Bx + B
(x + 1)(x + 2).
We need to find the constants A and B.Coefficients of the numerators:
A + B = 11ā 1
2A + B = 25ā 2
By doing some calculation, we have A = 14 andB = ā3.
Illustration
ā2Ć 1 + 2
ā2Aā 2B = ā22
2A + B = 25
āāāāāāāāāāB = 3
Step 4: Integrate the result of step 3.ā«q(x) dx =
ā«(2x ā 10) dx +
ā«14
x + 1dx +
ā«ā3
x + 2dx
= x2 ā 10x + 14 ln | x + 1 | ā3 ln | x + 2 | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 33 / 47
Remark
1 The number of constants A,B,C , etc. is equal to the degree of the denominatorg(x). Therefore, in the case of repeated factors of the denominator, we have tocheck the number of the constants and the degree of g(x).
2 If the denominator g(x) contains irreducible quadratic factors, the numerators ofthe partial fractions should be polynomials of degree one (see step 3 on page 59).
Example
Evaluate the integral
ā«2x2 ā 25x ā 33
(x + 1)2(x ā 5)dx .
Solution:Steps 1 and 2 can be skipped in this example.Step 3: Find the partial fraction decomposition.Since the denominator g(x) has repeated factors, then
2x2 ā 25x ā 33
(x + 1)2(x ā 5)=
A
x + 1+
B
(x + 1)2+
C
x ā 5=
A(x2 ā 4x ā 5) + B(x ā 5) + C(x2 + 2x + 1)
(x + 1)2(x ā 5).
Dr. M. Alghamdi MATH 106 January 2, 2019 34 / 47
Remark
1 The number of constants A,B,C , etc. is equal to the degree of the denominatorg(x). Therefore, in the case of repeated factors of the denominator, we have tocheck the number of the constants and the degree of g(x).
2 If the denominator g(x) contains irreducible quadratic factors, the numerators ofthe partial fractions should be polynomials of degree one (see step 3 on page 59).
Example
Evaluate the integral
ā«2x2 ā 25x ā 33
(x + 1)2(x ā 5)dx .
Solution:Steps 1 and 2 can be skipped in this example.Step 3: Find the partial fraction decomposition.Since the denominator g(x) has repeated factors, then
2x2 ā 25x ā 33
(x + 1)2(x ā 5)=
A
x + 1+
B
(x + 1)2+
C
x ā 5=
A(x2 ā 4x ā 5) + B(x ā 5) + C(x2 + 2x + 1)
(x + 1)2(x ā 5).
Dr. M. Alghamdi MATH 106 January 2, 2019 34 / 47
Remark
1 The number of constants A,B,C , etc. is equal to the degree of the denominatorg(x). Therefore, in the case of repeated factors of the denominator, we have tocheck the number of the constants and the degree of g(x).
2 If the denominator g(x) contains irreducible quadratic factors, the numerators ofthe partial fractions should be polynomials of degree one (see step 3 on page 59).
Example
Evaluate the integral
ā«2x2 ā 25x ā 33
(x + 1)2(x ā 5)dx .
Solution:Steps 1 and 2 can be skipped in this example.
Step 3: Find the partial fraction decomposition.Since the denominator g(x) has repeated factors, then
2x2 ā 25x ā 33
(x + 1)2(x ā 5)=
A
x + 1+
B
(x + 1)2+
C
x ā 5=
A(x2 ā 4x ā 5) + B(x ā 5) + C(x2 + 2x + 1)
(x + 1)2(x ā 5).
Dr. M. Alghamdi MATH 106 January 2, 2019 34 / 47
Remark
1 The number of constants A,B,C , etc. is equal to the degree of the denominatorg(x). Therefore, in the case of repeated factors of the denominator, we have tocheck the number of the constants and the degree of g(x).
2 If the denominator g(x) contains irreducible quadratic factors, the numerators ofthe partial fractions should be polynomials of degree one (see step 3 on page 59).
Example
Evaluate the integral
ā«2x2 ā 25x ā 33
(x + 1)2(x ā 5)dx .
Solution:Steps 1 and 2 can be skipped in this example.Step 3: Find the partial fraction decomposition.Since the denominator g(x) has repeated factors, then
2x2 ā 25x ā 33
(x + 1)2(x ā 5)=
A
x + 1+
B
(x + 1)2+
C
x ā 5=
A(x2 ā 4x ā 5) + B(x ā 5) + C(x2 + 2x + 1)
(x + 1)2(x ā 5).
Dr. M. Alghamdi MATH 106 January 2, 2019 34 / 47
Coefficients of the numerators:
A + C = 2ā 1
ā4A + B + 2C = ā25ā 2
ā5Aā 5B + C = ā33ā 3
Illustration
5Ć 2 + 3 =ā25A + 11C = ā158 ā 425Ć 1 + 4 =36C = ā108ā C = ā3
By solving the system of equations, we have A = 5, B = 1 and C = ā3.Step 4: Integrate the result of step 3.
ā«2x2 ā 25x ā 33
(x + 1)2(x ā 5)dx =
ā«5
x + 1dx +
ā«1
(x + 1)2dx +
ā«ā3
x ā 5dx
= 5 ln | x + 1 | +
ā«(x + 1)ā2 dx ā 3 ln | x ā 5 |
= 5 ln | x + 1 | ā 1
(x + 1)ā 3 ln | x ā 5 | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 35 / 47
Example
Evaluate the integral
ā«x + 1
x(x2 + 1)dx .
Solution:Steps 1 and 2 can be skipped in this example.Step 3: Find the partial fraction decomposition.
x + 1
x(x2 + 1)=
A
x+
Bx + C
x2 + 1=
Ax2 + A + Bx2 + Cx
x(x2 + 1).
Coefficients of the numerators:A + B = 0ā 1
C = 1ā 2
A = 1ā 3
We have A = 1, B = ā1 and C = 1.Step 4: Integrate the result of step 3.ā«
x + 1
x(x2 + 1)dx =
ā«1
xdx +
ā«āx + 1
x2 + 1dx
= ln | x | āā«
x
x2 + 1dx +
ā«1
x2 + 1dx
= ln | x | ā1
2ln(x2 + 1) + tanā1 x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 36 / 47
Example
Evaluate the integral
ā«x + 1
x(x2 + 1)dx .
Solution:Steps 1 and 2 can be skipped in this example.
Step 3: Find the partial fraction decomposition.
x + 1
x(x2 + 1)=
A
x+
Bx + C
x2 + 1=
Ax2 + A + Bx2 + Cx
x(x2 + 1).
Coefficients of the numerators:A + B = 0ā 1
C = 1ā 2
A = 1ā 3
We have A = 1, B = ā1 and C = 1.Step 4: Integrate the result of step 3.ā«
x + 1
x(x2 + 1)dx =
ā«1
xdx +
ā«āx + 1
x2 + 1dx
= ln | x | āā«
x
x2 + 1dx +
ā«1
x2 + 1dx
= ln | x | ā1
2ln(x2 + 1) + tanā1 x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 36 / 47
Example
Evaluate the integral
ā«x + 1
x(x2 + 1)dx .
Solution:Steps 1 and 2 can be skipped in this example.Step 3: Find the partial fraction decomposition.
x + 1
x(x2 + 1)=
A
x+
Bx + C
x2 + 1=
Ax2 + A + Bx2 + Cx
x(x2 + 1).
Coefficients of the numerators:A + B = 0ā 1
C = 1ā 2
A = 1ā 3
We have A = 1, B = ā1 and C = 1.
Step 4: Integrate the result of step 3.ā«x + 1
x(x2 + 1)dx =
ā«1
xdx +
ā«āx + 1
x2 + 1dx
= ln | x | āā«
x
x2 + 1dx +
ā«1
x2 + 1dx
= ln | x | ā1
2ln(x2 + 1) + tanā1 x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 36 / 47
Example
Evaluate the integral
ā«x + 1
x(x2 + 1)dx .
Solution:Steps 1 and 2 can be skipped in this example.Step 3: Find the partial fraction decomposition.
x + 1
x(x2 + 1)=
A
x+
Bx + C
x2 + 1=
Ax2 + A + Bx2 + Cx
x(x2 + 1).
Coefficients of the numerators:A + B = 0ā 1
C = 1ā 2
A = 1ā 3
We have A = 1, B = ā1 and C = 1.Step 4: Integrate the result of step 3.ā«
x + 1
x(x2 + 1)dx =
ā«1
xdx +
ā«āx + 1
x2 + 1dx
= ln | x | āā«
x
x2 + 1dx +
ā«1
x2 + 1dx
= ln | x | ā1
2ln(x2 + 1) + tanā1 x + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 36 / 47
Integrals Involving Quadratic FormsIn this section, we provide a new technique for integrals that contain irreduciblequadratic expressions ax2 + bx + c where b 6= 0. This technique is completing squaremethod: a2 Ā± 2ab + b2 = (aĀ± b)2.
Notes:If a quadratic polynomial has real roots, it is called reducible; otherwise it is called
irreducible.For the expression ax2 + bx + c, if b2 ā 4ac < 0, then the quadratic expression isirreducible.
To complete the square, we need to find(
b2a
)2, then add and subtract it.
Example
For the quadratic expression x2 ā 6x + 13, we have a = 1, b = ā6 and c = 13. Sinceb2 ā 4ac = ā16 < 0, then the quadratic expression is irreducible. To complete thesquare, we find ( b
2a)2 = 9, then we add and substrate it as follows:
x2 ā 6x + 13 = x2 ā 6x + 9ļøø ļø·ļø· ļøø=(xā3)2
ā 9 + 13ļøø ļø·ļø· ļøø=4
Hence, x2 ā 6x + 13 = (x ā 3)2 + 4.
Dr. M. Alghamdi MATH 106 January 2, 2019 37 / 47
Example
Evaluate the integral
ā«1
x2 ā 6x + 13dx .
Solution:The quadratic expression x2 ā 6x + 13 is irreducible. By completing the square, we havefrom the previous exampleā«
1
x2 ā 6x + 13dx =
ā«1
(x ā 3)2 + 4dx .
Let u = x ā 3, then du = dx . By substitution,ā«1
u2 + 4du =
1
2tanā1 u
2+ c =
1
2tanā1 (x ā 3
2
)+ c.
Example
Evaluate the integral
ā«x
x2 ā 4x + 8dx .
Solution:For the quadratic expression x2 ā 4x + 8, we have b2 ā 4ac < 0. Therefore, thequadratic expression x2 ā 4x + 8 is irreducible. By completing the square, we obtain
Dr. M. Alghamdi MATH 106 January 2, 2019 38 / 47
Example
Evaluate the integral
ā«1
x2 ā 6x + 13dx .
Solution:The quadratic expression x2 ā 6x + 13 is irreducible. By completing the square, we havefrom the previous exampleā«
1
x2 ā 6x + 13dx =
ā«1
(x ā 3)2 + 4dx .
Let u = x ā 3, then du = dx . By substitution,ā«1
u2 + 4du =
1
2tanā1 u
2+ c =
1
2tanā1 (x ā 3
2
)+ c.
Example
Evaluate the integral
ā«x
x2 ā 4x + 8dx .
Solution:For the quadratic expression x2 ā 4x + 8, we have b2 ā 4ac < 0. Therefore, thequadratic expression x2 ā 4x + 8 is irreducible. By completing the square, we obtain
Dr. M. Alghamdi MATH 106 January 2, 2019 38 / 47
Example
Evaluate the integral
ā«1
x2 ā 6x + 13dx .
Solution:The quadratic expression x2 ā 6x + 13 is irreducible. By completing the square, we havefrom the previous exampleā«
1
x2 ā 6x + 13dx =
ā«1
(x ā 3)2 + 4dx .
Let u = x ā 3, then du = dx . By substitution,ā«1
u2 + 4du =
1
2tanā1 u
2+ c =
1
2tanā1 (x ā 3
2
)+ c.
Example
Evaluate the integral
ā«x
x2 ā 4x + 8dx .
Solution:For the quadratic expression x2 ā 4x + 8, we have b2 ā 4ac < 0. Therefore, thequadratic expression x2 ā 4x + 8 is irreducible. By completing the square, we obtain
Dr. M. Alghamdi MATH 106 January 2, 2019 38 / 47
x2 ā 4x + 8 = (x2 ā 4x + 4) + 8ā 4
= (x ā 2)2 + 4.
Hence ā«x
x2 ā 4x + 8dx =
ā«x
(x ā 2)2 + 4dx .
Let u = x ā 2, then du = dx . By substitution,ā«u + 2
u2 + 4du =
ā«u
u2 + 4du +
ā«2
u2 + 4du
=1
2ln | u2 + 4 | + tanā1 u
2
=1
2ln((x ā 2)2 + 4
)+ tanā1 (x ā 2
2
)+ c
=1
2ln(x2 ā 4x + 8
)+ tanā1 (x ā 2
2
)+ c.
Example
Evaluate the integral
ā«1ā
2x ā x2dx .
Dr. M. Alghamdi MATH 106 January 2, 2019 39 / 47
Solution:By completing the square, we have2x ā x2 = ā(x2 ā 2x) = ā(x2 ā 2x + 1ā 1) = 1ā (x ā 1)2. Henceā«
1ā2x ā x2
dx =
ā«1ā
1ā (x ā 1)2dx .
Let u = x ā 1, then du = dx . By substitution, the integral becomesā«1ā
1ā u2du = sinā1 u + c = sinā1 (x ā 1) + c.
Example
Evaluate the integral
ā« āx2 + 2x ā 1 dx .
Solution:By completing the square, we have x2 + 2x ā 1 = (x2 + 2x + 1)ā 1ā 1 = (x + 1)2 ā 2.Hence, ā« ā
x2 + 2x ā 1 dx =
ā« ā(x + 1)2 ā 2 dx .
Let u = x + 1, then du = dx . The integral becomes
ā« āu2 ā 2 du.
Dr. M. Alghamdi MATH 106 January 2, 2019 40 / 47
Solution:By completing the square, we have2x ā x2 = ā(x2 ā 2x) = ā(x2 ā 2x + 1ā 1) = 1ā (x ā 1)2. Henceā«
1ā2x ā x2
dx =
ā«1ā
1ā (x ā 1)2dx .
Let u = x ā 1, then du = dx . By substitution, the integral becomesā«1ā
1ā u2du = sinā1 u + c = sinā1 (x ā 1) + c.
Example
Evaluate the integral
ā« āx2 + 2x ā 1 dx .
Solution:By completing the square, we have x2 + 2x ā 1 = (x2 + 2x + 1)ā 1ā 1 = (x + 1)2 ā 2.Hence, ā« ā
x2 + 2x ā 1 dx =
ā« ā(x + 1)2 ā 2 dx .
Let u = x + 1, then du = dx . The integral becomes
ā« āu2 ā 2 du.
Dr. M. Alghamdi MATH 106 January 2, 2019 40 / 47
Solution:By completing the square, we have2x ā x2 = ā(x2 ā 2x) = ā(x2 ā 2x + 1ā 1) = 1ā (x ā 1)2. Henceā«
1ā2x ā x2
dx =
ā«1ā
1ā (x ā 1)2dx .
Let u = x ā 1, then du = dx . By substitution, the integral becomesā«1ā
1ā u2du = sinā1 u + c = sinā1 (x ā 1) + c.
Example
Evaluate the integral
ā« āx2 + 2x ā 1 dx .
Solution:By completing the square, we have x2 + 2x ā 1 = (x2 + 2x + 1)ā 1ā 1 = (x + 1)2 ā 2.Hence, ā« ā
x2 + 2x ā 1 dx =
ā« ā(x + 1)2 ā 2 dx .
Let u = x + 1, then du = dx . The integral becomes
ā« āu2 ā 2 du.
Dr. M. Alghamdi MATH 106 January 2, 2019 40 / 47
Use the trigonometric substitutions, in par-ticular let
u =ā
2 sec Īø ā du =ā
2 sec Īø tan Īø dĪø
where Īø ā [0, Ļ/2) āŖ [Ļ, 3Ļ/2). By substitu-tion, we have
2
ā«tan2 Īø sec Īø dĪø = 2
ā«(sec3 Īø ā sec Īø) dĪø.
ā2
āu2 ā 2
u
Īø
From Example 8, we have
2
ā«(sec3 Īø ā sec Īø) dĪø = sec Īø tan Īø ā ln
ā£ā£ sec Īø + tan Īøā£ā£+ c.
By returning to the variable u and then to x ,ā« āu2 ā 2 du =
uāu2 ā 2
2āln
ā£ā£ā£u +āu2 ā 2ā2
ā£ā£ā£+c =(x + 1)
ā(x + 1)2 ā 2
2āln
ā£ā£ā£x + 1 +ā
(x + 1)2 ā 2ā2
ā£ā£ā£+c.
Dr. M. Alghamdi MATH 106 January 2, 2019 41 / 47
Miscellaneous Substitutions(A) Fractional Functions in sin x and cos xThe integrals that consist of rational expressions in sin x and cos x are treated by using the
substitution u = tan (x/2), āĻ < x < Ļ. This implies that du = sec2 (x/2)2
dx and since
sec2 x = tan2 x + 1, then du = u2+12
dx . Also,
sin x = sin 2(x
2)
= 2 sinx
2cos
x
2
= 2sin x
2
cos x2
cosx
2cos
x
2
= 2 tanx
2cos2 x
2
=2 tan x
2
sec2 x2
=2u
u2 + 1.
(multiply and divide bycos x
2)
(cos x = 1sec x
)
For cos x , we have cos x = cos 2( x2
) = cos2 x2ā sin2 x
2.
We can find that cos x2
= 1āu2+1
and sin x2
=
uāu2+1
.(use the identitiessec2 x
2= tan2 x
2+ 1
and cos2 x2
+ sin2 x2
= 1)
This implies cos x = 1āu2
1+u2 .Dr. M. Alghamdi MATH 106 January 2, 2019 42 / 47
TheoremFor an integral that contains a rational expression in sin x and cos x , we assume
sin x =2u
1 + u2, and cos x =
1ā u2
1 + u2.
to produce a rational expression in u where u = tan (x/2), and du = 1+u2
2dx .
ExampleEvaluate the integral.
1
ā«1
1 + sin xdx 2
ā«1
2 + cos xdx
3 ā«1
1 + sin x + cos xdx
Solution:1) Let u = tan x
2, then du = 1+u2
2dx and sin x = 2u
1+u2 . By substituting that into the integral,
we haveā«1
1 + 2u1+u2
.2
1 + u2du = 2
ā«1
u2 + 2u + 1du = 2
ā«(u + 1)ā2 du =
ā2
u + 1+ c
=ā2
tan x/2 + 1+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 43 / 47
TheoremFor an integral that contains a rational expression in sin x and cos x , we assume
sin x =2u
1 + u2, and cos x =
1ā u2
1 + u2.
to produce a rational expression in u where u = tan (x/2), and du = 1+u2
2dx .
ExampleEvaluate the integral.
1
ā«1
1 + sin xdx 2
ā«1
2 + cos xdx
3 ā«1
1 + sin x + cos xdx
Solution:1) Let u = tan x
2, then du = 1+u2
2dx and sin x = 2u
1+u2 . By substituting that into the integral,
we haveā«1
1 + 2u1+u2
.2
1 + u2du = 2
ā«1
u2 + 2u + 1du = 2
ā«(u + 1)ā2 du =
ā2
u + 1+ c
=ā2
tan x/2 + 1+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 43 / 47
2) Let u = tan x2
, then du = 1+u2
2dx and cos x = 1āu2
1+u2 . By substitution, we haveā«1
2 + 1āu2
1+u2
.2
1 + u2du = 2
ā«1
u2 + 3du
=2ā
3tanā1 u
ā3
+ c
=2ā
3tanā1
( tan x/2ā
3
)+ c.
3) Let u = tan x2
, this implies du = 1+u2
2dx , sin x = 2u
1+u2 and cos x = 1āu2
1+u2 . By substitution,
we have ā«1
1 + 2u1+u2 + 1āu2
1+u2
.2
1 + u2du =
ā«2
2 + 2udu
=
ā«1
1 + udu
= ln | 1 + u | +c
= lnā£ā£ā£1 + tan
x
2
ā£ā£ā£+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 44 / 47
2) Let u = tan x2
, then du = 1+u2
2dx and cos x = 1āu2
1+u2 . By substitution, we haveā«1
2 + 1āu2
1+u2
.2
1 + u2du = 2
ā«1
u2 + 3du
=2ā
3tanā1 u
ā3
+ c
=2ā
3tanā1
( tan x/2ā
3
)+ c.
3) Let u = tan x2
, this implies du = 1+u2
2dx , sin x = 2u
1+u2 and cos x = 1āu2
1+u2 . By substitution,
we have ā«1
1 + 2u1+u2 + 1āu2
1+u2
.2
1 + u2du =
ā«2
2 + 2udu
=
ā«1
1 + udu
= ln | 1 + u | +c
= lnā£ā£ā£1 + tan
x
2
ā£ā£ā£+ c.
Dr. M. Alghamdi MATH 106 January 2, 2019 44 / 47
Integrals of Fractional PowersIn the case of an integrand that consists of fractional powers, it is better to use the
substitution u = x1n where n is the least common multiple of the denominators of the
powers. In the following, we provide an example.
Example
Evaluate the integral
ā«1ā
x + 4āx
dx .
Solution:Let u = x
14 , we find x = u4 and dx = 4u3du. Therefore, x
12 = (x
14 )2 = u2.
By substitution, we haveā«1
u2 + u4u3 du = 4
ā«u2
u + 1du
= 4
ā«(u ā 1) du + 4
ā«1
1 + udu
= 2u2 ā 4u + 4 ln | u + 1 | +c
= 2āx ā 4 4
āx + 4 ln | 4
āx + 1 | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 45 / 47
Integrals of Fractional PowersIn the case of an integrand that consists of fractional powers, it is better to use the
substitution u = x1n where n is the least common multiple of the denominators of the
powers. In the following, we provide an example.
Example
Evaluate the integral
ā«1ā
x + 4āx
dx .
Solution:Let u = x
14 , we find x = u4 and dx = 4u3du. Therefore, x
12 = (x
14 )2 = u2.
By substitution, we haveā«1
u2 + u4u3 du = 4
ā«u2
u + 1du
= 4
ā«(u ā 1) du + 4
ā«1
1 + udu
= 2u2 ā 4u + 4 ln | u + 1 | +c
= 2āx ā 4 4
āx + 4 ln | 4
āx + 1 | +c.
Dr. M. Alghamdi MATH 106 January 2, 2019 45 / 47
Integrals of Form nā
f (x)
If the integrand is of from nā
f (x), it is useful to assume u = nā
f (x). This case differs
from that given in the substitution method in Chapter ?? i.e., nā
f (x) f ā²(x) and thedifference lies on the existence of the derivative of f (x).
Example
Evaluate the integral
ā« āex + 1 dx .
Solution:Let u =
āex + 1, we obtain du = ex
2ā
ex+1dx and u2 = ex + 1. By substitution, we haveā«
2u2
u2 ā 1du =
ā«2 du + 2
ā«1
u2 ā 1du
= 2u +
ā«1
u ā 1du +
ā«1
u + 1du
= 2u + ln | u ā 1 | ā ln | u + 1 | +c
= 2āex + 1 + ln(
āex + 1ā 1)ā ln(
āex + 1 + 1) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 46 / 47
Integrals of Form nā
f (x)
If the integrand is of from nā
f (x), it is useful to assume u = nā
f (x). This case differs
from that given in the substitution method in Chapter ?? i.e., nā
f (x) f ā²(x) and thedifference lies on the existence of the derivative of f (x).
Example
Evaluate the integral
ā« āex + 1 dx .
Solution:Let u =
āex + 1, we obtain du = ex
2āex+1
dx and u2 = ex + 1. By substitution, we haveā«2u2
u2 ā 1du =
ā«2 du + 2
ā«1
u2 ā 1du
= 2u +
ā«1
u ā 1du +
ā«1
u + 1du
= 2u + ln | u ā 1 | ā ln | u + 1 | +c
= 2āex + 1 + ln(
āex + 1ā 1)ā ln(
āex + 1 + 1) + c.
Dr. M. Alghamdi MATH 106 January 2, 2019 46 / 47
Dr. M. Alghamdi MATH 106 January 2, 2019 47 / 47