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1 | 3918-3921 | 2
2
2
1
5
4
3
x
x
+x
+
∫
17 2
3
4
0 (2sec
2)
x
x
dx
π
+
+
∫
18 2
2
0 (sin
–cos
)
2
2
x
x dx
π∫
19 2
2
0
6
3
4
x
dx
x
++
∫
20 |
1 | 3919-3922 | 2
3
4
0 (2sec
2)
x
x
dx
π
+
+
∫
18 2
2
0 (sin
–cos
)
2
2
x
x dx
π∫
19 2
2
0
6
3
4
x
dx
x
++
∫
20 1
0(
sin
)
4
x
x
x e
dx
π
+
∫
Choose the correct answer in Exercises 21 and 22 |
1 | 3920-3923 | 2
2
0 (sin
–cos
)
2
2
x
x dx
π∫
19 2
2
0
6
3
4
x
dx
x
++
∫
20 1
0(
sin
)
4
x
x
x e
dx
π
+
∫
Choose the correct answer in Exercises 21 and 22 21 |
1 | 3921-3924 | 2
2
0
6
3
4
x
dx
x
++
∫
20 1
0(
sin
)
4
x
x
x e
dx
π
+
∫
Choose the correct answer in Exercises 21 and 22 21 3
2
1 1
dx
+x
∫
equals
(A)
π3
(B)
π32
(C)
π6
(D) 12
π
22 |
1 | 3922-3925 | 1
0(
sin
)
4
x
x
x e
dx
π
+
∫
Choose the correct answer in Exercises 21 and 22 21 3
2
1 1
dx
+x
∫
equals
(A)
π3
(B)
π32
(C)
π6
(D) 12
π
22 2
3
2
0 4
9
dx
x
+
∫
equals
(A) 6
π
(B) 12
π
(C) 24
π
(D) 4
π
7 |
1 | 3923-3926 | 21 3
2
1 1
dx
+x
∫
equals
(A)
π3
(B)
π32
(C)
π6
(D) 12
π
22 2
3
2
0 4
9
dx
x
+
∫
equals
(A) 6
π
(B) 12
π
(C) 24
π
(D) 4
π
7 9 Evaluation of Definite Integrals by Substitution
In the previous sections, we have discussed several methods for finding the indefinite
integral |
1 | 3924-3927 | 3
2
1 1
dx
+x
∫
equals
(A)
π3
(B)
π32
(C)
π6
(D) 12
π
22 2
3
2
0 4
9
dx
x
+
∫
equals
(A) 6
π
(B) 12
π
(C) 24
π
(D) 4
π
7 9 Evaluation of Definite Integrals by Substitution
In the previous sections, we have discussed several methods for finding the indefinite
integral One of the important methods for finding the indefinite integral is the method
of substitution |
1 | 3925-3928 | 2
3
2
0 4
9
dx
x
+
∫
equals
(A) 6
π
(B) 12
π
(C) 24
π
(D) 4
π
7 9 Evaluation of Definite Integrals by Substitution
In the previous sections, we have discussed several methods for finding the indefinite
integral One of the important methods for finding the indefinite integral is the method
of substitution INTEGRALS 339
To evaluate
( )
b
∫a f x dx
, by substitution, the steps could be as follows:
1 |
1 | 3926-3929 | 9 Evaluation of Definite Integrals by Substitution
In the previous sections, we have discussed several methods for finding the indefinite
integral One of the important methods for finding the indefinite integral is the method
of substitution INTEGRALS 339
To evaluate
( )
b
∫a f x dx
, by substitution, the steps could be as follows:
1 Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce
the given integral to a known form |
1 | 3927-3930 | One of the important methods for finding the indefinite integral is the method
of substitution INTEGRALS 339
To evaluate
( )
b
∫a f x dx
, by substitution, the steps could be as follows:
1 Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce
the given integral to a known form 2 |
1 | 3928-3931 | INTEGRALS 339
To evaluate
( )
b
∫a f x dx
, by substitution, the steps could be as follows:
1 Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce
the given integral to a known form 2 Integrate the new integrand with respect to the new variable without mentioning
the constant of integration |
1 | 3929-3932 | Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce
the given integral to a known form 2 Integrate the new integrand with respect to the new variable without mentioning
the constant of integration 3 |
1 | 3930-3933 | 2 Integrate the new integrand with respect to the new variable without mentioning
the constant of integration 3 Resubstitute for the new variable and write the answer in terms of the original
variable |
1 | 3931-3934 | Integrate the new integrand with respect to the new variable without mentioning
the constant of integration 3 Resubstitute for the new variable and write the answer in terms of the original
variable 4 |
1 | 3932-3935 | 3 Resubstitute for the new variable and write the answer in terms of the original
variable 4 Find the values of answers obtained in (3) at the given limits of integral and find
the difference of the values at the upper and lower limits |
1 | 3933-3936 | Resubstitute for the new variable and write the answer in terms of the original
variable 4 Find the values of answers obtained in (3) at the given limits of integral and find
the difference of the values at the upper and lower limits �Note In order to quicken this method, we can proceed as follows: After
performing steps 1, and 2, there is no need of step 3 |
1 | 3934-3937 | 4 Find the values of answers obtained in (3) at the given limits of integral and find
the difference of the values at the upper and lower limits �Note In order to quicken this method, we can proceed as follows: After
performing steps 1, and 2, there is no need of step 3 Here, the integral will be kept
in the new variable itself, and the limits of the integral will accordingly be changed,
so that we can perform the last step |
1 | 3935-3938 | Find the values of answers obtained in (3) at the given limits of integral and find
the difference of the values at the upper and lower limits �Note In order to quicken this method, we can proceed as follows: After
performing steps 1, and 2, there is no need of step 3 Here, the integral will be kept
in the new variable itself, and the limits of the integral will accordingly be changed,
so that we can perform the last step Let us illustrate this by examples |
1 | 3936-3939 | �Note In order to quicken this method, we can proceed as follows: After
performing steps 1, and 2, there is no need of step 3 Here, the integral will be kept
in the new variable itself, and the limits of the integral will accordingly be changed,
so that we can perform the last step Let us illustrate this by examples Example 28 Evaluate
1
4
5
15
1
x
x
dx
−
+
∫ |
1 | 3937-3940 | Here, the integral will be kept
in the new variable itself, and the limits of the integral will accordingly be changed,
so that we can perform the last step Let us illustrate this by examples Example 28 Evaluate
1
4
5
15
1
x
x
dx
−
+
∫ Solution Put t = x5 + 1, then dt = 5x4 dx |
1 | 3938-3941 | Let us illustrate this by examples Example 28 Evaluate
1
4
5
15
1
x
x
dx
−
+
∫ Solution Put t = x5 + 1, then dt = 5x4 dx Therefore,
4
5
5
1
x
x
dx
+
∫
=
∫t dt
=
223
3
t =
3
5
2
2 (
1)
3
x +
Hence,
1
4
5
15
1
x
x
dx
−
+
∫
=
1
3
5
2
– 1
2 (
1)
3
x
+
=
(
)
3
3
5
5
2
2
2 (1
1) – (–1)
1
3
+
+
=
3
3
2
2
2 2
0
3
−
=
2
4
2
(2
2)
3
3
=
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits |
1 | 3939-3942 | Example 28 Evaluate
1
4
5
15
1
x
x
dx
−
+
∫ Solution Put t = x5 + 1, then dt = 5x4 dx Therefore,
4
5
5
1
x
x
dx
+
∫
=
∫t dt
=
223
3
t =
3
5
2
2 (
1)
3
x +
Hence,
1
4
5
15
1
x
x
dx
−
+
∫
=
1
3
5
2
– 1
2 (
1)
3
x
+
=
(
)
3
3
5
5
2
2
2 (1
1) – (–1)
1
3
+
+
=
3
3
2
2
2 2
0
3
−
=
2
4
2
(2
2)
3
3
=
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits 340
MATHEMATICS
Let
t = x5 + 1 |
1 | 3940-3943 | Solution Put t = x5 + 1, then dt = 5x4 dx Therefore,
4
5
5
1
x
x
dx
+
∫
=
∫t dt
=
223
3
t =
3
5
2
2 (
1)
3
x +
Hence,
1
4
5
15
1
x
x
dx
−
+
∫
=
1
3
5
2
– 1
2 (
1)
3
x
+
=
(
)
3
3
5
5
2
2
2 (1
1) – (–1)
1
3
+
+
=
3
3
2
2
2 2
0
3
−
=
2
4
2
(2
2)
3
3
=
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits 340
MATHEMATICS
Let
t = x5 + 1 Then dt = 5 x4 dx |
1 | 3941-3944 | Therefore,
4
5
5
1
x
x
dx
+
∫
=
∫t dt
=
223
3
t =
3
5
2
2 (
1)
3
x +
Hence,
1
4
5
15
1
x
x
dx
−
+
∫
=
1
3
5
2
– 1
2 (
1)
3
x
+
=
(
)
3
3
5
5
2
2
2 (1
1) – (–1)
1
3
+
+
=
3
3
2
2
2 2
0
3
−
=
2
4
2
(2
2)
3
3
=
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits 340
MATHEMATICS
Let
t = x5 + 1 Then dt = 5 x4 dx Note that, when
x = – 1, t = 0 and when x = 1, t = 2
Thus, as x varies from – 1 to 1, t varies from 0 to 2
Therefore
1
4
5
15
1
x
x
dx
−
+
∫
=
2
0
t dt
∫
=
2
3
3
3
2
2
2
0
2
2 2 – 0
3
3
t
=
=
2
4
2
(2
2)
3
3
=
Example 29 Evaluate
– 1
1
2
0
tan
1
x dx
+x
∫
Solution Let t = tan – 1x, then
12
1
dt
dx
x
=
+ |
1 | 3942-3945 | 340
MATHEMATICS
Let
t = x5 + 1 Then dt = 5 x4 dx Note that, when
x = – 1, t = 0 and when x = 1, t = 2
Thus, as x varies from – 1 to 1, t varies from 0 to 2
Therefore
1
4
5
15
1
x
x
dx
−
+
∫
=
2
0
t dt
∫
=
2
3
3
3
2
2
2
0
2
2 2 – 0
3
3
t
=
=
2
4
2
(2
2)
3
3
=
Example 29 Evaluate
– 1
1
2
0
tan
1
x dx
+x
∫
Solution Let t = tan – 1x, then
12
1
dt
dx
x
=
+ The new limits are, when x = 0, t = 0 and
when x = 1,
4
t
=π |
1 | 3943-3946 | Then dt = 5 x4 dx Note that, when
x = – 1, t = 0 and when x = 1, t = 2
Thus, as x varies from – 1 to 1, t varies from 0 to 2
Therefore
1
4
5
15
1
x
x
dx
−
+
∫
=
2
0
t dt
∫
=
2
3
3
3
2
2
2
0
2
2 2 – 0
3
3
t
=
=
2
4
2
(2
2)
3
3
=
Example 29 Evaluate
– 1
1
2
0
tan
1
x dx
+x
∫
Solution Let t = tan – 1x, then
12
1
dt
dx
x
=
+ The new limits are, when x = 0, t = 0 and
when x = 1,
4
t
=π Thus, as x varies from 0 to 1, t varies from 0 to
4
π |
1 | 3944-3947 | Note that, when
x = – 1, t = 0 and when x = 1, t = 2
Thus, as x varies from – 1 to 1, t varies from 0 to 2
Therefore
1
4
5
15
1
x
x
dx
−
+
∫
=
2
0
t dt
∫
=
2
3
3
3
2
2
2
0
2
2 2 – 0
3
3
t
=
=
2
4
2
(2
2)
3
3
=
Example 29 Evaluate
– 1
1
2
0
tan
1
x dx
+x
∫
Solution Let t = tan – 1x, then
12
1
dt
dx
x
=
+ The new limits are, when x = 0, t = 0 and
when x = 1,
4
t
=π Thus, as x varies from 0 to 1, t varies from 0 to
4
π Therefore
–1
1
2
0
tan
1
x dx
+x
∫
=
2
4
4
0
t20
t dt
π
π
∫
=
2
2
1
– 0
2 16
32
π
=π
EXERCISE 7 |
1 | 3945-3948 | The new limits are, when x = 0, t = 0 and
when x = 1,
4
t
=π Thus, as x varies from 0 to 1, t varies from 0 to
4
π Therefore
–1
1
2
0
tan
1
x dx
+x
∫
=
2
4
4
0
t20
t dt
π
π
∫
=
2
2
1
– 0
2 16
32
π
=π
EXERCISE 7 10
Evaluate the integrals in Exercises 1 to 8 using substitution |
1 | 3946-3949 | Thus, as x varies from 0 to 1, t varies from 0 to
4
π Therefore
–1
1
2
0
tan
1
x dx
+x
∫
=
2
4
4
0
t20
t dt
π
π
∫
=
2
2
1
– 0
2 16
32
π
=π
EXERCISE 7 10
Evaluate the integrals in Exercises 1 to 8 using substitution 1 |
1 | 3947-3950 | Therefore
–1
1
2
0
tan
1
x dx
+x
∫
=
2
4
4
0
t20
t dt
π
π
∫
=
2
2
1
– 0
2 16
32
π
=π
EXERCISE 7 10
Evaluate the integrals in Exercises 1 to 8 using substitution 1 1
2
0
1
x
dx
∫x +
2 |
1 | 3948-3951 | 10
Evaluate the integrals in Exercises 1 to 8 using substitution 1 1
2
0
1
x
dx
∫x +
2 5
02
sin
cos
d
π
φ
φ
φ
∫
3 |
1 | 3949-3952 | 1 1
2
0
1
x
dx
∫x +
2 5
02
sin
cos
d
π
φ
φ
φ
∫
3 1
– 1
2
0
2
sin
1
x
dx
x
+
∫
4 |
1 | 3950-3953 | 1
2
0
1
x
dx
∫x +
2 5
02
sin
cos
d
π
φ
φ
φ
∫
3 1
– 1
2
0
2
sin
1
x
dx
x
+
∫
4 2
0
2
x
x +
∫
(Put x + 2 = t2)
5 |
1 | 3951-3954 | 5
02
sin
cos
d
π
φ
φ
φ
∫
3 1
– 1
2
0
2
sin
1
x
dx
x
+
∫
4 2
0
2
x
x +
∫
(Put x + 2 = t2)
5 2
2
0
1sin
cos
x
dx
x
π
+
∫
6 |
1 | 3952-3955 | 1
– 1
2
0
2
sin
1
x
dx
x
+
∫
4 2
0
2
x
x +
∫
(Put x + 2 = t2)
5 2
2
0
1sin
cos
x
dx
x
π
+
∫
6 2
2
0
4 –
dx
x
x
+
∫
7 |
1 | 3953-3956 | 2
0
2
x
x +
∫
(Put x + 2 = t2)
5 2
2
0
1sin
cos
x
dx
x
π
+
∫
6 2
2
0
4 –
dx
x
x
+
∫
7 1
2
1
2
5
dx
x
x
−
+
+
∫
8 |
1 | 3954-3957 | 2
2
0
1sin
cos
x
dx
x
π
+
∫
6 2
2
0
4 –
dx
x
x
+
∫
7 1
2
1
2
5
dx
x
x
−
+
+
∫
8 2
2
2
1
1
–1
2
x
e dx
x
x
∫
Choose the correct answer in Exercises 9 and 10 |
1 | 3955-3958 | 2
2
0
4 –
dx
x
x
+
∫
7 1
2
1
2
5
dx
x
x
−
+
+
∫
8 2
2
2
1
1
–1
2
x
e dx
x
x
∫
Choose the correct answer in Exercises 9 and 10 9 |
1 | 3956-3959 | 1
2
1
2
5
dx
x
x
−
+
+
∫
8 2
2
2
1
1
–1
2
x
e dx
x
x
∫
Choose the correct answer in Exercises 9 and 10 9 The value of the integral
3 31
1
1
4
3
(
)
x
x
dx
−x
∫
is
(A) 6
(B) 0
(C) 3
(D) 4
10 |
1 | 3957-3960 | 2
2
2
1
1
–1
2
x
e dx
x
x
∫
Choose the correct answer in Exercises 9 and 10 9 The value of the integral
3 31
1
1
4
3
(
)
x
x
dx
−x
∫
is
(A) 6
(B) 0
(C) 3
(D) 4
10 If f (x) =
0
x tsin
t dt
∫
, then f′(x) is
(A) cosx + x sin x
(B) x sinx
(C) x cosx
(D) sinx + x cosx
INTEGRALS 341
7 |
1 | 3958-3961 | 9 The value of the integral
3 31
1
1
4
3
(
)
x
x
dx
−x
∫
is
(A) 6
(B) 0
(C) 3
(D) 4
10 If f (x) =
0
x tsin
t dt
∫
, then f′(x) is
(A) cosx + x sin x
(B) x sinx
(C) x cosx
(D) sinx + x cosx
INTEGRALS 341
7 10 Some Properties of Definite Integrals
We list below some important properties of definite integrals |
1 | 3959-3962 | The value of the integral
3 31
1
1
4
3
(
)
x
x
dx
−x
∫
is
(A) 6
(B) 0
(C) 3
(D) 4
10 If f (x) =
0
x tsin
t dt
∫
, then f′(x) is
(A) cosx + x sin x
(B) x sinx
(C) x cosx
(D) sinx + x cosx
INTEGRALS 341
7 10 Some Properties of Definite Integrals
We list below some important properties of definite integrals These will be useful in
evaluating the definite integrals more easily |
1 | 3960-3963 | If f (x) =
0
x tsin
t dt
∫
, then f′(x) is
(A) cosx + x sin x
(B) x sinx
(C) x cosx
(D) sinx + x cosx
INTEGRALS 341
7 10 Some Properties of Definite Integrals
We list below some important properties of definite integrals These will be useful in
evaluating the definite integrals more easily P0 :
( )
( )
b
b
a
a
f x dx
f t dt
=
∫
∫
P1 :
( )
–
( )
b
a
a
b
f x dx
f x dx
=
∫
∫ |
1 | 3961-3964 | 10 Some Properties of Definite Integrals
We list below some important properties of definite integrals These will be useful in
evaluating the definite integrals more easily P0 :
( )
( )
b
b
a
a
f x dx
f t dt
=
∫
∫
P1 :
( )
–
( )
b
a
a
b
f x dx
f x dx
=
∫
∫ In particular,
( )
0
a
a f x dx =
∫
P2 :
( )
( )
( )
b
c
b
a
a
c
f x dx
f x dx
f x dx
=
+
∫
∫
∫
P3 :
( )
(
)
b
b
a
a
f x dx
f a
b
x dx
=
+
−
∫
∫
P4 :
0
0
( )
(
)
a
a
f x dx
f a
x dx
=
−
∫
∫
(Note that P4 is a particular case of P3)
P5 :
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫
P6 :
2
0
0
( )
2
( )
, if
(2
)
( )
a
a
f x dx
f x dx
f
a
x
f x
=
−
=
∫
∫
and
0 if f (2a – x) = – f (x)
P7 :
(i)
0
( )
2
( )
a
a
a f x dx
f x dx
−
=
∫
∫
, if f is an even function, i |
1 | 3962-3965 | These will be useful in
evaluating the definite integrals more easily P0 :
( )
( )
b
b
a
a
f x dx
f t dt
=
∫
∫
P1 :
( )
–
( )
b
a
a
b
f x dx
f x dx
=
∫
∫ In particular,
( )
0
a
a f x dx =
∫
P2 :
( )
( )
( )
b
c
b
a
a
c
f x dx
f x dx
f x dx
=
+
∫
∫
∫
P3 :
( )
(
)
b
b
a
a
f x dx
f a
b
x dx
=
+
−
∫
∫
P4 :
0
0
( )
(
)
a
a
f x dx
f a
x dx
=
−
∫
∫
(Note that P4 is a particular case of P3)
P5 :
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫
P6 :
2
0
0
( )
2
( )
, if
(2
)
( )
a
a
f x dx
f x dx
f
a
x
f x
=
−
=
∫
∫
and
0 if f (2a – x) = – f (x)
P7 :
(i)
0
( )
2
( )
a
a
a f x dx
f x dx
−
=
∫
∫
, if f is an even function, i e |
1 | 3963-3966 | P0 :
( )
( )
b
b
a
a
f x dx
f t dt
=
∫
∫
P1 :
( )
–
( )
b
a
a
b
f x dx
f x dx
=
∫
∫ In particular,
( )
0
a
a f x dx =
∫
P2 :
( )
( )
( )
b
c
b
a
a
c
f x dx
f x dx
f x dx
=
+
∫
∫
∫
P3 :
( )
(
)
b
b
a
a
f x dx
f a
b
x dx
=
+
−
∫
∫
P4 :
0
0
( )
(
)
a
a
f x dx
f a
x dx
=
−
∫
∫
(Note that P4 is a particular case of P3)
P5 :
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫
P6 :
2
0
0
( )
2
( )
, if
(2
)
( )
a
a
f x dx
f x dx
f
a
x
f x
=
−
=
∫
∫
and
0 if f (2a – x) = – f (x)
P7 :
(i)
0
( )
2
( )
a
a
a f x dx
f x dx
−
=
∫
∫
, if f is an even function, i e , if f (– x) = f (x) |
1 | 3964-3967 | In particular,
( )
0
a
a f x dx =
∫
P2 :
( )
( )
( )
b
c
b
a
a
c
f x dx
f x dx
f x dx
=
+
∫
∫
∫
P3 :
( )
(
)
b
b
a
a
f x dx
f a
b
x dx
=
+
−
∫
∫
P4 :
0
0
( )
(
)
a
a
f x dx
f a
x dx
=
−
∫
∫
(Note that P4 is a particular case of P3)
P5 :
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫
P6 :
2
0
0
( )
2
( )
, if
(2
)
( )
a
a
f x dx
f x dx
f
a
x
f x
=
−
=
∫
∫
and
0 if f (2a – x) = – f (x)
P7 :
(i)
0
( )
2
( )
a
a
a f x dx
f x dx
−
=
∫
∫
, if f is an even function, i e , if f (– x) = f (x) (ii)
( )
0
a
−a f x dx
=
∫
, if f is an odd function, i |
1 | 3965-3968 | e , if f (– x) = f (x) (ii)
( )
0
a
−a f x dx
=
∫
, if f is an odd function, i e |
1 | 3966-3969 | , if f (– x) = f (x) (ii)
( )
0
a
−a f x dx
=
∫
, if f is an odd function, i e , if f (– x) = – f (x) |
1 | 3967-3970 | (ii)
( )
0
a
−a f x dx
=
∫
, if f is an odd function, i e , if f (– x) = – f (x) We give the proofs of these properties one by one |
1 | 3968-3971 | e , if f (– x) = – f (x) We give the proofs of these properties one by one Proof of P0 It follows directly by making the substitution x = t |
1 | 3969-3972 | , if f (– x) = – f (x) We give the proofs of these properties one by one Proof of P0 It follows directly by making the substitution x = t Proof of P1 Let F be anti derivative of f |
1 | 3970-3973 | We give the proofs of these properties one by one Proof of P0 It follows directly by making the substitution x = t Proof of P1 Let F be anti derivative of f Then, by the second fundamental theorem of
calculus, we have
( )
F ( ) – F ( )
–[F ( )
F ( )]
( )
b
a
a
b
f x dx
b
a
a
b
f x dx
=
=
−
= −
∫
∫
Here, we observe that, if a = b, then
( )
0
a
∫a f x dx = |
1 | 3971-3974 | Proof of P0 It follows directly by making the substitution x = t Proof of P1 Let F be anti derivative of f Then, by the second fundamental theorem of
calculus, we have
( )
F ( ) – F ( )
–[F ( )
F ( )]
( )
b
a
a
b
f x dx
b
a
a
b
f x dx
=
=
−
= −
∫
∫
Here, we observe that, if a = b, then
( )
0
a
∫a f x dx = Proof of P2 Let F be anti derivative of f |
1 | 3972-3975 | Proof of P1 Let F be anti derivative of f Then, by the second fundamental theorem of
calculus, we have
( )
F ( ) – F ( )
–[F ( )
F ( )]
( )
b
a
a
b
f x dx
b
a
a
b
f x dx
=
=
−
= −
∫
∫
Here, we observe that, if a = b, then
( )
0
a
∫a f x dx = Proof of P2 Let F be anti derivative of f Then
( )
b
∫a f x dx
= F(b) – F(a) |
1 | 3973-3976 | Then, by the second fundamental theorem of
calculus, we have
( )
F ( ) – F ( )
–[F ( )
F ( )]
( )
b
a
a
b
f x dx
b
a
a
b
f x dx
=
=
−
= −
∫
∫
Here, we observe that, if a = b, then
( )
0
a
∫a f x dx = Proof of P2 Let F be anti derivative of f Then
( )
b
∫a f x dx
= F(b) – F(a) (1)
( )
c
∫a f x dx
= F(c) – F(a) |
1 | 3974-3977 | Proof of P2 Let F be anti derivative of f Then
( )
b
∫a f x dx
= F(b) – F(a) (1)
( )
c
∫a f x dx
= F(c) – F(a) (2)
and
( )
b
∫c f x dx
= F(b) – F(c) |
1 | 3975-3978 | Then
( )
b
∫a f x dx
= F(b) – F(a) (1)
( )
c
∫a f x dx
= F(c) – F(a) (2)
and
( )
b
∫c f x dx
= F(b) – F(c) (3)
342
MATHEMATICS
Adding (2) and (3), we get
( )
( )
F( ) – F( )
( )
c
b
b
a
c
a
f x dx
f x dx
b
a
f x dx
+
=
=
∫
∫
∫
This proves the property P2 |
1 | 3976-3979 | (1)
( )
c
∫a f x dx
= F(c) – F(a) (2)
and
( )
b
∫c f x dx
= F(b) – F(c) (3)
342
MATHEMATICS
Adding (2) and (3), we get
( )
( )
F( ) – F( )
( )
c
b
b
a
c
a
f x dx
f x dx
b
a
f x dx
+
=
=
∫
∫
∫
This proves the property P2 Proof of P3 Let t = a + b – x |
1 | 3977-3980 | (2)
and
( )
b
∫c f x dx
= F(b) – F(c) (3)
342
MATHEMATICS
Adding (2) and (3), we get
( )
( )
F( ) – F( )
( )
c
b
b
a
c
a
f x dx
f x dx
b
a
f x dx
+
=
=
∫
∫
∫
This proves the property P2 Proof of P3 Let t = a + b – x Then dt = – dx |
1 | 3978-3981 | (3)
342
MATHEMATICS
Adding (2) and (3), we get
( )
( )
F( ) – F( )
( )
c
b
b
a
c
a
f x dx
f x dx
b
a
f x dx
+
=
=
∫
∫
∫
This proves the property P2 Proof of P3 Let t = a + b – x Then dt = – dx When x = a, t = b and when x = b, t = a |
1 | 3979-3982 | Proof of P3 Let t = a + b – x Then dt = – dx When x = a, t = b and when x = b, t = a Therefore
( )
b
∫a f x dx
=
(
– )
a
b f a
b
t dt
−
+
∫
=
(
– )
b
a f a
b
t dt
+
∫
(by P1)
=
(
– )
b
a f a
b
x
+
∫
dx by P0
Proof of P4 Put t = a – x |
1 | 3980-3983 | Then dt = – dx When x = a, t = b and when x = b, t = a Therefore
( )
b
∫a f x dx
=
(
– )
a
b f a
b
t dt
−
+
∫
=
(
– )
b
a f a
b
t dt
+
∫
(by P1)
=
(
– )
b
a f a
b
x
+
∫
dx by P0
Proof of P4 Put t = a – x Then dt = – dx |
1 | 3981-3984 | When x = a, t = b and when x = b, t = a Therefore
( )
b
∫a f x dx
=
(
– )
a
b f a
b
t dt
−
+
∫
=
(
– )
b
a f a
b
t dt
+
∫
(by P1)
=
(
– )
b
a f a
b
x
+
∫
dx by P0
Proof of P4 Put t = a – x Then dt = – dx When x = 0, t = a and when x = a, t = 0 |
1 | 3982-3985 | Therefore
( )
b
∫a f x dx
=
(
– )
a
b f a
b
t dt
−
+
∫
=
(
– )
b
a f a
b
t dt
+
∫
(by P1)
=
(
– )
b
a f a
b
x
+
∫
dx by P0
Proof of P4 Put t = a – x Then dt = – dx When x = 0, t = a and when x = a, t = 0 Now
proceed as in P3 |
1 | 3983-3986 | Then dt = – dx When x = 0, t = a and when x = a, t = 0 Now
proceed as in P3 Proof of P5 Using P2, we have
2
2
0
0
( )
( )
( )
a
a
a
a
f x dx
f x dx
f x dx
=
+
∫
∫
∫ |
1 | 3984-3987 | When x = 0, t = a and when x = a, t = 0 Now
proceed as in P3 Proof of P5 Using P2, we have
2
2
0
0
( )
( )
( )
a
a
a
a
f x dx
f x dx
f x dx
=
+
∫
∫
∫ Let
t = 2a – x in the second integral on the right hand side |
1 | 3985-3988 | Now
proceed as in P3 Proof of P5 Using P2, we have
2
2
0
0
( )
( )
( )
a
a
a
a
f x dx
f x dx
f x dx
=
+
∫
∫
∫ Let
t = 2a – x in the second integral on the right hand side Then
dt = – dx |
1 | 3986-3989 | Proof of P5 Using P2, we have
2
2
0
0
( )
( )
( )
a
a
a
a
f x dx
f x dx
f x dx
=
+
∫
∫
∫ Let
t = 2a – x in the second integral on the right hand side Then
dt = – dx When x = a, t = a and when x = 2a, t = 0 |
1 | 3987-3990 | Let
t = 2a – x in the second integral on the right hand side Then
dt = – dx When x = a, t = a and when x = 2a, t = 0 Also x = 2a – t |
1 | 3988-3991 | Then
dt = – dx When x = a, t = a and when x = 2a, t = 0 Also x = 2a – t Therefore, the second integral becomes
2
( )
a
a
f x dx
∫
=
–0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
x dx
∫
Hence
2
0
( )
∫a f x dx
=
0
0
( )
(2
)
a
a
f x dx
f
a
x dx
+
−
∫
∫
Proof of P6 Using P5, we have
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫ |
1 | 3989-3992 | When x = a, t = a and when x = 2a, t = 0 Also x = 2a – t Therefore, the second integral becomes
2
( )
a
a
f x dx
∫
=
–0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
x dx
∫
Hence
2
0
( )
∫a f x dx
=
0
0
( )
(2
)
a
a
f x dx
f
a
x dx
+
−
∫
∫
Proof of P6 Using P5, we have
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫ (1)
Now, if
f (2a – x) = f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
0
( )
( )
2
( )
,
a
a
a
f x dx
f x dx
f x dx
+
=
∫
∫
∫
and if
f(2a – x) = – f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
( )
( )
0
a
a
f x dx
f x dx
−
=
∫
∫
Proof of P7 Using P2, we have
( )
a
a f x dx
−∫
=
0
0
( )
( )
a
a f x dx
f x dx
−
+
∫
∫ |
1 | 3990-3993 | Also x = 2a – t Therefore, the second integral becomes
2
( )
a
a
f x dx
∫
=
–0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
x dx
∫
Hence
2
0
( )
∫a f x dx
=
0
0
( )
(2
)
a
a
f x dx
f
a
x dx
+
−
∫
∫
Proof of P6 Using P5, we have
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫ (1)
Now, if
f (2a – x) = f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
0
( )
( )
2
( )
,
a
a
a
f x dx
f x dx
f x dx
+
=
∫
∫
∫
and if
f(2a – x) = – f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
( )
( )
0
a
a
f x dx
f x dx
−
=
∫
∫
Proof of P7 Using P2, we have
( )
a
a f x dx
−∫
=
0
0
( )
( )
a
a f x dx
f x dx
−
+
∫
∫ Then
Let
t = – x in the first integral on the right hand side |
1 | 3991-3994 | Therefore, the second integral becomes
2
( )
a
a
f x dx
∫
=
–0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
t dt
∫
=
0
(2 – )
a f
a
x dx
∫
Hence
2
0
( )
∫a f x dx
=
0
0
( )
(2
)
a
a
f x dx
f
a
x dx
+
−
∫
∫
Proof of P6 Using P5, we have
2
0
0
0
( )
( )
(2
)
a
a
a
f x dx
f x dx
f
a
x dx
=
+
−
∫
∫
∫ (1)
Now, if
f (2a – x) = f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
0
( )
( )
2
( )
,
a
a
a
f x dx
f x dx
f x dx
+
=
∫
∫
∫
and if
f(2a – x) = – f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
( )
( )
0
a
a
f x dx
f x dx
−
=
∫
∫
Proof of P7 Using P2, we have
( )
a
a f x dx
−∫
=
0
0
( )
( )
a
a f x dx
f x dx
−
+
∫
∫ Then
Let
t = – x in the first integral on the right hand side dt = – dx |
1 | 3992-3995 | (1)
Now, if
f (2a – x) = f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
0
( )
( )
2
( )
,
a
a
a
f x dx
f x dx
f x dx
+
=
∫
∫
∫
and if
f(2a – x) = – f (x), then (1) becomes
2
0
( )
∫a f x dx
=
0
0
( )
( )
0
a
a
f x dx
f x dx
−
=
∫
∫
Proof of P7 Using P2, we have
( )
a
a f x dx
−∫
=
0
0
( )
( )
a
a f x dx
f x dx
−
+
∫
∫ Then
Let
t = – x in the first integral on the right hand side dt = – dx When x = – a, t = a and when
x = 0, t = 0 |
1 | 3993-3996 | Then
Let
t = – x in the first integral on the right hand side dt = – dx When x = – a, t = a and when
x = 0, t = 0 Also x = – t |
1 | 3994-3997 | dt = – dx When x = – a, t = a and when
x = 0, t = 0 Also x = – t INTEGRALS 343
Therefore
( )
a
a f x dx
−∫
=
0
0
–
(– )
( )
a
a f
t dt
f x dx
+
∫
∫
=
0
0
(– )
( )
a
a
f
x dx
f x dx
+
∫
∫
(by P0) |
1 | 3995-3998 | When x = – a, t = a and when
x = 0, t = 0 Also x = – t INTEGRALS 343
Therefore
( )
a
a f x dx
−∫
=
0
0
–
(– )
( )
a
a f
t dt
f x dx
+
∫
∫
=
0
0
(– )
( )
a
a
f
x dx
f x dx
+
∫
∫
(by P0) (1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
0
0
0
( )
( )
( )
2
( )
a
a
a
a
a f x dx
f x dx
f x dx
f x dx
−
=
+
=
∫
∫
∫
∫
(ii) If f is an odd function, then f (–x) = – f(x) and so (1) becomes
0
0
( )
( )
( )
0
a
a
a
a f x dx
f x dx
f x dx
−
= −
+
=
∫
∫
∫
Example 30 Evaluate
2
3
1
x–
x dx
−∫
Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that
x3 – x ≥ 0 on [1, 2] |
1 | 3996-3999 | Also x = – t INTEGRALS 343
Therefore
( )
a
a f x dx
−∫
=
0
0
–
(– )
( )
a
a f
t dt
f x dx
+
∫
∫
=
0
0
(– )
( )
a
a
f
x dx
f x dx
+
∫
∫
(by P0) (1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
0
0
0
( )
( )
( )
2
( )
a
a
a
a
a f x dx
f x dx
f x dx
f x dx
−
=
+
=
∫
∫
∫
∫
(ii) If f is an odd function, then f (–x) = – f(x) and so (1) becomes
0
0
( )
( )
( )
0
a
a
a
a f x dx
f x dx
f x dx
−
= −
+
=
∫
∫
∫
Example 30 Evaluate
2
3
1
x–
x dx
−∫
Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that
x3 – x ≥ 0 on [1, 2] So by P2 we write
2
3
1
x–
x dx
−∫
=
0
1
2
3
3
3
1
0
1
(
– )
– (
– )
(
– )
x
x dx
x
x dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
3
3
3
1
0
1
(
– )
( –
)
(
– )
x
x dx
x
x
dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
4
2
2
4
4
2
–1
0
1
–
–
–
4
2
2
4
4
2
x
x
x
x
x
x
+
+
=
(
)
1
1
1
1
1
1
–
–
–
4 – 2 –
–
4
2
2
4
4
2
+
+
=
1
1
1
1
1
1
–
2
4
2
2
4
4
2
+
+
−
+
−
+
= 3
3
11
2
2
4
4
−
+
=
Example 31 Evaluate
2
–4
4
sin x dx
π
π
∫
Solution We observe that sin2 x is an even function |
1 | 3997-4000 | INTEGRALS 343
Therefore
( )
a
a f x dx
−∫
=
0
0
–
(– )
( )
a
a f
t dt
f x dx
+
∫
∫
=
0
0
(– )
( )
a
a
f
x dx
f x dx
+
∫
∫
(by P0) (1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
0
0
0
( )
( )
( )
2
( )
a
a
a
a
a f x dx
f x dx
f x dx
f x dx
−
=
+
=
∫
∫
∫
∫
(ii) If f is an odd function, then f (–x) = – f(x) and so (1) becomes
0
0
( )
( )
( )
0
a
a
a
a f x dx
f x dx
f x dx
−
= −
+
=
∫
∫
∫
Example 30 Evaluate
2
3
1
x–
x dx
−∫
Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that
x3 – x ≥ 0 on [1, 2] So by P2 we write
2
3
1
x–
x dx
−∫
=
0
1
2
3
3
3
1
0
1
(
– )
– (
– )
(
– )
x
x dx
x
x dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
3
3
3
1
0
1
(
– )
( –
)
(
– )
x
x dx
x
x
dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
4
2
2
4
4
2
–1
0
1
–
–
–
4
2
2
4
4
2
x
x
x
x
x
x
+
+
=
(
)
1
1
1
1
1
1
–
–
–
4 – 2 –
–
4
2
2
4
4
2
+
+
=
1
1
1
1
1
1
–
2
4
2
2
4
4
2
+
+
−
+
−
+
= 3
3
11
2
2
4
4
−
+
=
Example 31 Evaluate
2
–4
4
sin x dx
π
π
∫
Solution We observe that sin2 x is an even function Therefore, by P7 (i), we get
2
–4
4
sin x dx
π
∫π
=
2
204
sin x dx
∫π
344
MATHEMATICS
=
4
0
(1 cos 2 )
2
2
x dx
π
−
∫
=
4
0 (1
cos 2 )x dx
π
−
∫
=
4
0
–1
sin 2
2
x
x
π
=
1
1
–
sin
–0
–
4
2
2
4
2
π
π
π
=
Example 32 Evaluate
2
0
sin
1
xcos
x
dx
x
π
+
∫
Solution Let I =
2
0
1sin
xcos
x
dx
x
π
+
∫ |
1 | 3998-4001 | (1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
0
0
0
( )
( )
( )
2
( )
a
a
a
a
a f x dx
f x dx
f x dx
f x dx
−
=
+
=
∫
∫
∫
∫
(ii) If f is an odd function, then f (–x) = – f(x) and so (1) becomes
0
0
( )
( )
( )
0
a
a
a
a f x dx
f x dx
f x dx
−
= −
+
=
∫
∫
∫
Example 30 Evaluate
2
3
1
x–
x dx
−∫
Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that
x3 – x ≥ 0 on [1, 2] So by P2 we write
2
3
1
x–
x dx
−∫
=
0
1
2
3
3
3
1
0
1
(
– )
– (
– )
(
– )
x
x dx
x
x dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
3
3
3
1
0
1
(
– )
( –
)
(
– )
x
x dx
x
x
dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
4
2
2
4
4
2
–1
0
1
–
–
–
4
2
2
4
4
2
x
x
x
x
x
x
+
+
=
(
)
1
1
1
1
1
1
–
–
–
4 – 2 –
–
4
2
2
4
4
2
+
+
=
1
1
1
1
1
1
–
2
4
2
2
4
4
2
+
+
−
+
−
+
= 3
3
11
2
2
4
4
−
+
=
Example 31 Evaluate
2
–4
4
sin x dx
π
π
∫
Solution We observe that sin2 x is an even function Therefore, by P7 (i), we get
2
–4
4
sin x dx
π
∫π
=
2
204
sin x dx
∫π
344
MATHEMATICS
=
4
0
(1 cos 2 )
2
2
x dx
π
−
∫
=
4
0 (1
cos 2 )x dx
π
−
∫
=
4
0
–1
sin 2
2
x
x
π
=
1
1
–
sin
–0
–
4
2
2
4
2
π
π
π
=
Example 32 Evaluate
2
0
sin
1
xcos
x
dx
x
π
+
∫
Solution Let I =
2
0
1sin
xcos
x
dx
x
π
+
∫ Then, by P4, we have
I =
2
0
(
)sin (
)
1
cos (
)
x
x dx
x
π π −
π −
+
π −
∫
=
2
0
(
)sin
1 cos
x
x dx
x
π π −
+
∫
=
2
0
sin
I
1
cos
x dx
x
π
π
−
+
∫
or
2 I =
2
0
1sin
cos
x dx
x
π
π
+
∫
or
I =
2
0
sin
2
1
cos
x dx
x
π
π
+
∫
Put cos x = t so that – sin x dx = dt |
1 | 3999-4002 | So by P2 we write
2
3
1
x–
x dx
−∫
=
0
1
2
3
3
3
1
0
1
(
– )
– (
– )
(
– )
x
x dx
x
x dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
3
3
3
1
0
1
(
– )
( –
)
(
– )
x
x dx
x
x
dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
4
2
2
4
4
2
–1
0
1
–
–
–
4
2
2
4
4
2
x
x
x
x
x
x
+
+
=
(
)
1
1
1
1
1
1
–
–
–
4 – 2 –
–
4
2
2
4
4
2
+
+
=
1
1
1
1
1
1
–
2
4
2
2
4
4
2
+
+
−
+
−
+
= 3
3
11
2
2
4
4
−
+
=
Example 31 Evaluate
2
–4
4
sin x dx
π
π
∫
Solution We observe that sin2 x is an even function Therefore, by P7 (i), we get
2
–4
4
sin x dx
π
∫π
=
2
204
sin x dx
∫π
344
MATHEMATICS
=
4
0
(1 cos 2 )
2
2
x dx
π
−
∫
=
4
0 (1
cos 2 )x dx
π
−
∫
=
4
0
–1
sin 2
2
x
x
π
=
1
1
–
sin
–0
–
4
2
2
4
2
π
π
π
=
Example 32 Evaluate
2
0
sin
1
xcos
x
dx
x
π
+
∫
Solution Let I =
2
0
1sin
xcos
x
dx
x
π
+
∫ Then, by P4, we have
I =
2
0
(
)sin (
)
1
cos (
)
x
x dx
x
π π −
π −
+
π −
∫
=
2
0
(
)sin
1 cos
x
x dx
x
π π −
+
∫
=
2
0
sin
I
1
cos
x dx
x
π
π
−
+
∫
or
2 I =
2
0
1sin
cos
x dx
x
π
π
+
∫
or
I =
2
0
sin
2
1
cos
x dx
x
π
π
+
∫
Put cos x = t so that – sin x dx = dt When x = 0, t = 1 and when x = π, t = – 1 |
1 | 4000-4003 | Therefore, by P7 (i), we get
2
–4
4
sin x dx
π
∫π
=
2
204
sin x dx
∫π
344
MATHEMATICS
=
4
0
(1 cos 2 )
2
2
x dx
π
−
∫
=
4
0 (1
cos 2 )x dx
π
−
∫
=
4
0
–1
sin 2
2
x
x
π
=
1
1
–
sin
–0
–
4
2
2
4
2
π
π
π
=
Example 32 Evaluate
2
0
sin
1
xcos
x
dx
x
π
+
∫
Solution Let I =
2
0
1sin
xcos
x
dx
x
π
+
∫ Then, by P4, we have
I =
2
0
(
)sin (
)
1
cos (
)
x
x dx
x
π π −
π −
+
π −
∫
=
2
0
(
)sin
1 cos
x
x dx
x
π π −
+
∫
=
2
0
sin
I
1
cos
x dx
x
π
π
−
+
∫
or
2 I =
2
0
1sin
cos
x dx
x
π
π
+
∫
or
I =
2
0
sin
2
1
cos
x dx
x
π
π
+
∫
Put cos x = t so that – sin x dx = dt When x = 0, t = 1 and when x = π, t = – 1 Therefore, (by P1) we get
I =
1
2
1
–
2
1
dt
t
−
π
+
∫
=
1
2
1
2
1
dt
t
−
π
+
∫
=
1
2
0 1
dt
t
π
+
∫
(by P7,
2
1
since
1 t
+
is even function)
=
2
1
– 1
– 1
1
0
tan
tan
1– tan
0
– 0
4
4
t
−
π
π
π
= π
= π
=
Example 33 Evaluate
1
5
4
1 sin
xcos
x dx
−∫
Solution Let I =
1
5
4
1sin
xcos
x dx
−∫ |
1 | 4001-4004 | Then, by P4, we have
I =
2
0
(
)sin (
)
1
cos (
)
x
x dx
x
π π −
π −
+
π −
∫
=
2
0
(
)sin
1 cos
x
x dx
x
π π −
+
∫
=
2
0
sin
I
1
cos
x dx
x
π
π
−
+
∫
or
2 I =
2
0
1sin
cos
x dx
x
π
π
+
∫
or
I =
2
0
sin
2
1
cos
x dx
x
π
π
+
∫
Put cos x = t so that – sin x dx = dt When x = 0, t = 1 and when x = π, t = – 1 Therefore, (by P1) we get
I =
1
2
1
–
2
1
dt
t
−
π
+
∫
=
1
2
1
2
1
dt
t
−
π
+
∫
=
1
2
0 1
dt
t
π
+
∫
(by P7,
2
1
since
1 t
+
is even function)
=
2
1
– 1
– 1
1
0
tan
tan
1– tan
0
– 0
4
4
t
−
π
π
π
= π
= π
=
Example 33 Evaluate
1
5
4
1 sin
xcos
x dx
−∫
Solution Let I =
1
5
4
1sin
xcos
x dx
−∫ Let f(x) = sin5 x cos4 x |
1 | 4002-4005 | When x = 0, t = 1 and when x = π, t = – 1 Therefore, (by P1) we get
I =
1
2
1
–
2
1
dt
t
−
π
+
∫
=
1
2
1
2
1
dt
t
−
π
+
∫
=
1
2
0 1
dt
t
π
+
∫
(by P7,
2
1
since
1 t
+
is even function)
=
2
1
– 1
– 1
1
0
tan
tan
1– tan
0
– 0
4
4
t
−
π
π
π
= π
= π
=
Example 33 Evaluate
1
5
4
1 sin
xcos
x dx
−∫
Solution Let I =
1
5
4
1sin
xcos
x dx
−∫ Let f(x) = sin5 x cos4 x Then
f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i |
1 | 4003-4006 | Therefore, (by P1) we get
I =
1
2
1
–
2
1
dt
t
−
π
+
∫
=
1
2
1
2
1
dt
t
−
π
+
∫
=
1
2
0 1
dt
t
π
+
∫
(by P7,
2
1
since
1 t
+
is even function)
=
2
1
– 1
– 1
1
0
tan
tan
1– tan
0
– 0
4
4
t
−
π
π
π
= π
= π
=
Example 33 Evaluate
1
5
4
1 sin
xcos
x dx
−∫
Solution Let I =
1
5
4
1sin
xcos
x dx
−∫ Let f(x) = sin5 x cos4 x Then
f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i e |
1 | 4004-4007 | Let f(x) = sin5 x cos4 x Then
f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i e , f is an odd function |
1 | 4005-4008 | Then
f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i e , f is an odd function Therefore, by P7 (ii), I = 0
INTEGRALS 345
Example 34 Evaluate
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫
Solution Let I =
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫ |
1 | 4006-4009 | e , f is an odd function Therefore, by P7 (ii), I = 0
INTEGRALS 345
Example 34 Evaluate
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫
Solution Let I =
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫ (1)
Then, by P4
I =
4
2
0
4
4
sin (
)
2
sin (
)
cos (
)
2
2
x
dx
x
x
π
π −
π
π
−
+
−
∫
=
4
2
4
4
0
coscos
sin
x
dx
x
x
π
+
∫ |
1 | 4007-4010 | , f is an odd function Therefore, by P7 (ii), I = 0
INTEGRALS 345
Example 34 Evaluate
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫
Solution Let I =
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫ (1)
Then, by P4
I =
4
2
0
4
4
sin (
)
2
sin (
)
cos (
)
2
2
x
dx
x
x
π
π −
π
π
−
+
−
∫
=
4
2
4
4
0
coscos
sin
x
dx
x
x
π
+
∫ (2)
Adding (1) and (2), we get
2I =
4
4
2
2
2
4
4
0
0
0
sin
cos
[ ]
2
sin
xcos
x dx
dx
x
x
x
π
π
π
+
π
=
=
=
+
∫
∫
Hence
I =
4
π
Example 35 Evaluate
3
6 1
tan
dx
x
π
π
+
∫
Solution Let I =
3
3
6
6
cos
1
tan
cos
sin
x dx
dx
x
x
x
π
π
π
π
=
+
+
∫
∫ |
1 | 4008-4011 | Therefore, by P7 (ii), I = 0
INTEGRALS 345
Example 34 Evaluate
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫
Solution Let I =
4
2
4
4
0
sinsin
cos
x
dx
x
x
π
+
∫ (1)
Then, by P4
I =
4
2
0
4
4
sin (
)
2
sin (
)
cos (
)
2
2
x
dx
x
x
π
π −
π
π
−
+
−
∫
=
4
2
4
4
0
coscos
sin
x
dx
x
x
π
+
∫ (2)
Adding (1) and (2), we get
2I =
4
4
2
2
2
4
4
0
0
0
sin
cos
[ ]
2
sin
xcos
x dx
dx
x
x
x
π
π
π
+
π
=
=
=
+
∫
∫
Hence
I =
4
π
Example 35 Evaluate
3
6 1
tan
dx
x
π
π
+
∫
Solution Let I =
3
3
6
6
cos
1
tan
cos
sin
x dx
dx
x
x
x
π
π
π
π
=
+
+
∫
∫ (1)
Then, by P3
I =
3
6
cos
3
6
cos
sin
3
6
3
6
x
dx
x
x
π
π
π
π
+
−
π
π
π
π
+
−
+
+
−
∫
=
3
6
sin
sin
cos
x
dx
x
x
π
π
+
∫ |
1 | 4009-4012 | (1)
Then, by P4
I =
4
2
0
4
4
sin (
)
2
sin (
)
cos (
)
2
2
x
dx
x
x
π
π −
π
π
−
+
−
∫
=
4
2
4
4
0
coscos
sin
x
dx
x
x
π
+
∫ (2)
Adding (1) and (2), we get
2I =
4
4
2
2
2
4
4
0
0
0
sin
cos
[ ]
2
sin
xcos
x dx
dx
x
x
x
π
π
π
+
π
=
=
=
+
∫
∫
Hence
I =
4
π
Example 35 Evaluate
3
6 1
tan
dx
x
π
π
+
∫
Solution Let I =
3
3
6
6
cos
1
tan
cos
sin
x dx
dx
x
x
x
π
π
π
π
=
+
+
∫
∫ (1)
Then, by P3
I =
3
6
cos
3
6
cos
sin
3
6
3
6
x
dx
x
x
π
π
π
π
+
−
π
π
π
π
+
−
+
+
−
∫
=
3
6
sin
sin
cos
x
dx
x
x
π
π
+
∫ (2)
Adding (1) and (2), we get
2I =
[ ]
3
3
6
6
3
6
6
dx
x
π
π
π
π
π
π
π
=
=
−
=
∫ |
1 | 4010-4013 | (2)
Adding (1) and (2), we get
2I =
4
4
2
2
2
4
4
0
0
0
sin
cos
[ ]
2
sin
xcos
x dx
dx
x
x
x
π
π
π
+
π
=
=
=
+
∫
∫
Hence
I =
4
π
Example 35 Evaluate
3
6 1
tan
dx
x
π
π
+
∫
Solution Let I =
3
3
6
6
cos
1
tan
cos
sin
x dx
dx
x
x
x
π
π
π
π
=
+
+
∫
∫ (1)
Then, by P3
I =
3
6
cos
3
6
cos
sin
3
6
3
6
x
dx
x
x
π
π
π
π
+
−
π
π
π
π
+
−
+
+
−
∫
=
3
6
sin
sin
cos
x
dx
x
x
π
π
+
∫ (2)
Adding (1) and (2), we get
2I =
[ ]
3
3
6
6
3
6
6
dx
x
π
π
π
π
π
π
π
=
=
−
=
∫ Hence I
12
=π
346
MATHEMATICS
Example 36 Evaluate
2
0 log sin x dx
π
∫
Solution Let I =
2
0 logsinx dx
π
∫
Then, by P4
I =
2
2
0
0
log sin
log cos
2
x dx
x dx
π
π
π
−
=
∫
∫
Adding the two values of I, we get
2I =
(
)
02
log sin
logcos
x
x dx
π
+
∫
=
(
)
02
log sin
cos
log2
log2
x
x
dx
π
+
−
∫
(by adding and subtracting log2)
=
2
2
0
0
log sin2
log2
x dx
dx
π
π
−
∫
∫
(Why |
1 | 4011-4014 | (1)
Then, by P3
I =
3
6
cos
3
6
cos
sin
3
6
3
6
x
dx
x
x
π
π
π
π
+
−
π
π
π
π
+
−
+
+
−
∫
=
3
6
sin
sin
cos
x
dx
x
x
π
π
+
∫ (2)
Adding (1) and (2), we get
2I =
[ ]
3
3
6
6
3
6
6
dx
x
π
π
π
π
π
π
π
=
=
−
=
∫ Hence I
12
=π
346
MATHEMATICS
Example 36 Evaluate
2
0 log sin x dx
π
∫
Solution Let I =
2
0 logsinx dx
π
∫
Then, by P4
I =
2
2
0
0
log sin
log cos
2
x dx
x dx
π
π
π
−
=
∫
∫
Adding the two values of I, we get
2I =
(
)
02
log sin
logcos
x
x dx
π
+
∫
=
(
)
02
log sin
cos
log2
log2
x
x
dx
π
+
−
∫
(by adding and subtracting log2)
=
2
2
0
0
log sin2
log2
x dx
dx
π
π
−
∫
∫
(Why )
Put 2x = t in the first integral |
1 | 4012-4015 | (2)
Adding (1) and (2), we get
2I =
[ ]
3
3
6
6
3
6
6
dx
x
π
π
π
π
π
π
π
=
=
−
=
∫ Hence I
12
=π
346
MATHEMATICS
Example 36 Evaluate
2
0 log sin x dx
π
∫
Solution Let I =
2
0 logsinx dx
π
∫
Then, by P4
I =
2
2
0
0
log sin
log cos
2
x dx
x dx
π
π
π
−
=
∫
∫
Adding the two values of I, we get
2I =
(
)
02
log sin
logcos
x
x dx
π
+
∫
=
(
)
02
log sin
cos
log2
log2
x
x
dx
π
+
−
∫
(by adding and subtracting log2)
=
2
2
0
0
log sin2
log2
x dx
dx
π
π
−
∫
∫
(Why )
Put 2x = t in the first integral Then 2 dx = dt, when x = 0, t = 0 and when
2
x
=π
,
t = π |
1 | 4013-4016 | Hence I
12
=π
346
MATHEMATICS
Example 36 Evaluate
2
0 log sin x dx
π
∫
Solution Let I =
2
0 logsinx dx
π
∫
Then, by P4
I =
2
2
0
0
log sin
log cos
2
x dx
x dx
π
π
π
−
=
∫
∫
Adding the two values of I, we get
2I =
(
)
02
log sin
logcos
x
x dx
π
+
∫
=
(
)
02
log sin
cos
log2
log2
x
x
dx
π
+
−
∫
(by adding and subtracting log2)
=
2
2
0
0
log sin2
log2
x dx
dx
π
π
−
∫
∫
(Why )
Put 2x = t in the first integral Then 2 dx = dt, when x = 0, t = 0 and when
2
x
=π
,
t = π Therefore
2I =
0
1
log sin
log2
2
2
t dt
π
−π
∫
=
02
2
log sin
log2
2
2
t dt
π
−π
∫
[by P6 as sin (π – t) = sin t)
=
2
0 log sin
log2
2
x dx
π
−π
∫
(by changing variable t to x)
= I
log2
π2
−
Hence
2
0 log sin x dx
∫π
= –
2log2
π |
1 | 4014-4017 | )
Put 2x = t in the first integral Then 2 dx = dt, when x = 0, t = 0 and when
2
x
=π
,
t = π Therefore
2I =
0
1
log sin
log2
2
2
t dt
π
−π
∫
=
02
2
log sin
log2
2
2
t dt
π
−π
∫
[by P6 as sin (π – t) = sin t)
=
2
0 log sin
log2
2
x dx
π
−π
∫
(by changing variable t to x)
= I
log2
π2
−
Hence
2
0 log sin x dx
∫π
= –
2log2
π INTEGRALS 347
EXERCISE 7 |
1 | 4015-4018 | Then 2 dx = dt, when x = 0, t = 0 and when
2
x
=π
,
t = π Therefore
2I =
0
1
log sin
log2
2
2
t dt
π
−π
∫
=
02
2
log sin
log2
2
2
t dt
π
−π
∫
[by P6 as sin (π – t) = sin t)
=
2
0 log sin
log2
2
x dx
π
−π
∫
(by changing variable t to x)
= I
log2
π2
−
Hence
2
0 log sin x dx
∫π
= –
2log2
π INTEGRALS 347
EXERCISE 7 11
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19 |
1 | 4016-4019 | Therefore
2I =
0
1
log sin
log2
2
2
t dt
π
−π
∫
=
02
2
log sin
log2
2
2
t dt
π
−π
∫
[by P6 as sin (π – t) = sin t)
=
2
0 log sin
log2
2
x dx
π
−π
∫
(by changing variable t to x)
= I
log2
π2
−
Hence
2
0 log sin x dx
∫π
= –
2log2
π INTEGRALS 347
EXERCISE 7 11
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19 1 |
1 | 4017-4020 | INTEGRALS 347
EXERCISE 7 11
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19 1 2
2
0 cos x dx
∫π
2 |
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