Chapter
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1 | 4018-4021 | 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 02
sin
sin
cos
x
dx
x
x
π
+
∫
3 |
1 | 4019-4022 | 1 2
2
0 cos x dx
∫π
2 02
sin
sin
cos
x
dx
x
x
π
+
∫
3 23
2
3
3
0
2
2
sin
sin
cos
x dx
x
x
π
+
∫
4 |
1 | 4020-4023 | 2
2
0 cos x dx
∫π
2 02
sin
sin
cos
x
dx
x
x
π
+
∫
3 23
2
3
3
0
2
2
sin
sin
cos
x dx
x
x
π
+
∫
4 5
2
5
5
0
cos
sin
cos
x dx
x
x
π
+
∫
5 |
1 | 4021-4024 | 02
sin
sin
cos
x
dx
x
x
π
+
∫
3 23
2
3
3
0
2
2
sin
sin
cos
x dx
x
x
π
+
∫
4 5
2
5
5
0
cos
sin
cos
x dx
x
x
π
+
∫
5 5
5 |
2|
x
dx
−
+
∫
6 |
1 | 4022-4025 | 23
2
3
3
0
2
2
sin
sin
cos
x dx
x
x
π
+
∫
4 5
2
5
5
0
cos
sin
cos
x dx
x
x
π
+
∫
5 5
5 |
2|
x
dx
−
+
∫
6 8
2
5
x
dx
−
∫
7 |
1 | 4023-4026 | 5
2
5
5
0
cos
sin
cos
x dx
x
x
π
+
∫
5 5
5 |
2|
x
dx
−
+
∫
6 8
2
5
x
dx
−
∫
7 1
0
(1
)n
x
x
dx
−
∫
8 |
1 | 4024-4027 | 5
5 |
2|
x
dx
−
+
∫
6 8
2
5
x
dx
−
∫
7 1
0
(1
)n
x
x
dx
−
∫
8 4
0 log (1
tan )x dx
π
+
∫
9 |
1 | 4025-4028 | 8
2
5
x
dx
−
∫
7 1
0
(1
)n
x
x
dx
−
∫
8 4
0 log (1
tan )x dx
π
+
∫
9 2
0
2
x
−x dx
∫
10 |
1 | 4026-4029 | 1
0
(1
)n
x
x
dx
−
∫
8 4
0 log (1
tan )x dx
π
+
∫
9 2
0
2
x
−x dx
∫
10 2
0 (2log sin
logsin2 )
x
x dx
π
−
∫
11 |
1 | 4027-4030 | 4
0 log (1
tan )x dx
π
+
∫
9 2
0
2
x
−x dx
∫
10 2
0 (2log sin
logsin2 )
x
x dx
π
−
∫
11 2
–2
2
sin x dx
π
π
∫
12 |
1 | 4028-4031 | 2
0
2
x
−x dx
∫
10 2
0 (2log sin
logsin2 )
x
x dx
π
−
∫
11 2
–2
2
sin x dx
π
π
∫
12 0 1
sin
x dx
x
π
+
∫
13 |
1 | 4029-4032 | 2
0 (2log sin
logsin2 )
x
x dx
π
−
∫
11 2
–2
2
sin x dx
π
π
∫
12 0 1
sin
x dx
x
π
+
∫
13 7
–2
2
sin x dx
π
∫π
14 |
1 | 4030-4033 | 2
–2
2
sin x dx
π
π
∫
12 0 1
sin
x dx
x
π
+
∫
13 7
–2
2
sin x dx
π
∫π
14 2
5
0 cos x dx
π
∫
15 |
1 | 4031-4034 | 0 1
sin
x dx
x
π
+
∫
13 7
–2
2
sin x dx
π
∫π
14 2
5
0 cos x dx
π
∫
15 02
sin
cos
1
sin
cos
x
x dx
x
x
π
−
+
∫
16 |
1 | 4032-4035 | 7
–2
2
sin x dx
π
∫π
14 2
5
0 cos x dx
π
∫
15 02
sin
cos
1
sin
cos
x
x dx
x
x
π
−
+
∫
16 0 log (1
cos )
x dx
π
+
∫
17 |
1 | 4033-4036 | 2
5
0 cos x dx
π
∫
15 02
sin
cos
1
sin
cos
x
x dx
x
x
π
−
+
∫
16 0 log (1
cos )
x dx
π
+
∫
17 0
a
x
dx
x
a
x
+
−
∫
18 |
1 | 4034-4037 | 02
sin
cos
1
sin
cos
x
x dx
x
x
π
−
+
∫
16 0 log (1
cos )
x dx
π
+
∫
17 0
a
x
dx
x
a
x
+
−
∫
18 4
0
1
x
dx
−
∫
19 |
1 | 4035-4038 | 0 log (1
cos )
x dx
π
+
∫
17 0
a
x
dx
x
a
x
+
−
∫
18 4
0
1
x
dx
−
∫
19 Show that
0
0
( ) ( )
2
( )
a
a
f x g x dx
f x dx
=
∫
∫
, if f and g are defined as f(x) = f(a – x)
and g(x) + g(a – x) = 4
Choose the correct answer in Exercises 20 and 21 |
1 | 4036-4039 | 0
a
x
dx
x
a
x
+
−
∫
18 4
0
1
x
dx
−
∫
19 Show that
0
0
( ) ( )
2
( )
a
a
f x g x dx
f x dx
=
∫
∫
, if f and g are defined as f(x) = f(a – x)
and g(x) + g(a – x) = 4
Choose the correct answer in Exercises 20 and 21 20 |
1 | 4037-4040 | 4
0
1
x
dx
−
∫
19 Show that
0
0
( ) ( )
2
( )
a
a
f x g x dx
f x dx
=
∫
∫
, if f and g are defined as f(x) = f(a – x)
and g(x) + g(a – x) = 4
Choose the correct answer in Exercises 20 and 21 20 The value of
3
5
2
2
(
cos
tan
1)
x
x
x
x
dx
π
−π
+
+
+
∫
is
(A) 0
(B) 2
(C) π
(D) 1
21 |
1 | 4038-4041 | Show that
0
0
( ) ( )
2
( )
a
a
f x g x dx
f x dx
=
∫
∫
, if f and g are defined as f(x) = f(a – x)
and g(x) + g(a – x) = 4
Choose the correct answer in Exercises 20 and 21 20 The value of
3
5
2
2
(
cos
tan
1)
x
x
x
x
dx
π
−π
+
+
+
∫
is
(A) 0
(B) 2
(C) π
(D) 1
21 The value of
02
4
3sin
log
4
3cos
x
dx
x
π
+
+
∫
is
(A) 2
(B)
43
(C) 0
(D) –2
348
MATHEMATICS
Miscellaneous Examples
Example 37 Find cos 6
1 sin 6
x
x dx
+
∫
Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore
21
1
cos 6
1 sin 6
6
x
x dx
t dt
+
=
∫
∫
=
3
3
2
2
1
2
1
( )
C =
(1 sin 6 )
C
6
3
9
t
x
×
+
+
+
Example 38 Find
1
4
4
5
(
)
x
x
dx
−x
∫
Solution We have
1
1
4
4
4
3
5
4
1
(1
)
(
)
x
x
x
dx
dx
x
x
−
−
=
∫
∫
Put
– 3
3
4
1
3
1
1–
, so that
x
t
dx
dt
x
x
−
=
=
=
Therefore
1
1
4
4
4
5
(
)
31
x
x
dx
t
dt
−x
=
∫
∫
=
5
5
4
4
3
1
4
4
1
C =
1
C
3
5
15
t
x
×
+
−
+
Example 39 Find
4
2
(
1) (
1)
x dx
x
x
−
+
∫
Solution We have
4
2
(
1)(
1)
x
x
x
−
+
=
3
12
(
1)
1
x
x
x
x
+
+
−
+
−
=
12
(
1)
(
1) (
1)
x
x
x
+
+
−
+ |
1 | 4039-4042 | 20 The value of
3
5
2
2
(
cos
tan
1)
x
x
x
x
dx
π
−π
+
+
+
∫
is
(A) 0
(B) 2
(C) π
(D) 1
21 The value of
02
4
3sin
log
4
3cos
x
dx
x
π
+
+
∫
is
(A) 2
(B)
43
(C) 0
(D) –2
348
MATHEMATICS
Miscellaneous Examples
Example 37 Find cos 6
1 sin 6
x
x dx
+
∫
Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore
21
1
cos 6
1 sin 6
6
x
x dx
t dt
+
=
∫
∫
=
3
3
2
2
1
2
1
( )
C =
(1 sin 6 )
C
6
3
9
t
x
×
+
+
+
Example 38 Find
1
4
4
5
(
)
x
x
dx
−x
∫
Solution We have
1
1
4
4
4
3
5
4
1
(1
)
(
)
x
x
x
dx
dx
x
x
−
−
=
∫
∫
Put
– 3
3
4
1
3
1
1–
, so that
x
t
dx
dt
x
x
−
=
=
=
Therefore
1
1
4
4
4
5
(
)
31
x
x
dx
t
dt
−x
=
∫
∫
=
5
5
4
4
3
1
4
4
1
C =
1
C
3
5
15
t
x
×
+
−
+
Example 39 Find
4
2
(
1) (
1)
x dx
x
x
−
+
∫
Solution We have
4
2
(
1)(
1)
x
x
x
−
+
=
3
12
(
1)
1
x
x
x
x
+
+
−
+
−
=
12
(
1)
(
1) (
1)
x
x
x
+
+
−
+ (1)
Now express
2
1
(
1)(
1)
x
x
−
+
=
2
A
B
C
(
1)
(
1)
x
x
x
+
+
−
+ |
1 | 4040-4043 | The value of
3
5
2
2
(
cos
tan
1)
x
x
x
x
dx
π
−π
+
+
+
∫
is
(A) 0
(B) 2
(C) π
(D) 1
21 The value of
02
4
3sin
log
4
3cos
x
dx
x
π
+
+
∫
is
(A) 2
(B)
43
(C) 0
(D) –2
348
MATHEMATICS
Miscellaneous Examples
Example 37 Find cos 6
1 sin 6
x
x dx
+
∫
Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore
21
1
cos 6
1 sin 6
6
x
x dx
t dt
+
=
∫
∫
=
3
3
2
2
1
2
1
( )
C =
(1 sin 6 )
C
6
3
9
t
x
×
+
+
+
Example 38 Find
1
4
4
5
(
)
x
x
dx
−x
∫
Solution We have
1
1
4
4
4
3
5
4
1
(1
)
(
)
x
x
x
dx
dx
x
x
−
−
=
∫
∫
Put
– 3
3
4
1
3
1
1–
, so that
x
t
dx
dt
x
x
−
=
=
=
Therefore
1
1
4
4
4
5
(
)
31
x
x
dx
t
dt
−x
=
∫
∫
=
5
5
4
4
3
1
4
4
1
C =
1
C
3
5
15
t
x
×
+
−
+
Example 39 Find
4
2
(
1) (
1)
x dx
x
x
−
+
∫
Solution We have
4
2
(
1)(
1)
x
x
x
−
+
=
3
12
(
1)
1
x
x
x
x
+
+
−
+
−
=
12
(
1)
(
1) (
1)
x
x
x
+
+
−
+ (1)
Now express
2
1
(
1)(
1)
x
x
−
+
=
2
A
B
C
(
1)
(
1)
x
x
x
+
+
−
+ (2)
INTEGRALS 349
So
1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give
1
1
A
, B
C
–
2
2
=
=
= |
1 | 4041-4044 | The value of
02
4
3sin
log
4
3cos
x
dx
x
π
+
+
∫
is
(A) 2
(B)
43
(C) 0
(D) –2
348
MATHEMATICS
Miscellaneous Examples
Example 37 Find cos 6
1 sin 6
x
x dx
+
∫
Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore
21
1
cos 6
1 sin 6
6
x
x dx
t dt
+
=
∫
∫
=
3
3
2
2
1
2
1
( )
C =
(1 sin 6 )
C
6
3
9
t
x
×
+
+
+
Example 38 Find
1
4
4
5
(
)
x
x
dx
−x
∫
Solution We have
1
1
4
4
4
3
5
4
1
(1
)
(
)
x
x
x
dx
dx
x
x
−
−
=
∫
∫
Put
– 3
3
4
1
3
1
1–
, so that
x
t
dx
dt
x
x
−
=
=
=
Therefore
1
1
4
4
4
5
(
)
31
x
x
dx
t
dt
−x
=
∫
∫
=
5
5
4
4
3
1
4
4
1
C =
1
C
3
5
15
t
x
×
+
−
+
Example 39 Find
4
2
(
1) (
1)
x dx
x
x
−
+
∫
Solution We have
4
2
(
1)(
1)
x
x
x
−
+
=
3
12
(
1)
1
x
x
x
x
+
+
−
+
−
=
12
(
1)
(
1) (
1)
x
x
x
+
+
−
+ (1)
Now express
2
1
(
1)(
1)
x
x
−
+
=
2
A
B
C
(
1)
(
1)
x
x
x
+
+
−
+ (2)
INTEGRALS 349
So
1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give
1
1
A
, B
C
–
2
2
=
=
= Substituting values of A, B and C in (2), we get
2
1
(
1) (
1)
x
x
−
+
=
2
2
1
1
1
2(
1)
2 (
1)
2(
1)
x
x
x
x
−
−
−
+
+ |
1 | 4042-4045 | (1)
Now express
2
1
(
1)(
1)
x
x
−
+
=
2
A
B
C
(
1)
(
1)
x
x
x
+
+
−
+ (2)
INTEGRALS 349
So
1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give
1
1
A
, B
C
–
2
2
=
=
= Substituting values of A, B and C in (2), we get
2
1
(
1) (
1)
x
x
−
+
=
2
2
1
1
1
2(
1)
2 (
1)
2(
1)
x
x
x
x
−
−
−
+
+ (3)
Again, substituting (3) in (1), we have
4
2
(
1) (
1)
x
x
x
x
−
+
+
=
2
2
1
1
1
(
1)
2(
1)
2 (
1)
2(
1)
x
x
x
x
x
+
+
−
−
−
+
+
Therefore
4
2
2
– 1
2
1
1
1
log
1 –
log (
1) –
tan
C
2
2
4
2
(
1) (
1)
x
x
dx
x
x
x
x
x
x
x
=
+
+
−
+
+
−
+
+
∫
Example 40 Find
2
1
log (log )
(log )
x
dx
x
+
∫
Solution Let
2
1
I
log (log )
(log )
x
dx
x
=
+
∫
=
2
1
log (log )
(log )
x dx
dx
x
+
∫
∫
In the first integral, let us take 1 as the second function |
1 | 4043-4046 | (2)
INTEGRALS 349
So
1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give
1
1
A
, B
C
–
2
2
=
=
= Substituting values of A, B and C in (2), we get
2
1
(
1) (
1)
x
x
−
+
=
2
2
1
1
1
2(
1)
2 (
1)
2(
1)
x
x
x
x
−
−
−
+
+ (3)
Again, substituting (3) in (1), we have
4
2
(
1) (
1)
x
x
x
x
−
+
+
=
2
2
1
1
1
(
1)
2(
1)
2 (
1)
2(
1)
x
x
x
x
x
+
+
−
−
−
+
+
Therefore
4
2
2
– 1
2
1
1
1
log
1 –
log (
1) –
tan
C
2
2
4
2
(
1) (
1)
x
x
dx
x
x
x
x
x
x
x
=
+
+
−
+
+
−
+
+
∫
Example 40 Find
2
1
log (log )
(log )
x
dx
x
+
∫
Solution Let
2
1
I
log (log )
(log )
x
dx
x
=
+
∫
=
2
1
log (log )
(log )
x dx
dx
x
+
∫
∫
In the first integral, let us take 1 as the second function Then integrating it by
parts, we get
I =
2
1
log (log )
log
(log )
dx
x
x
x dx
x
x
x
−
+
∫
∫
=
2
log (log )
log
(log )
dx
dx
x
x
x
x
−
+
∫
∫ |
1 | 4044-4047 | Substituting values of A, B and C in (2), we get
2
1
(
1) (
1)
x
x
−
+
=
2
2
1
1
1
2(
1)
2 (
1)
2(
1)
x
x
x
x
−
−
−
+
+ (3)
Again, substituting (3) in (1), we have
4
2
(
1) (
1)
x
x
x
x
−
+
+
=
2
2
1
1
1
(
1)
2(
1)
2 (
1)
2(
1)
x
x
x
x
x
+
+
−
−
−
+
+
Therefore
4
2
2
– 1
2
1
1
1
log
1 –
log (
1) –
tan
C
2
2
4
2
(
1) (
1)
x
x
dx
x
x
x
x
x
x
x
=
+
+
−
+
+
−
+
+
∫
Example 40 Find
2
1
log (log )
(log )
x
dx
x
+
∫
Solution Let
2
1
I
log (log )
(log )
x
dx
x
=
+
∫
=
2
1
log (log )
(log )
x dx
dx
x
+
∫
∫
In the first integral, let us take 1 as the second function Then integrating it by
parts, we get
I =
2
1
log (log )
log
(log )
dx
x
x
x dx
x
x
x
−
+
∫
∫
=
2
log (log )
log
(log )
dx
dx
x
x
x
x
−
+
∫
∫ (1)
Again, consider
log
dx
x
∫
, take 1 as the second function and integrate it by parts,
we have
2
1
1
–
–
log
log
(log )
dx
x
x
dx
x
x
x
x
=
∫
∫ |
1 | 4045-4048 | (3)
Again, substituting (3) in (1), we have
4
2
(
1) (
1)
x
x
x
x
−
+
+
=
2
2
1
1
1
(
1)
2(
1)
2 (
1)
2(
1)
x
x
x
x
x
+
+
−
−
−
+
+
Therefore
4
2
2
– 1
2
1
1
1
log
1 –
log (
1) –
tan
C
2
2
4
2
(
1) (
1)
x
x
dx
x
x
x
x
x
x
x
=
+
+
−
+
+
−
+
+
∫
Example 40 Find
2
1
log (log )
(log )
x
dx
x
+
∫
Solution Let
2
1
I
log (log )
(log )
x
dx
x
=
+
∫
=
2
1
log (log )
(log )
x dx
dx
x
+
∫
∫
In the first integral, let us take 1 as the second function Then integrating it by
parts, we get
I =
2
1
log (log )
log
(log )
dx
x
x
x dx
x
x
x
−
+
∫
∫
=
2
log (log )
log
(log )
dx
dx
x
x
x
x
−
+
∫
∫ (1)
Again, consider
log
dx
x
∫
, take 1 as the second function and integrate it by parts,
we have
2
1
1
–
–
log
log
(log )
dx
x
x
dx
x
x
x
x
=
∫
∫ (2)
350
MATHEMATICS
Putting (2) in (1), we get
2
2
I
log (log )
log
(log )
(log )
x
dx
dx
x
x
x
x
x
=
−
−
+
∫
∫
=
log (log )
C
log
x
x
x
x
−
+
Example 41 Find
cot
tan
x
x dx
+
∫
Solution We have
I =
cot
tan
x
x
dx
+
∫
tan (1
cot )
x
x dx
=
+
∫
Put tan x = t2, so that sec2 x dx = 2t dt
or
dx =
4
2
1
t dt
t
+
Then
I =
2
4
1
2
1
(1
)
t
t
dt
t
t
+
+
∫
=
2
2
2
4
2
2
2
1
1
1
1
(
1)
2
= 2
= 2
1
1
1
2
dt
dt
t
t
t
dt
t
t
t
t
t
+
+
+
+
+
−
+
∫
∫
∫
Put
1
t
t
− = y, so that
12
1
t
+
dt = dy |
1 | 4046-4049 | Then integrating it by
parts, we get
I =
2
1
log (log )
log
(log )
dx
x
x
x dx
x
x
x
−
+
∫
∫
=
2
log (log )
log
(log )
dx
dx
x
x
x
x
−
+
∫
∫ (1)
Again, consider
log
dx
x
∫
, take 1 as the second function and integrate it by parts,
we have
2
1
1
–
–
log
log
(log )
dx
x
x
dx
x
x
x
x
=
∫
∫ (2)
350
MATHEMATICS
Putting (2) in (1), we get
2
2
I
log (log )
log
(log )
(log )
x
dx
dx
x
x
x
x
x
=
−
−
+
∫
∫
=
log (log )
C
log
x
x
x
x
−
+
Example 41 Find
cot
tan
x
x dx
+
∫
Solution We have
I =
cot
tan
x
x
dx
+
∫
tan (1
cot )
x
x dx
=
+
∫
Put tan x = t2, so that sec2 x dx = 2t dt
or
dx =
4
2
1
t dt
t
+
Then
I =
2
4
1
2
1
(1
)
t
t
dt
t
t
+
+
∫
=
2
2
2
4
2
2
2
1
1
1
1
(
1)
2
= 2
= 2
1
1
1
2
dt
dt
t
t
t
dt
t
t
t
t
t
+
+
+
+
+
−
+
∫
∫
∫
Put
1
t
t
− = y, so that
12
1
t
+
dt = dy Then
I =
(
)
– 1
– 1
2
2
1
2
2 tan
C =
2 tan
C
2
2
2
t
dy
y
t
y
−
=
+
+
+
∫
=
2
– 1
– 1
1
tan
1
2 tan
C =
2 tan
C
2
2tan
t
x
t
x
−
−
+
+
Example 42 Find
4
sin 2 cos 2
9– cos (2 )
x
x dx
x
∫
Solution Let
4
sin 2 cos 2
I
9 – cos 2
x
x dx
x
=∫
INTEGRALS 351
Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt
Therefore
–1
1
2
2
1
1
1
1
I
–
–
sin
C
sin
cos 2
C
4
4
3
4
3
9 –
dt
t
x
t
−
=
=
+
= −
+
∫
Example 43 Evaluate
123
sin (
)
x
x dx
−
π
∫
Solution Here f (x) = | x sin πx | =
sin
for
1
1
3
sin
for 1
2
x
x
x
x
x
x
π
− ≤
≤
−
π
≤
≤
Therefore
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
+
−
π
∫
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
−
π
∫
∫
Integrating both integrals on righthand side, we get
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
2
2
1
1
–
cos
sin
cos
sin
x
x
x
x
x
x
−
π
π
−
π
π
+
−
+
π
π
π
π
=
2
2
2
1
1
3
1
− −
−
=
+
π
π
π
π
π
Example 44 Evaluate
2
2
2
2
0
cos
sin
x dx
a
x
b
x
π
+
∫
Solution Let I =
2
2
2
2
2
2
2
2
0
0
(
)
cos
sin
cos (
)
sin (
)
x dx
x dx
a
x
b
x
a
x
b
x
π
π
π −
=
+
π −
+
π −
∫
∫
(using P4)
=
2
2
2
2
2
2
2
2
0
0
cos
sin
cos
sin
dx
x dx
a
x
b
x
a
x
b
x
π
π
π
−
+
+
∫
∫
=
2
2
2
2
0
I
cos
sin
dx
a
x
b
x
π
π
−
+
∫
Thus
2I =
2
2
2
2
0
cos
sin
dx
a
x
b
x
π
π
+
∫
352
MATHEMATICS
or
I =
2
2
2
2
2
2
2
2
2
0
20
2
2
cos
sin
cos
sin
dx
dx
a
x
b
x
a
x
b
x
π
π
π
=π
⋅
+
+
∫
∫
(using P6)
=
2
4
2
2
2
2
2
2
2
2
0
4
cos
sin
cos
sin
π
π
π
π
+
+
+
∫
∫
dx
dx
a
x
b
x
a
x
b
x
=
2
2
2
4
2
2
2
2
2
2
0
4
sec
cosec
tan
cot
π
π
π
π
+
+
+
∫
∫
xdx
xdx
a
b
x
a
x
b
=
(
)
0
1
2
2 2
2
2
2
0
1
tan
t
cot
π
−
=
=
+
+
∫
∫
dt
du
put
x
and
x
u
a
b t
a u
b
=
1
0
–1
–1
0
1
tan
–
tan
π
π
bt
au
ab
a
ab
b
=
–1
–1
tan
tan
π
+
b
a
ab
a
b =
2
2
π
ab
Miscellaneous Exercise on Chapter 7
Integrate the functions in Exercises 1 to 24 |
1 | 4047-4050 | (1)
Again, consider
log
dx
x
∫
, take 1 as the second function and integrate it by parts,
we have
2
1
1
–
–
log
log
(log )
dx
x
x
dx
x
x
x
x
=
∫
∫ (2)
350
MATHEMATICS
Putting (2) in (1), we get
2
2
I
log (log )
log
(log )
(log )
x
dx
dx
x
x
x
x
x
=
−
−
+
∫
∫
=
log (log )
C
log
x
x
x
x
−
+
Example 41 Find
cot
tan
x
x dx
+
∫
Solution We have
I =
cot
tan
x
x
dx
+
∫
tan (1
cot )
x
x dx
=
+
∫
Put tan x = t2, so that sec2 x dx = 2t dt
or
dx =
4
2
1
t dt
t
+
Then
I =
2
4
1
2
1
(1
)
t
t
dt
t
t
+
+
∫
=
2
2
2
4
2
2
2
1
1
1
1
(
1)
2
= 2
= 2
1
1
1
2
dt
dt
t
t
t
dt
t
t
t
t
t
+
+
+
+
+
−
+
∫
∫
∫
Put
1
t
t
− = y, so that
12
1
t
+
dt = dy Then
I =
(
)
– 1
– 1
2
2
1
2
2 tan
C =
2 tan
C
2
2
2
t
dy
y
t
y
−
=
+
+
+
∫
=
2
– 1
– 1
1
tan
1
2 tan
C =
2 tan
C
2
2tan
t
x
t
x
−
−
+
+
Example 42 Find
4
sin 2 cos 2
9– cos (2 )
x
x dx
x
∫
Solution Let
4
sin 2 cos 2
I
9 – cos 2
x
x dx
x
=∫
INTEGRALS 351
Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt
Therefore
–1
1
2
2
1
1
1
1
I
–
–
sin
C
sin
cos 2
C
4
4
3
4
3
9 –
dt
t
x
t
−
=
=
+
= −
+
∫
Example 43 Evaluate
123
sin (
)
x
x dx
−
π
∫
Solution Here f (x) = | x sin πx | =
sin
for
1
1
3
sin
for 1
2
x
x
x
x
x
x
π
− ≤
≤
−
π
≤
≤
Therefore
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
+
−
π
∫
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
−
π
∫
∫
Integrating both integrals on righthand side, we get
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
2
2
1
1
–
cos
sin
cos
sin
x
x
x
x
x
x
−
π
π
−
π
π
+
−
+
π
π
π
π
=
2
2
2
1
1
3
1
− −
−
=
+
π
π
π
π
π
Example 44 Evaluate
2
2
2
2
0
cos
sin
x dx
a
x
b
x
π
+
∫
Solution Let I =
2
2
2
2
2
2
2
2
0
0
(
)
cos
sin
cos (
)
sin (
)
x dx
x dx
a
x
b
x
a
x
b
x
π
π
π −
=
+
π −
+
π −
∫
∫
(using P4)
=
2
2
2
2
2
2
2
2
0
0
cos
sin
cos
sin
dx
x dx
a
x
b
x
a
x
b
x
π
π
π
−
+
+
∫
∫
=
2
2
2
2
0
I
cos
sin
dx
a
x
b
x
π
π
−
+
∫
Thus
2I =
2
2
2
2
0
cos
sin
dx
a
x
b
x
π
π
+
∫
352
MATHEMATICS
or
I =
2
2
2
2
2
2
2
2
2
0
20
2
2
cos
sin
cos
sin
dx
dx
a
x
b
x
a
x
b
x
π
π
π
=π
⋅
+
+
∫
∫
(using P6)
=
2
4
2
2
2
2
2
2
2
2
0
4
cos
sin
cos
sin
π
π
π
π
+
+
+
∫
∫
dx
dx
a
x
b
x
a
x
b
x
=
2
2
2
4
2
2
2
2
2
2
0
4
sec
cosec
tan
cot
π
π
π
π
+
+
+
∫
∫
xdx
xdx
a
b
x
a
x
b
=
(
)
0
1
2
2 2
2
2
2
0
1
tan
t
cot
π
−
=
=
+
+
∫
∫
dt
du
put
x
and
x
u
a
b t
a u
b
=
1
0
–1
–1
0
1
tan
–
tan
π
π
bt
au
ab
a
ab
b
=
–1
–1
tan
tan
π
+
b
a
ab
a
b =
2
2
π
ab
Miscellaneous Exercise on Chapter 7
Integrate the functions in Exercises 1 to 24 1 |
1 | 4048-4051 | (2)
350
MATHEMATICS
Putting (2) in (1), we get
2
2
I
log (log )
log
(log )
(log )
x
dx
dx
x
x
x
x
x
=
−
−
+
∫
∫
=
log (log )
C
log
x
x
x
x
−
+
Example 41 Find
cot
tan
x
x dx
+
∫
Solution We have
I =
cot
tan
x
x
dx
+
∫
tan (1
cot )
x
x dx
=
+
∫
Put tan x = t2, so that sec2 x dx = 2t dt
or
dx =
4
2
1
t dt
t
+
Then
I =
2
4
1
2
1
(1
)
t
t
dt
t
t
+
+
∫
=
2
2
2
4
2
2
2
1
1
1
1
(
1)
2
= 2
= 2
1
1
1
2
dt
dt
t
t
t
dt
t
t
t
t
t
+
+
+
+
+
−
+
∫
∫
∫
Put
1
t
t
− = y, so that
12
1
t
+
dt = dy Then
I =
(
)
– 1
– 1
2
2
1
2
2 tan
C =
2 tan
C
2
2
2
t
dy
y
t
y
−
=
+
+
+
∫
=
2
– 1
– 1
1
tan
1
2 tan
C =
2 tan
C
2
2tan
t
x
t
x
−
−
+
+
Example 42 Find
4
sin 2 cos 2
9– cos (2 )
x
x dx
x
∫
Solution Let
4
sin 2 cos 2
I
9 – cos 2
x
x dx
x
=∫
INTEGRALS 351
Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt
Therefore
–1
1
2
2
1
1
1
1
I
–
–
sin
C
sin
cos 2
C
4
4
3
4
3
9 –
dt
t
x
t
−
=
=
+
= −
+
∫
Example 43 Evaluate
123
sin (
)
x
x dx
−
π
∫
Solution Here f (x) = | x sin πx | =
sin
for
1
1
3
sin
for 1
2
x
x
x
x
x
x
π
− ≤
≤
−
π
≤
≤
Therefore
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
+
−
π
∫
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
−
π
∫
∫
Integrating both integrals on righthand side, we get
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
2
2
1
1
–
cos
sin
cos
sin
x
x
x
x
x
x
−
π
π
−
π
π
+
−
+
π
π
π
π
=
2
2
2
1
1
3
1
− −
−
=
+
π
π
π
π
π
Example 44 Evaluate
2
2
2
2
0
cos
sin
x dx
a
x
b
x
π
+
∫
Solution Let I =
2
2
2
2
2
2
2
2
0
0
(
)
cos
sin
cos (
)
sin (
)
x dx
x dx
a
x
b
x
a
x
b
x
π
π
π −
=
+
π −
+
π −
∫
∫
(using P4)
=
2
2
2
2
2
2
2
2
0
0
cos
sin
cos
sin
dx
x dx
a
x
b
x
a
x
b
x
π
π
π
−
+
+
∫
∫
=
2
2
2
2
0
I
cos
sin
dx
a
x
b
x
π
π
−
+
∫
Thus
2I =
2
2
2
2
0
cos
sin
dx
a
x
b
x
π
π
+
∫
352
MATHEMATICS
or
I =
2
2
2
2
2
2
2
2
2
0
20
2
2
cos
sin
cos
sin
dx
dx
a
x
b
x
a
x
b
x
π
π
π
=π
⋅
+
+
∫
∫
(using P6)
=
2
4
2
2
2
2
2
2
2
2
0
4
cos
sin
cos
sin
π
π
π
π
+
+
+
∫
∫
dx
dx
a
x
b
x
a
x
b
x
=
2
2
2
4
2
2
2
2
2
2
0
4
sec
cosec
tan
cot
π
π
π
π
+
+
+
∫
∫
xdx
xdx
a
b
x
a
x
b
=
(
)
0
1
2
2 2
2
2
2
0
1
tan
t
cot
π
−
=
=
+
+
∫
∫
dt
du
put
x
and
x
u
a
b t
a u
b
=
1
0
–1
–1
0
1
tan
–
tan
π
π
bt
au
ab
a
ab
b
=
–1
–1
tan
tan
π
+
b
a
ab
a
b =
2
2
π
ab
Miscellaneous Exercise on Chapter 7
Integrate the functions in Exercises 1 to 24 1 3
1
x
−x
2 |
1 | 4049-4052 | Then
I =
(
)
– 1
– 1
2
2
1
2
2 tan
C =
2 tan
C
2
2
2
t
dy
y
t
y
−
=
+
+
+
∫
=
2
– 1
– 1
1
tan
1
2 tan
C =
2 tan
C
2
2tan
t
x
t
x
−
−
+
+
Example 42 Find
4
sin 2 cos 2
9– cos (2 )
x
x dx
x
∫
Solution Let
4
sin 2 cos 2
I
9 – cos 2
x
x dx
x
=∫
INTEGRALS 351
Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt
Therefore
–1
1
2
2
1
1
1
1
I
–
–
sin
C
sin
cos 2
C
4
4
3
4
3
9 –
dt
t
x
t
−
=
=
+
= −
+
∫
Example 43 Evaluate
123
sin (
)
x
x dx
−
π
∫
Solution Here f (x) = | x sin πx | =
sin
for
1
1
3
sin
for 1
2
x
x
x
x
x
x
π
− ≤
≤
−
π
≤
≤
Therefore
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
+
−
π
∫
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
−
π
∫
∫
Integrating both integrals on righthand side, we get
23
1 |
sin
|
x
x dx
−
π
∫
=
3
1
2
2
2
1
1
–
cos
sin
cos
sin
x
x
x
x
x
x
−
π
π
−
π
π
+
−
+
π
π
π
π
=
2
2
2
1
1
3
1
− −
−
=
+
π
π
π
π
π
Example 44 Evaluate
2
2
2
2
0
cos
sin
x dx
a
x
b
x
π
+
∫
Solution Let I =
2
2
2
2
2
2
2
2
0
0
(
)
cos
sin
cos (
)
sin (
)
x dx
x dx
a
x
b
x
a
x
b
x
π
π
π −
=
+
π −
+
π −
∫
∫
(using P4)
=
2
2
2
2
2
2
2
2
0
0
cos
sin
cos
sin
dx
x dx
a
x
b
x
a
x
b
x
π
π
π
−
+
+
∫
∫
=
2
2
2
2
0
I
cos
sin
dx
a
x
b
x
π
π
−
+
∫
Thus
2I =
2
2
2
2
0
cos
sin
dx
a
x
b
x
π
π
+
∫
352
MATHEMATICS
or
I =
2
2
2
2
2
2
2
2
2
0
20
2
2
cos
sin
cos
sin
dx
dx
a
x
b
x
a
x
b
x
π
π
π
=π
⋅
+
+
∫
∫
(using P6)
=
2
4
2
2
2
2
2
2
2
2
0
4
cos
sin
cos
sin
π
π
π
π
+
+
+
∫
∫
dx
dx
a
x
b
x
a
x
b
x
=
2
2
2
4
2
2
2
2
2
2
0
4
sec
cosec
tan
cot
π
π
π
π
+
+
+
∫
∫
xdx
xdx
a
b
x
a
x
b
=
(
)
0
1
2
2 2
2
2
2
0
1
tan
t
cot
π
−
=
=
+
+
∫
∫
dt
du
put
x
and
x
u
a
b t
a u
b
=
1
0
–1
–1
0
1
tan
–
tan
π
π
bt
au
ab
a
ab
b
=
–1
–1
tan
tan
π
+
b
a
ab
a
b =
2
2
π
ab
Miscellaneous Exercise on Chapter 7
Integrate the functions in Exercises 1 to 24 1 3
1
x
−x
2 1
x
a
x
b
+
+
+
3 |
1 | 4050-4053 | 1 3
1
x
−x
2 1
x
a
x
b
+
+
+
3 2
1
x
ax
x
−
[Hint: Put x = a
t ]
4 |
1 | 4051-4054 | 3
1
x
−x
2 1
x
a
x
b
+
+
+
3 2
1
x
ax
x
−
[Hint: Put x = a
t ]
4 3
2
4
4
1
(
1)
x
x +
5 |
1 | 4052-4055 | 1
x
a
x
b
+
+
+
3 2
1
x
ax
x
−
[Hint: Put x = a
t ]
4 3
2
4
4
1
(
1)
x
x +
5 1
1
3
2
1
x
x
+
[Hint:
1
1
1
1
3
2
3
6
1
1
1
x
x
x
x
=
+
+
, put x = t6]
6 |
1 | 4053-4056 | 2
1
x
ax
x
−
[Hint: Put x = a
t ]
4 3
2
4
4
1
(
1)
x
x +
5 1
1
3
2
1
x
x
+
[Hint:
1
1
1
1
3
2
3
6
1
1
1
x
x
x
x
=
+
+
, put x = t6]
6 2
5
(
1) (
9)
x
x
x
+
+
7 |
1 | 4054-4057 | 3
2
4
4
1
(
1)
x
x +
5 1
1
3
2
1
x
x
+
[Hint:
1
1
1
1
3
2
3
6
1
1
1
x
x
x
x
=
+
+
, put x = t6]
6 2
5
(
1) (
9)
x
x
x
+
+
7 sin
sin (
)
x
x
−a
8 |
1 | 4055-4058 | 1
1
3
2
1
x
x
+
[Hint:
1
1
1
1
3
2
3
6
1
1
1
x
x
x
x
=
+
+
, put x = t6]
6 2
5
(
1) (
9)
x
x
x
+
+
7 sin
sin (
)
x
x
−a
8 5 log
4 log
3 log
2 log
x
x
x
x
e
e
e
−e
−
9 |
1 | 4056-4059 | 2
5
(
1) (
9)
x
x
x
+
+
7 sin
sin (
)
x
x
−a
8 5 log
4 log
3 log
2 log
x
x
x
x
e
e
e
−e
−
9 2
cos
4
sin
x
x
−
10 |
1 | 4057-4060 | sin
sin (
)
x
x
−a
8 5 log
4 log
3 log
2 log
x
x
x
x
e
e
e
−e
−
9 2
cos
4
sin
x
x
−
10 8
8
2
2
sin
cos
1 2sin
cos
x
x
x
−
−
11 |
1 | 4058-4061 | 5 log
4 log
3 log
2 log
x
x
x
x
e
e
e
−e
−
9 2
cos
4
sin
x
x
−
10 8
8
2
2
sin
cos
1 2sin
cos
x
x
x
−
−
11 1
cos (
) cos (
)
x
a
x
b
+
+
12 |
1 | 4059-4062 | 2
cos
4
sin
x
x
−
10 8
8
2
2
sin
cos
1 2sin
cos
x
x
x
−
−
11 1
cos (
) cos (
)
x
a
x
b
+
+
12 3
8
1
x
x
−
13 |
1 | 4060-4063 | 8
8
2
2
sin
cos
1 2sin
cos
x
x
x
−
−
11 1
cos (
) cos (
)
x
a
x
b
+
+
12 3
8
1
x
x
−
13 (1
) (2
)
x
x
x
e
e
e
+
+
14 |
1 | 4061-4064 | 1
cos (
) cos (
)
x
a
x
b
+
+
12 3
8
1
x
x
−
13 (1
) (2
)
x
x
x
e
e
e
+
+
14 2
2
1
(
1) (
4)
x
x
+
+
15 |
1 | 4062-4065 | 3
8
1
x
x
−
13 (1
) (2
)
x
x
x
e
e
e
+
+
14 2
2
1
(
1) (
4)
x
x
+
+
15 cos 3x elog sinx
16 |
1 | 4063-4066 | (1
) (2
)
x
x
x
e
e
e
+
+
14 2
2
1
(
1) (
4)
x
x
+
+
15 cos 3x elog sinx
16 e3 logx (x4 + 1)– 1
17 |
1 | 4064-4067 | 2
2
1
(
1) (
4)
x
x
+
+
15 cos 3x elog sinx
16 e3 logx (x4 + 1)– 1
17 f ′ (ax + b) [f (ax + b)] n
18 |
1 | 4065-4068 | cos 3x elog sinx
16 e3 logx (x4 + 1)– 1
17 f ′ (ax + b) [f (ax + b)] n
18 3
1
sin
sin (
)
x
x + α
19 |
1 | 4066-4069 | e3 logx (x4 + 1)– 1
17 f ′ (ax + b) [f (ax + b)] n
18 3
1
sin
sin (
)
x
x + α
19 1
1
1
1
sin
cos
sin
cos
x
x
x
x
−
−
−
−
−
+
, x ∈ [0, 1]
INTEGRALS 353
20 |
1 | 4067-4070 | f ′ (ax + b) [f (ax + b)] n
18 3
1
sin
sin (
)
x
x + α
19 1
1
1
1
sin
cos
sin
cos
x
x
x
x
−
−
−
−
−
+
, x ∈ [0, 1]
INTEGRALS 353
20 1
1
x
x
−
+
21 |
1 | 4068-4071 | 3
1
sin
sin (
)
x
x + α
19 1
1
1
1
sin
cos
sin
cos
x
x
x
x
−
−
−
−
−
+
, x ∈ [0, 1]
INTEGRALS 353
20 1
1
x
x
−
+
21 2
sin 2
1
cos2
x ex
x
++
22 |
1 | 4069-4072 | 1
1
1
1
sin
cos
sin
cos
x
x
x
x
−
−
−
−
−
+
, x ∈ [0, 1]
INTEGRALS 353
20 1
1
x
x
−
+
21 2
sin 2
1
cos2
x ex
x
++
22 2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
23 |
1 | 4070-4073 | 1
1
x
x
−
+
21 2
sin 2
1
cos2
x ex
x
++
22 2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
23 – 1
1
tan
1
x
x
+−
24 |
1 | 4071-4074 | 2
sin 2
1
cos2
x ex
x
++
22 2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
23 – 1
1
tan
1
x
x
+−
24 2
2
4
1 log (
1)
2 log
x
x
x
x
+
+
−
Evaluate the definite integrals in Exercises 25 to 33 |
1 | 4072-4075 | 2
2
1
(
1) (
2)
x
x
x
+x
+
+
+
23 – 1
1
tan
1
x
x
+−
24 2
2
4
1 log (
1)
2 log
x
x
x
x
+
+
−
Evaluate the definite integrals in Exercises 25 to 33 25 |
1 | 4073-4076 | – 1
1
tan
1
x
x
+−
24 2
2
4
1 log (
1)
2 log
x
x
x
x
+
+
−
Evaluate the definite integrals in Exercises 25 to 33 25 2
1
sin
1 cos
π
π
−
−
∫
x
x
e
xdx
26 |
1 | 4074-4077 | 2
2
4
1 log (
1)
2 log
x
x
x
x
+
+
−
Evaluate the definite integrals in Exercises 25 to 33 25 2
1
sin
1 cos
π
π
−
−
∫
x
x
e
xdx
26 4
4
4
0
sin
cos
cos
sin
x
x
dx
x
x
π
+
∫
27 |
1 | 4075-4078 | 25 2
1
sin
1 cos
π
π
−
−
∫
x
x
e
xdx
26 4
4
4
0
sin
cos
cos
sin
x
x
dx
x
x
π
+
∫
27 2
2
2
2
0
cos
cos
4 sin
x dx
x
x
π
+
∫
28 |
1 | 4076-4079 | 2
1
sin
1 cos
π
π
−
−
∫
x
x
e
xdx
26 4
4
4
0
sin
cos
cos
sin
x
x
dx
x
x
π
+
∫
27 2
2
2
2
0
cos
cos
4 sin
x dx
x
x
π
+
∫
28 3
6
sin
cos
sin 2
x
x dx
x
π
π
+
∫
29 |
1 | 4077-4080 | 4
4
4
0
sin
cos
cos
sin
x
x
dx
x
x
π
+
∫
27 2
2
2
2
0
cos
cos
4 sin
x dx
x
x
π
+
∫
28 3
6
sin
cos
sin 2
x
x dx
x
π
π
+
∫
29 1
0
1
dx
x
x
+
−
∫
30 |
1 | 4078-4081 | 2
2
2
2
0
cos
cos
4 sin
x dx
x
x
π
+
∫
28 3
6
sin
cos
sin 2
x
x dx
x
π
π
+
∫
29 1
0
1
dx
x
x
+
−
∫
30 04
sin
cos
9
x16 sin 2
x dx
x
π
+
+
∫
31 |
1 | 4079-4082 | 3
6
sin
cos
sin 2
x
x dx
x
π
π
+
∫
29 1
0
1
dx
x
x
+
−
∫
30 04
sin
cos
9
x16 sin 2
x dx
x
π
+
+
∫
31 1
2
0 sin 2 tan
(sin )
x
x dx
π
−
∫
32 |
1 | 4080-4083 | 1
0
1
dx
x
x
+
−
∫
30 04
sin
cos
9
x16 sin 2
x dx
x
π
+
+
∫
31 1
2
0 sin 2 tan
(sin )
x
x dx
π
−
∫
32 0
tan
sec
tan
x
x
dx
x
x
π
+
∫
33 |
1 | 4081-4084 | 04
sin
cos
9
x16 sin 2
x dx
x
π
+
+
∫
31 1
2
0 sin 2 tan
(sin )
x
x dx
π
−
∫
32 0
tan
sec
tan
x
x
dx
x
x
π
+
∫
33 4
1 [
1|
|
2|
|
3|]
x
x
x
dx
−
+
−
+
−
∫
Prove the following (Exercises 34 to 39)
34 |
1 | 4082-4085 | 1
2
0 sin 2 tan
(sin )
x
x dx
π
−
∫
32 0
tan
sec
tan
x
x
dx
x
x
π
+
∫
33 4
1 [
1|
|
2|
|
3|]
x
x
x
dx
−
+
−
+
−
∫
Prove the following (Exercises 34 to 39)
34 3
2
1
2
2
log
3
3
(
xdx1)
x
=
+
+
∫
35 |
1 | 4083-4086 | 0
tan
sec
tan
x
x
dx
x
x
π
+
∫
33 4
1 [
1|
|
2|
|
3|]
x
x
x
dx
−
+
−
+
−
∫
Prove the following (Exercises 34 to 39)
34 3
2
1
2
2
log
3
3
(
xdx1)
x
=
+
+
∫
35 1
0
1
x
x e dx =
∫
36 |
1 | 4084-4087 | 4
1 [
1|
|
2|
|
3|]
x
x
x
dx
−
+
−
+
−
∫
Prove the following (Exercises 34 to 39)
34 3
2
1
2
2
log
3
3
(
xdx1)
x
=
+
+
∫
35 1
0
1
x
x e dx =
∫
36 1
17
4
1
cos
0
x
x dx
−
=
∫
37 |
1 | 4085-4088 | 3
2
1
2
2
log
3
3
(
xdx1)
x
=
+
+
∫
35 1
0
1
x
x e dx =
∫
36 1
17
4
1
cos
0
x
x dx
−
=
∫
37 3
2
0
2
sin
3
x dx
π
=
∫
38 |
1 | 4086-4089 | 1
0
1
x
x e dx =
∫
36 1
17
4
1
cos
0
x
x dx
−
=
∫
37 3
2
0
2
sin
3
x dx
π
=
∫
38 3
4
0 2 tan
1
log2
x dx
π
= −
∫
39 |
1 | 4087-4090 | 1
17
4
1
cos
0
x
x dx
−
=
∫
37 3
2
0
2
sin
3
x dx
π
=
∫
38 3
4
0 2 tan
1
log2
x dx
π
= −
∫
39 1
1
0sin
1
2
x dx
−
=π
−
∫
40 |
1 | 4088-4091 | 3
2
0
2
sin
3
x dx
π
=
∫
38 3
4
0 2 tan
1
log2
x dx
π
= −
∫
39 1
1
0sin
1
2
x dx
−
=π
−
∫
40 Evaluate
1
2 3
0
x
e
dx
−
∫
as a limit of a sum |
1 | 4089-4092 | 3
4
0 2 tan
1
log2
x dx
π
= −
∫
39 1
1
0sin
1
2
x dx
−
=π
−
∫
40 Evaluate
1
2 3
0
x
e
dx
−
∫
as a limit of a sum Choose the correct answers in Exercises 41 to 44 |
1 | 4090-4093 | 1
1
0sin
1
2
x dx
−
=π
−
∫
40 Evaluate
1
2 3
0
x
e
dx
−
∫
as a limit of a sum Choose the correct answers in Exercises 41 to 44 41 |
1 | 4091-4094 | Evaluate
1
2 3
0
x
e
dx
−
∫
as a limit of a sum Choose the correct answers in Exercises 41 to 44 41 x
x
dx
e
+e−
∫
is equal to
(A) tan–1 (ex) + C
(B) tan–1 (e–x) + C
(C) log (ex – e–x) + C
(D) log (ex + e–x) + C
42 |
1 | 4092-4095 | Choose the correct answers in Exercises 41 to 44 41 x
x
dx
e
+e−
∫
is equal to
(A) tan–1 (ex) + C
(B) tan–1 (e–x) + C
(C) log (ex – e–x) + C
(D) log (ex + e–x) + C
42 2
(sincos2
cos )
x
dx
x
x
+
∫
is equal to
(A)
–1
C
sin
cos
x
x
+
+
(B) log |sin
cos |
C
x
x
+
+
(C) log |sin
cos |
C
x
x
−
+
(D)
2
1
(sin
cos )
x
x
+
354
MATHEMATICS
43 |
1 | 4093-4096 | 41 x
x
dx
e
+e−
∫
is equal to
(A) tan–1 (ex) + C
(B) tan–1 (e–x) + C
(C) log (ex – e–x) + C
(D) log (ex + e–x) + C
42 2
(sincos2
cos )
x
dx
x
x
+
∫
is equal to
(A)
–1
C
sin
cos
x
x
+
+
(B) log |sin
cos |
C
x
x
+
+
(C) log |sin
cos |
C
x
x
−
+
(D)
2
1
(sin
cos )
x
x
+
354
MATHEMATICS
43 If f (a + b – x) = f (x), then
( )
b
∫a x f x dx
is equal to
(A)
(
)
2
b
a
a
b
f b
x dx
+
−
∫
(B)
(
)
2
b
a
a
b
f b
x dx
+
+
∫
(C)
( )
2
b
a
b
a
− ∫f x dx
(D)
( )
2
b
a
a
b
f x dx
+ ∫
44 |
1 | 4094-4097 | x
x
dx
e
+e−
∫
is equal to
(A) tan–1 (ex) + C
(B) tan–1 (e–x) + C
(C) log (ex – e–x) + C
(D) log (ex + e–x) + C
42 2
(sincos2
cos )
x
dx
x
x
+
∫
is equal to
(A)
–1
C
sin
cos
x
x
+
+
(B) log |sin
cos |
C
x
x
+
+
(C) log |sin
cos |
C
x
x
−
+
(D)
2
1
(sin
cos )
x
x
+
354
MATHEMATICS
43 If f (a + b – x) = f (x), then
( )
b
∫a x f x dx
is equal to
(A)
(
)
2
b
a
a
b
f b
x dx
+
−
∫
(B)
(
)
2
b
a
a
b
f b
x dx
+
+
∫
(C)
( )
2
b
a
b
a
− ∫f x dx
(D)
( )
2
b
a
a
b
f x dx
+ ∫
44 The value of
1
1
2
0
2
1
tan
1
x
dx
x
x
−
−
+
−
∫
is
(A) 1
(B) 0
(C) –1
(D)
4
π
Summary
� Integration is the inverse process of differentiation |
1 | 4095-4098 | 2
(sincos2
cos )
x
dx
x
x
+
∫
is equal to
(A)
–1
C
sin
cos
x
x
+
+
(B) log |sin
cos |
C
x
x
+
+
(C) log |sin
cos |
C
x
x
−
+
(D)
2
1
(sin
cos )
x
x
+
354
MATHEMATICS
43 If f (a + b – x) = f (x), then
( )
b
∫a x f x dx
is equal to
(A)
(
)
2
b
a
a
b
f b
x dx
+
−
∫
(B)
(
)
2
b
a
a
b
f b
x dx
+
+
∫
(C)
( )
2
b
a
b
a
− ∫f x dx
(D)
( )
2
b
a
a
b
f x dx
+ ∫
44 The value of
1
1
2
0
2
1
tan
1
x
dx
x
x
−
−
+
−
∫
is
(A) 1
(B) 0
(C) –1
(D)
4
π
Summary
� Integration is the inverse process of differentiation In the differential calculus,
we are given a function and we have to find the derivative or differential of
this function, but in the integral calculus, we are to find a function whose
differential is given |
1 | 4096-4099 | If f (a + b – x) = f (x), then
( )
b
∫a x f x dx
is equal to
(A)
(
)
2
b
a
a
b
f b
x dx
+
−
∫
(B)
(
)
2
b
a
a
b
f b
x dx
+
+
∫
(C)
( )
2
b
a
b
a
− ∫f x dx
(D)
( )
2
b
a
a
b
f x dx
+ ∫
44 The value of
1
1
2
0
2
1
tan
1
x
dx
x
x
−
−
+
−
∫
is
(A) 1
(B) 0
(C) –1
(D)
4
π
Summary
� Integration is the inverse process of differentiation In the differential calculus,
we are given a function and we have to find the derivative or differential of
this function, but in the integral calculus, we are to find a function whose
differential is given Thus, integration is a process which is the inverse of
differentiation |
1 | 4097-4100 | The value of
1
1
2
0
2
1
tan
1
x
dx
x
x
−
−
+
−
∫
is
(A) 1
(B) 0
(C) –1
(D)
4
π
Summary
� Integration is the inverse process of differentiation In the differential calculus,
we are given a function and we have to find the derivative or differential of
this function, but in the integral calculus, we are to find a function whose
differential is given Thus, integration is a process which is the inverse of
differentiation Let
F( )
( )
d
x
f x
dx
= |
1 | 4098-4101 | In the differential calculus,
we are given a function and we have to find the derivative or differential of
this function, but in the integral calculus, we are to find a function whose
differential is given Thus, integration is a process which is the inverse of
differentiation Let
F( )
( )
d
x
f x
dx
= Then we write
( )
F ( )
C
f x dx
x
=
+
∫ |
1 | 4099-4102 | Thus, integration is a process which is the inverse of
differentiation Let
F( )
( )
d
x
f x
dx
= Then we write
( )
F ( )
C
f x dx
x
=
+
∫ These integrals
are called indefinite integrals or general integrals, C is called constant of
integration |
1 | 4100-4103 | Let
F( )
( )
d
x
f x
dx
= Then we write
( )
F ( )
C
f x dx
x
=
+
∫ These integrals
are called indefinite integrals or general integrals, C is called constant of
integration All these integrals differ by a constant |
1 | 4101-4104 | Then we write
( )
F ( )
C
f x dx
x
=
+
∫ These integrals
are called indefinite integrals or general integrals, C is called constant of
integration All these integrals differ by a constant � From the geometric point of view, an indefinite integral is collection of family
of curves, each of which is obtained by translating one of the curves parallel
to itself upwards or downwards along the y-axis |
1 | 4102-4105 | These integrals
are called indefinite integrals or general integrals, C is called constant of
integration All these integrals differ by a constant � From the geometric point of view, an indefinite integral is collection of family
of curves, each of which is obtained by translating one of the curves parallel
to itself upwards or downwards along the y-axis � Some properties of indefinite integrals are as follows:
1 |
1 | 4103-4106 | All these integrals differ by a constant � From the geometric point of view, an indefinite integral is collection of family
of curves, each of which is obtained by translating one of the curves parallel
to itself upwards or downwards along the y-axis � Some properties of indefinite integrals are as follows:
1 [ ( )
( )]
( )
( )
f x
g x
dx
f x dx
g x dx
+
=
+
∫
∫
∫
2 |
1 | 4104-4107 | � From the geometric point of view, an indefinite integral is collection of family
of curves, each of which is obtained by translating one of the curves parallel
to itself upwards or downwards along the y-axis � Some properties of indefinite integrals are as follows:
1 [ ( )
( )]
( )
( )
f x
g x
dx
f x dx
g x dx
+
=
+
∫
∫
∫
2 For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
More generally, if f1, f2, f3, |
1 | 4105-4108 | � Some properties of indefinite integrals are as follows:
1 [ ( )
( )]
( )
( )
f x
g x
dx
f x dx
g x dx
+
=
+
∫
∫
∫
2 For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
More generally, if f1, f2, f3, , fn are functions and k1, k2, |
1 | 4106-4109 | [ ( )
( )]
( )
( )
f x
g x
dx
f x dx
g x dx
+
=
+
∫
∫
∫
2 For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
More generally, if f1, f2, f3, , fn are functions and k1, k2, ,kn are real
numbers |
1 | 4107-4110 | For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
More generally, if f1, f2, f3, , fn are functions and k1, k2, ,kn are real
numbers Then
1
1
2
2
[
( )
( ) |
1 | 4108-4111 | , fn are functions and k1, k2, ,kn are real
numbers Then
1
1
2
2
[
( )
( ) ( )]
n
n
k f
x
k f
x
k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( ) |
1 | 4109-4112 | ,kn are real
numbers Then
1
1
2
2
[
( )
( ) ( )]
n
n
k f
x
k f
x
k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( ) ( )
n
n
k
f
x dx
k
f
x dx
k
f
x dx
+
+
+
∫
∫
∫
INTEGRALS 355
� Some standard integrals
(i)
1
C
1
n
n
x
x dx
n
+
=
+
+
∫
, n ≠ – 1 |
1 | 4110-4113 | Then
1
1
2
2
[
( )
( ) ( )]
n
n
k f
x
k f
x
k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( ) ( )
n
n
k
f
x dx
k
f
x dx
k
f
x dx
+
+
+
∫
∫
∫
INTEGRALS 355
� Some standard integrals
(i)
1
C
1
n
n
x
x dx
n
+
=
+
+
∫
, n ≠ – 1 Particularly,
C
dx
=x
+
∫
(ii)
cos
sin
C
x dx
x
=
+
∫
(iii)
sin
– cos
C
x dx
x
=
+
∫
(iv)
sec2
tan
C
x dx
x
=
+
∫
(v)
cosec2
– cot
C
x dx
x
=
+
∫
(vi)
sec
tan
sec
C
x
x dx
x
=
+
∫
(vii)
cosec
cot
– cosec
C
x
x dx
x
=
+
∫
(viii)
1
2
sin
C
1
dx
x
x
−
=
+
−
∫
(ix)
1
2
cos
C
1
dx
x
x
−
= −
+
−
∫
(x)
1
2
tan
C
1
dx
x
x
−
=
+
+
∫
(xi)
1
2
cot
C
1
dx
x
x
−
= −
+
∫+
(xii)
C
x
x
e dx
=e
+
∫
(xiii)
C
log
x
x
a
a dx
a
=
+
∫
(xiv)
1
2
sec
C
1
dx
x
x x
−
=
+
−
∫
(xv)
1
2
cosec
C
1
dx
x
x x
−
= −
+
−
∫
(xvi)
1
log|
|
C
dx
x
x
=
+
∫
� Integration by partial fractions
Recall that a rational function is ratio of two polynomials of the form P( )
Q( )
x
x
,
where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0 |
1 | 4111-4114 | ( )]
n
n
k f
x
k f
x
k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( ) ( )
n
n
k
f
x dx
k
f
x dx
k
f
x dx
+
+
+
∫
∫
∫
INTEGRALS 355
� Some standard integrals
(i)
1
C
1
n
n
x
x dx
n
+
=
+
+
∫
, n ≠ – 1 Particularly,
C
dx
=x
+
∫
(ii)
cos
sin
C
x dx
x
=
+
∫
(iii)
sin
– cos
C
x dx
x
=
+
∫
(iv)
sec2
tan
C
x dx
x
=
+
∫
(v)
cosec2
– cot
C
x dx
x
=
+
∫
(vi)
sec
tan
sec
C
x
x dx
x
=
+
∫
(vii)
cosec
cot
– cosec
C
x
x dx
x
=
+
∫
(viii)
1
2
sin
C
1
dx
x
x
−
=
+
−
∫
(ix)
1
2
cos
C
1
dx
x
x
−
= −
+
−
∫
(x)
1
2
tan
C
1
dx
x
x
−
=
+
+
∫
(xi)
1
2
cot
C
1
dx
x
x
−
= −
+
∫+
(xii)
C
x
x
e dx
=e
+
∫
(xiii)
C
log
x
x
a
a dx
a
=
+
∫
(xiv)
1
2
sec
C
1
dx
x
x x
−
=
+
−
∫
(xv)
1
2
cosec
C
1
dx
x
x x
−
= −
+
−
∫
(xvi)
1
log|
|
C
dx
x
x
=
+
∫
� Integration by partial fractions
Recall that a rational function is ratio of two polynomials of the form P( )
Q( )
x
x
,
where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0 If degree of the
polynomial P (x) is greater than the degree of the polynomial Q (x), then we
may divide P (x) by Q (x) so that
1P ( )
P( )
T ( )
Q( )
Q( )
x
x
x
x
x
=
+
, where T(x) is a
polynomial in x and degree of P1(x) is less than the degree of Q(x) |
1 | 4112-4115 | ( )
n
n
k
f
x dx
k
f
x dx
k
f
x dx
+
+
+
∫
∫
∫
INTEGRALS 355
� Some standard integrals
(i)
1
C
1
n
n
x
x dx
n
+
=
+
+
∫
, n ≠ – 1 Particularly,
C
dx
=x
+
∫
(ii)
cos
sin
C
x dx
x
=
+
∫
(iii)
sin
– cos
C
x dx
x
=
+
∫
(iv)
sec2
tan
C
x dx
x
=
+
∫
(v)
cosec2
– cot
C
x dx
x
=
+
∫
(vi)
sec
tan
sec
C
x
x dx
x
=
+
∫
(vii)
cosec
cot
– cosec
C
x
x dx
x
=
+
∫
(viii)
1
2
sin
C
1
dx
x
x
−
=
+
−
∫
(ix)
1
2
cos
C
1
dx
x
x
−
= −
+
−
∫
(x)
1
2
tan
C
1
dx
x
x
−
=
+
+
∫
(xi)
1
2
cot
C
1
dx
x
x
−
= −
+
∫+
(xii)
C
x
x
e dx
=e
+
∫
(xiii)
C
log
x
x
a
a dx
a
=
+
∫
(xiv)
1
2
sec
C
1
dx
x
x x
−
=
+
−
∫
(xv)
1
2
cosec
C
1
dx
x
x x
−
= −
+
−
∫
(xvi)
1
log|
|
C
dx
x
x
=
+
∫
� Integration by partial fractions
Recall that a rational function is ratio of two polynomials of the form P( )
Q( )
x
x
,
where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0 If degree of the
polynomial P (x) is greater than the degree of the polynomial Q (x), then we
may divide P (x) by Q (x) so that
1P ( )
P( )
T ( )
Q( )
Q( )
x
x
x
x
x
=
+
, where T(x) is a
polynomial in x and degree of P1(x) is less than the degree of Q(x) T(x)
being polynomial can be easily integrated |
1 | 4113-4116 | Particularly,
C
dx
=x
+
∫
(ii)
cos
sin
C
x dx
x
=
+
∫
(iii)
sin
– cos
C
x dx
x
=
+
∫
(iv)
sec2
tan
C
x dx
x
=
+
∫
(v)
cosec2
– cot
C
x dx
x
=
+
∫
(vi)
sec
tan
sec
C
x
x dx
x
=
+
∫
(vii)
cosec
cot
– cosec
C
x
x dx
x
=
+
∫
(viii)
1
2
sin
C
1
dx
x
x
−
=
+
−
∫
(ix)
1
2
cos
C
1
dx
x
x
−
= −
+
−
∫
(x)
1
2
tan
C
1
dx
x
x
−
=
+
+
∫
(xi)
1
2
cot
C
1
dx
x
x
−
= −
+
∫+
(xii)
C
x
x
e dx
=e
+
∫
(xiii)
C
log
x
x
a
a dx
a
=
+
∫
(xiv)
1
2
sec
C
1
dx
x
x x
−
=
+
−
∫
(xv)
1
2
cosec
C
1
dx
x
x x
−
= −
+
−
∫
(xvi)
1
log|
|
C
dx
x
x
=
+
∫
� Integration by partial fractions
Recall that a rational function is ratio of two polynomials of the form P( )
Q( )
x
x
,
where P(x) and Q (x) are polynomials in x and Q (x) ≠ 0 If degree of the
polynomial P (x) is greater than the degree of the polynomial Q (x), then we
may divide P (x) by Q (x) so that
1P ( )
P( )
T ( )
Q( )
Q( )
x
x
x
x
x
=
+
, where T(x) is a
polynomial in x and degree of P1(x) is less than the degree of Q(x) T(x)
being polynomial can be easily integrated 1P ( )
Q( )
x
x can be integrated by
356
MATHEMATICS
expressing
1P ( )
Q( )
x
x as the sum of partial fractions of the following type:
1 |
1 | 4114-4117 | If degree of the
polynomial P (x) is greater than the degree of the polynomial Q (x), then we
may divide P (x) by Q (x) so that
1P ( )
P( )
T ( )
Q( )
Q( )
x
x
x
x
x
=
+
, where T(x) is a
polynomial in x and degree of P1(x) is less than the degree of Q(x) T(x)
being polynomial can be easily integrated 1P ( )
Q( )
x
x can be integrated by
356
MATHEMATICS
expressing
1P ( )
Q( )
x
x as the sum of partial fractions of the following type:
1 (
) (
)
px
q
x
a
x
b
+
−
−
=
A
B
x
a
x
b
+
−
−
, a ≠ b
2 |
1 | 4115-4118 | T(x)
being polynomial can be easily integrated 1P ( )
Q( )
x
x can be integrated by
356
MATHEMATICS
expressing
1P ( )
Q( )
x
x as the sum of partial fractions of the following type:
1 (
) (
)
px
q
x
a
x
b
+
−
−
=
A
B
x
a
x
b
+
−
−
, a ≠ b
2 2
(
)
px
q
x
a
+
−
=
2
A
B
(
)
x
a
x
a
+
−
−
3 |
1 | 4116-4119 | 1P ( )
Q( )
x
x can be integrated by
356
MATHEMATICS
expressing
1P ( )
Q( )
x
x as the sum of partial fractions of the following type:
1 (
) (
)
px
q
x
a
x
b
+
−
−
=
A
B
x
a
x
b
+
−
−
, a ≠ b
2 2
(
)
px
q
x
a
+
−
=
2
A
B
(
)
x
a
x
a
+
−
−
3 2
(
) (
) (
)
px
qx
r
x
a
x
b
x
c
+
+
−
−
−
=
A
B
C
x
a
x
b
x
c
+
+
−
−
−
4 |
1 | 4117-4120 | (
) (
)
px
q
x
a
x
b
+
−
−
=
A
B
x
a
x
b
+
−
−
, a ≠ b
2 2
(
)
px
q
x
a
+
−
=
2
A
B
(
)
x
a
x
a
+
−
−
3 2
(
) (
) (
)
px
qx
r
x
a
x
b
x
c
+
+
−
−
−
=
A
B
C
x
a
x
b
x
c
+
+
−
−
−
4 2
2
(
) (
)
px
qx
r
x
a
x
b
+
+
−
−
=
2
A
B
C
(
)
x
a
x
b
x
a
+
+
−
−
−
5 |
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