Chapter
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1 | 3518-3521 | 1
1
–tan
x
34 tan
sin
cos
x
x
x
35 (
)
2
1
log x
x
+
36 (
)
2
(
1)
log
x
x
x
x
+
+
37 |
1 | 3519-3522 | tan
sin
cos
x
x
x
35 (
)
2
1
log x
x
+
36 (
)
2
(
1)
log
x
x
x
x
+
+
37 (
)
3
1 4
sin tan
1
–
x
x
x8
+
Choose the correct answer in Exercises 38 and 39 |
1 | 3520-3523 | (
)
2
1
log x
x
+
36 (
)
2
(
1)
log
x
x
x
x
+
+
37 (
)
3
1 4
sin tan
1
–
x
x
x8
+
Choose the correct answer in Exercises 38 and 39 38 |
1 | 3521-3524 | (
)
2
(
1)
log
x
x
x
x
+
+
37 (
)
3
1 4
sin tan
1
–
x
x
x8
+
Choose the correct answer in Exercises 38 and 39 38 10
9
10
10
10 log
10
x
e
x
x
dx
x
+
+
∫
equals
(A) 10x – x10 + C
(B) 10x + x10 + C
(C) (10x – x10)–1 + C
(D) log (10x + x10) + C
39 |
1 | 3522-3525 | (
)
3
1 4
sin tan
1
–
x
x
x8
+
Choose the correct answer in Exercises 38 and 39 38 10
9
10
10
10 log
10
x
e
x
x
dx
x
+
+
∫
equals
(A) 10x – x10 + C
(B) 10x + x10 + C
(C) (10x – x10)–1 + C
(D) log (10x + x10) + C
39 2
2
equals
sin
xdxcos
x
∫
(A)
tan x + cot x + C
(B) tan x – cot x + C
(C)
tan x cot x + C
(D) tan x – cot 2x + C
7 |
1 | 3523-3526 | 38 10
9
10
10
10 log
10
x
e
x
x
dx
x
+
+
∫
equals
(A) 10x – x10 + C
(B) 10x + x10 + C
(C) (10x – x10)–1 + C
(D) log (10x + x10) + C
39 2
2
equals
sin
xdxcos
x
∫
(A)
tan x + cot x + C
(B) tan x – cot x + C
(C)
tan x cot x + C
(D) tan x – cot 2x + C
7 3 |
1 | 3524-3527 | 10
9
10
10
10 log
10
x
e
x
x
dx
x
+
+
∫
equals
(A) 10x – x10 + C
(B) 10x + x10 + C
(C) (10x – x10)–1 + C
(D) log (10x + x10) + C
39 2
2
equals
sin
xdxcos
x
∫
(A)
tan x + cot x + C
(B) tan x – cot x + C
(C)
tan x cot x + C
(D) tan x – cot 2x + C
7 3 2 Integration using trigonometric identities
When the integrand involves some trigonometric functions, we use some known identities
to find the integral as illustrated through the following example |
1 | 3525-3528 | 2
2
equals
sin
xdxcos
x
∫
(A)
tan x + cot x + C
(B) tan x – cot x + C
(C)
tan x cot x + C
(D) tan x – cot 2x + C
7 3 2 Integration using trigonometric identities
When the integrand involves some trigonometric functions, we use some known identities
to find the integral as illustrated through the following example Example 7 Find (i)
2
∫cos x dx
(ii) sin 2 cos 3
x
x dx
∫
(iii)
3
∫sin x dx
306
MATHEMATICS
Solution
(i)
Recall the identity cos 2x = 2 cos2 x – 1, which gives
cos2x = 1
cos 2
2
x
+
Therefore,
∫cos2
x dx = 1 (1+ cos 2 )
2
x dx
∫
= 1
1
cos 2
2
2
dx
x dx
+
∫
∫
=
1 sin 2
C
2
4
x
x
+
+
(ii)
Recall the identity sin x cos y = 1
2 [sin (x + y) + sin (x – y)]
(Why |
1 | 3526-3529 | 3 2 Integration using trigonometric identities
When the integrand involves some trigonometric functions, we use some known identities
to find the integral as illustrated through the following example Example 7 Find (i)
2
∫cos x dx
(ii) sin 2 cos 3
x
x dx
∫
(iii)
3
∫sin x dx
306
MATHEMATICS
Solution
(i)
Recall the identity cos 2x = 2 cos2 x – 1, which gives
cos2x = 1
cos 2
2
x
+
Therefore,
∫cos2
x dx = 1 (1+ cos 2 )
2
x dx
∫
= 1
1
cos 2
2
2
dx
x dx
+
∫
∫
=
1 sin 2
C
2
4
x
x
+
+
(ii)
Recall the identity sin x cos y = 1
2 [sin (x + y) + sin (x – y)]
(Why )
Then sin 2 cos 3
∫
x
xdx =
1
sin 5
sin
2
•
∫
∫
x dx
x dx
=
1
1 cos 5
cos
C
2
–5
x
x
+
+
=
1
1
cos 5
cos
C
10
2
–
x
x
+
+
(iii)
From the identity sin 3x = 3 sin x – 4 sin3 x, we find that
sin3x = 3sin
sin 3
4
x –
x
Therefore,
3
∫sin x dx
= 3
1
sin
sin 3
4
4
x dx –
x dx
∫
∫
=
3
1
–
cos
cos 3
C
4
12
x
x
+
+
Alternatively,
3
2
sin
sin
sin
x dx
x
x dx
=
∫
∫
=
(1– cos2
x) sin
x dx
∫
Put cos x = t so that – sin x dx = dt
Therefore,
3
∫sin x dx
=
(
)
1 – t2
dt
−∫
=
3
2
C
t3
–
dt
t dt
– t
+
=
+
+
∫
∫
=
3
1
cos
cos
C
3
–
x
x
+
+
Remark It can be shown using trigonometric identities that both answers are equivalent |
1 | 3527-3530 | 2 Integration using trigonometric identities
When the integrand involves some trigonometric functions, we use some known identities
to find the integral as illustrated through the following example Example 7 Find (i)
2
∫cos x dx
(ii) sin 2 cos 3
x
x dx
∫
(iii)
3
∫sin x dx
306
MATHEMATICS
Solution
(i)
Recall the identity cos 2x = 2 cos2 x – 1, which gives
cos2x = 1
cos 2
2
x
+
Therefore,
∫cos2
x dx = 1 (1+ cos 2 )
2
x dx
∫
= 1
1
cos 2
2
2
dx
x dx
+
∫
∫
=
1 sin 2
C
2
4
x
x
+
+
(ii)
Recall the identity sin x cos y = 1
2 [sin (x + y) + sin (x – y)]
(Why )
Then sin 2 cos 3
∫
x
xdx =
1
sin 5
sin
2
•
∫
∫
x dx
x dx
=
1
1 cos 5
cos
C
2
–5
x
x
+
+
=
1
1
cos 5
cos
C
10
2
–
x
x
+
+
(iii)
From the identity sin 3x = 3 sin x – 4 sin3 x, we find that
sin3x = 3sin
sin 3
4
x –
x
Therefore,
3
∫sin x dx
= 3
1
sin
sin 3
4
4
x dx –
x dx
∫
∫
=
3
1
–
cos
cos 3
C
4
12
x
x
+
+
Alternatively,
3
2
sin
sin
sin
x dx
x
x dx
=
∫
∫
=
(1– cos2
x) sin
x dx
∫
Put cos x = t so that – sin x dx = dt
Therefore,
3
∫sin x dx
=
(
)
1 – t2
dt
−∫
=
3
2
C
t3
–
dt
t dt
– t
+
=
+
+
∫
∫
=
3
1
cos
cos
C
3
–
x
x
+
+
Remark It can be shown using trigonometric identities that both answers are equivalent INTEGRALS 307
EXERCISE 7 |
1 | 3528-3531 | Example 7 Find (i)
2
∫cos x dx
(ii) sin 2 cos 3
x
x dx
∫
(iii)
3
∫sin x dx
306
MATHEMATICS
Solution
(i)
Recall the identity cos 2x = 2 cos2 x – 1, which gives
cos2x = 1
cos 2
2
x
+
Therefore,
∫cos2
x dx = 1 (1+ cos 2 )
2
x dx
∫
= 1
1
cos 2
2
2
dx
x dx
+
∫
∫
=
1 sin 2
C
2
4
x
x
+
+
(ii)
Recall the identity sin x cos y = 1
2 [sin (x + y) + sin (x – y)]
(Why )
Then sin 2 cos 3
∫
x
xdx =
1
sin 5
sin
2
•
∫
∫
x dx
x dx
=
1
1 cos 5
cos
C
2
–5
x
x
+
+
=
1
1
cos 5
cos
C
10
2
–
x
x
+
+
(iii)
From the identity sin 3x = 3 sin x – 4 sin3 x, we find that
sin3x = 3sin
sin 3
4
x –
x
Therefore,
3
∫sin x dx
= 3
1
sin
sin 3
4
4
x dx –
x dx
∫
∫
=
3
1
–
cos
cos 3
C
4
12
x
x
+
+
Alternatively,
3
2
sin
sin
sin
x dx
x
x dx
=
∫
∫
=
(1– cos2
x) sin
x dx
∫
Put cos x = t so that – sin x dx = dt
Therefore,
3
∫sin x dx
=
(
)
1 – t2
dt
−∫
=
3
2
C
t3
–
dt
t dt
– t
+
=
+
+
∫
∫
=
3
1
cos
cos
C
3
–
x
x
+
+
Remark It can be shown using trigonometric identities that both answers are equivalent INTEGRALS 307
EXERCISE 7 3
Find the integrals of the functions in Exercises 1 to 22:
1 |
1 | 3529-3532 | )
Then sin 2 cos 3
∫
x
xdx =
1
sin 5
sin
2
•
∫
∫
x dx
x dx
=
1
1 cos 5
cos
C
2
–5
x
x
+
+
=
1
1
cos 5
cos
C
10
2
–
x
x
+
+
(iii)
From the identity sin 3x = 3 sin x – 4 sin3 x, we find that
sin3x = 3sin
sin 3
4
x –
x
Therefore,
3
∫sin x dx
= 3
1
sin
sin 3
4
4
x dx –
x dx
∫
∫
=
3
1
–
cos
cos 3
C
4
12
x
x
+
+
Alternatively,
3
2
sin
sin
sin
x dx
x
x dx
=
∫
∫
=
(1– cos2
x) sin
x dx
∫
Put cos x = t so that – sin x dx = dt
Therefore,
3
∫sin x dx
=
(
)
1 – t2
dt
−∫
=
3
2
C
t3
–
dt
t dt
– t
+
=
+
+
∫
∫
=
3
1
cos
cos
C
3
–
x
x
+
+
Remark It can be shown using trigonometric identities that both answers are equivalent INTEGRALS 307
EXERCISE 7 3
Find the integrals of the functions in Exercises 1 to 22:
1 sin2 (2x + 5)
2 |
1 | 3530-3533 | INTEGRALS 307
EXERCISE 7 3
Find the integrals of the functions in Exercises 1 to 22:
1 sin2 (2x + 5)
2 sin 3x cos 4x
3 |
1 | 3531-3534 | 3
Find the integrals of the functions in Exercises 1 to 22:
1 sin2 (2x + 5)
2 sin 3x cos 4x
3 cos 2x cos 4x cos 6x
4 |
1 | 3532-3535 | sin2 (2x + 5)
2 sin 3x cos 4x
3 cos 2x cos 4x cos 6x
4 sin3 (2x + 1)
5 |
1 | 3533-3536 | sin 3x cos 4x
3 cos 2x cos 4x cos 6x
4 sin3 (2x + 1)
5 sin3 x cos3 x
6 |
1 | 3534-3537 | cos 2x cos 4x cos 6x
4 sin3 (2x + 1)
5 sin3 x cos3 x
6 sin x sin 2x sin 3x
7 |
1 | 3535-3538 | sin3 (2x + 1)
5 sin3 x cos3 x
6 sin x sin 2x sin 3x
7 sin 4x sin 8x
8 |
1 | 3536-3539 | sin3 x cos3 x
6 sin x sin 2x sin 3x
7 sin 4x sin 8x
8 1
cos
1
cos
–
xx
+
9 |
1 | 3537-3540 | sin x sin 2x sin 3x
7 sin 4x sin 8x
8 1
cos
1
cos
–
xx
+
9 cos
1
cos
x
x
+
10 |
1 | 3538-3541 | sin 4x sin 8x
8 1
cos
1
cos
–
xx
+
9 cos
1
cos
x
x
+
10 sin4 x
11 |
1 | 3539-3542 | 1
cos
1
cos
–
xx
+
9 cos
1
cos
x
x
+
10 sin4 x
11 cos4 2x
12 |
1 | 3540-3543 | cos
1
cos
x
x
+
10 sin4 x
11 cos4 2x
12 2
sin
1
cos
x
x
+
13 |
1 | 3541-3544 | sin4 x
11 cos4 2x
12 2
sin
1
cos
x
x
+
13 cos 2
cos 2
cos
x –cos
x –
α
α
14 |
1 | 3542-3545 | cos4 2x
12 2
sin
1
cos
x
x
+
13 cos 2
cos 2
cos
x –cos
x –
α
α
14 cos
sin
1
sin 2
x –
x
x
+
15 |
1 | 3543-3546 | 2
sin
1
cos
x
x
+
13 cos 2
cos 2
cos
x –cos
x –
α
α
14 cos
sin
1
sin 2
x –
x
x
+
15 tan3 2x sec 2x
16 |
1 | 3544-3547 | cos 2
cos 2
cos
x –cos
x –
α
α
14 cos
sin
1
sin 2
x –
x
x
+
15 tan3 2x sec 2x
16 tan4x
17 |
1 | 3545-3548 | cos
sin
1
sin 2
x –
x
x
+
15 tan3 2x sec 2x
16 tan4x
17 3
3
2
2
sin
cos
sin
cos
x
x
x
x
+
18 |
1 | 3546-3549 | tan3 2x sec 2x
16 tan4x
17 3
3
2
2
sin
cos
sin
cos
x
x
x
x
+
18 2
2
cos 2
2sin
xcos
x
x
+
19 |
1 | 3547-3550 | tan4x
17 3
3
2
2
sin
cos
sin
cos
x
x
x
x
+
18 2
2
cos 2
2sin
xcos
x
x
+
19 3
1
sin
xcos
x
20 |
1 | 3548-3551 | 3
3
2
2
sin
cos
sin
cos
x
x
x
x
+
18 2
2
cos 2
2sin
xcos
x
x
+
19 3
1
sin
xcos
x
20 (
)
2
cos 2
cos
sin
x
x
x
+
21 |
1 | 3549-3552 | 2
2
cos 2
2sin
xcos
x
x
+
19 3
1
sin
xcos
x
20 (
)
2
cos 2
cos
sin
x
x
x
+
21 sin – 1 (cos x)
22 |
1 | 3550-3553 | 3
1
sin
xcos
x
20 (
)
2
cos 2
cos
sin
x
x
x
+
21 sin – 1 (cos x)
22 1
cos (
) cos (
)
x – a
x – b
Choose the correct answer in Exercises 23 and 24 |
1 | 3551-3554 | (
)
2
cos 2
cos
sin
x
x
x
+
21 sin – 1 (cos x)
22 1
cos (
) cos (
)
x – a
x – b
Choose the correct answer in Exercises 23 and 24 23 |
1 | 3552-3555 | sin – 1 (cos x)
22 1
cos (
) cos (
)
x – a
x – b
Choose the correct answer in Exercises 23 and 24 23 2
2
2
2
sin
cos
is equal to
sin
xcos
x dx
x
x
−
∫
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) – tan x + cot x + C
(D) tan x + sec x + C
24 |
1 | 3553-3556 | 1
cos (
) cos (
)
x – a
x – b
Choose the correct answer in Exercises 23 and 24 23 2
2
2
2
sin
cos
is equal to
sin
xcos
x dx
x
x
−
∫
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) – tan x + cot x + C
(D) tan x + sec x + C
24 (12
)
equals
cos (
)
x
x
e
x
+e xdx
∫
(A) – cot (exx) + C
(B) tan (xex) + C
(C) tan (ex) + C
(D) cot (ex) + C
7 |
1 | 3554-3557 | 23 2
2
2
2
sin
cos
is equal to
sin
xcos
x dx
x
x
−
∫
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) – tan x + cot x + C
(D) tan x + sec x + C
24 (12
)
equals
cos (
)
x
x
e
x
+e xdx
∫
(A) – cot (exx) + C
(B) tan (xex) + C
(C) tan (ex) + C
(D) cot (ex) + C
7 4 Integrals of Some Particular Functions
In this section, we mention below some important formulae of integrals and apply them
for integrating many other related standard integrals:
(1) ∫
2
2
1
–
=
log
+ C
2
+
dx–
x
a
a
x
a
x
a
308
MATHEMATICS
(2) ∫
2
2
1
+
=
log
+ C
2
–
dx–
a
x
a
a
x
a
x
(3) ∫
– 1
2
2
1 tan
C
dx
x
=
+
a
a
x + a
(4) ∫
2
2
2
2 = log
+
–
+C
–
dx
x
x
a
x
a
(5) ∫
– 1
2
2 = sin
+ C
–
dx
ax
a
x
(6) ∫
2
2
2
2 = log
+
+
+ C
+
dx
x
x
a
x
a
We now prove the above results:
(1)
We have
2
12
1
(
) (
)
x – a
x
a
x – a
=
+
=
1
(
) – (
)
1
1
1
2
(
) (
)
2
x
a
x – a
–
a
x – a
x
a
a
x – a
x
a
+
=
+
+
Therefore,
2
2
1
2
dx
dx
dx
–
a
x – a
x
a
x – a
=
+
∫
∫
∫
=
[
]
1
log (
)|
log (
)|
C
2
| x – a –
| x
a
a
+
+
= 1 log
C
2
x – a
a
x
a +
+
(2)
In view of (1) above, we have
2
2
1
1
(
)
(
)
2
(
) (
)
–
a
x
a
x
a
a
x
a
x
a
x
+
+
−
=
+
−
= 1
1
1
2a
a
x
a
x
+
−
+
INTEGRALS 309
Therefore,
2
–2
dx
a
x
∫
=
1
2
dx
dx
a
a
x
a
x
+
−
+
∫
∫
= 1 [ log |
|
log |
|]
C
2
a
x
a
x
a −
−
+
+
+
=
1 log
C
2
a
x
a
a
+x
+
−
�Note The technique used in (1) will be explained in Section 7 |
1 | 3555-3558 | 2
2
2
2
sin
cos
is equal to
sin
xcos
x dx
x
x
−
∫
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) – tan x + cot x + C
(D) tan x + sec x + C
24 (12
)
equals
cos (
)
x
x
e
x
+e xdx
∫
(A) – cot (exx) + C
(B) tan (xex) + C
(C) tan (ex) + C
(D) cot (ex) + C
7 4 Integrals of Some Particular Functions
In this section, we mention below some important formulae of integrals and apply them
for integrating many other related standard integrals:
(1) ∫
2
2
1
–
=
log
+ C
2
+
dx–
x
a
a
x
a
x
a
308
MATHEMATICS
(2) ∫
2
2
1
+
=
log
+ C
2
–
dx–
a
x
a
a
x
a
x
(3) ∫
– 1
2
2
1 tan
C
dx
x
=
+
a
a
x + a
(4) ∫
2
2
2
2 = log
+
–
+C
–
dx
x
x
a
x
a
(5) ∫
– 1
2
2 = sin
+ C
–
dx
ax
a
x
(6) ∫
2
2
2
2 = log
+
+
+ C
+
dx
x
x
a
x
a
We now prove the above results:
(1)
We have
2
12
1
(
) (
)
x – a
x
a
x – a
=
+
=
1
(
) – (
)
1
1
1
2
(
) (
)
2
x
a
x – a
–
a
x – a
x
a
a
x – a
x
a
+
=
+
+
Therefore,
2
2
1
2
dx
dx
dx
–
a
x – a
x
a
x – a
=
+
∫
∫
∫
=
[
]
1
log (
)|
log (
)|
C
2
| x – a –
| x
a
a
+
+
= 1 log
C
2
x – a
a
x
a +
+
(2)
In view of (1) above, we have
2
2
1
1
(
)
(
)
2
(
) (
)
–
a
x
a
x
a
a
x
a
x
a
x
+
+
−
=
+
−
= 1
1
1
2a
a
x
a
x
+
−
+
INTEGRALS 309
Therefore,
2
–2
dx
a
x
∫
=
1
2
dx
dx
a
a
x
a
x
+
−
+
∫
∫
= 1 [ log |
|
log |
|]
C
2
a
x
a
x
a −
−
+
+
+
=
1 log
C
2
a
x
a
a
+x
+
−
�Note The technique used in (1) will be explained in Section 7 5 |
1 | 3556-3559 | (12
)
equals
cos (
)
x
x
e
x
+e xdx
∫
(A) – cot (exx) + C
(B) tan (xex) + C
(C) tan (ex) + C
(D) cot (ex) + C
7 4 Integrals of Some Particular Functions
In this section, we mention below some important formulae of integrals and apply them
for integrating many other related standard integrals:
(1) ∫
2
2
1
–
=
log
+ C
2
+
dx–
x
a
a
x
a
x
a
308
MATHEMATICS
(2) ∫
2
2
1
+
=
log
+ C
2
–
dx–
a
x
a
a
x
a
x
(3) ∫
– 1
2
2
1 tan
C
dx
x
=
+
a
a
x + a
(4) ∫
2
2
2
2 = log
+
–
+C
–
dx
x
x
a
x
a
(5) ∫
– 1
2
2 = sin
+ C
–
dx
ax
a
x
(6) ∫
2
2
2
2 = log
+
+
+ C
+
dx
x
x
a
x
a
We now prove the above results:
(1)
We have
2
12
1
(
) (
)
x – a
x
a
x – a
=
+
=
1
(
) – (
)
1
1
1
2
(
) (
)
2
x
a
x – a
–
a
x – a
x
a
a
x – a
x
a
+
=
+
+
Therefore,
2
2
1
2
dx
dx
dx
–
a
x – a
x
a
x – a
=
+
∫
∫
∫
=
[
]
1
log (
)|
log (
)|
C
2
| x – a –
| x
a
a
+
+
= 1 log
C
2
x – a
a
x
a +
+
(2)
In view of (1) above, we have
2
2
1
1
(
)
(
)
2
(
) (
)
–
a
x
a
x
a
a
x
a
x
a
x
+
+
−
=
+
−
= 1
1
1
2a
a
x
a
x
+
−
+
INTEGRALS 309
Therefore,
2
–2
dx
a
x
∫
=
1
2
dx
dx
a
a
x
a
x
+
−
+
∫
∫
= 1 [ log |
|
log |
|]
C
2
a
x
a
x
a −
−
+
+
+
=
1 log
C
2
a
x
a
a
+x
+
−
�Note The technique used in (1) will be explained in Section 7 5 (3) Put x = a tan θ |
1 | 3557-3560 | 4 Integrals of Some Particular Functions
In this section, we mention below some important formulae of integrals and apply them
for integrating many other related standard integrals:
(1) ∫
2
2
1
–
=
log
+ C
2
+
dx–
x
a
a
x
a
x
a
308
MATHEMATICS
(2) ∫
2
2
1
+
=
log
+ C
2
–
dx–
a
x
a
a
x
a
x
(3) ∫
– 1
2
2
1 tan
C
dx
x
=
+
a
a
x + a
(4) ∫
2
2
2
2 = log
+
–
+C
–
dx
x
x
a
x
a
(5) ∫
– 1
2
2 = sin
+ C
–
dx
ax
a
x
(6) ∫
2
2
2
2 = log
+
+
+ C
+
dx
x
x
a
x
a
We now prove the above results:
(1)
We have
2
12
1
(
) (
)
x – a
x
a
x – a
=
+
=
1
(
) – (
)
1
1
1
2
(
) (
)
2
x
a
x – a
–
a
x – a
x
a
a
x – a
x
a
+
=
+
+
Therefore,
2
2
1
2
dx
dx
dx
–
a
x – a
x
a
x – a
=
+
∫
∫
∫
=
[
]
1
log (
)|
log (
)|
C
2
| x – a –
| x
a
a
+
+
= 1 log
C
2
x – a
a
x
a +
+
(2)
In view of (1) above, we have
2
2
1
1
(
)
(
)
2
(
) (
)
–
a
x
a
x
a
a
x
a
x
a
x
+
+
−
=
+
−
= 1
1
1
2a
a
x
a
x
+
−
+
INTEGRALS 309
Therefore,
2
–2
dx
a
x
∫
=
1
2
dx
dx
a
a
x
a
x
+
−
+
∫
∫
= 1 [ log |
|
log |
|]
C
2
a
x
a
x
a −
−
+
+
+
=
1 log
C
2
a
x
a
a
+x
+
−
�Note The technique used in (1) will be explained in Section 7 5 (3) Put x = a tan θ Then dx = a sec2 θ dθ |
1 | 3558-3561 | 5 (3) Put x = a tan θ Then dx = a sec2 θ dθ Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
secθ
tan
a
d
a
+a
∫
=
1
1
1
1
θ
θ
C
tan
C
– x
d
a
a
a
a
=
+
=
+
∫
(4) Let x = a secθ |
1 | 3559-3562 | (3) Put x = a tan θ Then dx = a sec2 θ dθ Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
secθ
tan
a
d
a
+a
∫
=
1
1
1
1
θ
θ
C
tan
C
– x
d
a
a
a
a
=
+
=
+
∫
(4) Let x = a secθ Then dx = a secθ tan θ d θ |
1 | 3560-3563 | Then dx = a sec2 θ dθ Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
secθ
tan
a
d
a
+a
∫
=
1
1
1
1
θ
θ
C
tan
C
– x
d
a
a
a
a
=
+
=
+
∫
(4) Let x = a secθ Then dx = a secθ tan θ d θ Therefore,
2
2
dx
x
−a
∫
=
2
2
2
secθ tanθ θ
sec θ
a
d
a
−a
∫
=
1
secθ θ
d =log secθ + tanθ + C
∫
=
2
1
2
log
1
C
x
x –
a
a
+
+
=
2
2
1
log
log
C
x
x – a
a
+
−
+
=
2
2
log
+ C
x
x – a
+
, where C = C1 – log |a|
(5) Let x = a sinθ |
1 | 3561-3564 | Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
secθ
tan
a
d
a
+a
∫
=
1
1
1
1
θ
θ
C
tan
C
– x
d
a
a
a
a
=
+
=
+
∫
(4) Let x = a secθ Then dx = a secθ tan θ d θ Therefore,
2
2
dx
x
−a
∫
=
2
2
2
secθ tanθ θ
sec θ
a
d
a
−a
∫
=
1
secθ θ
d =log secθ + tanθ + C
∫
=
2
1
2
log
1
C
x
x –
a
a
+
+
=
2
2
1
log
log
C
x
x – a
a
+
−
+
=
2
2
log
+ C
x
x – a
+
, where C = C1 – log |a|
(5) Let x = a sinθ Then dx = a cosθ dθ |
1 | 3562-3565 | Then dx = a secθ tan θ d θ Therefore,
2
2
dx
x
−a
∫
=
2
2
2
secθ tanθ θ
sec θ
a
d
a
−a
∫
=
1
secθ θ
d =log secθ + tanθ + C
∫
=
2
1
2
log
1
C
x
x –
a
a
+
+
=
2
2
1
log
log
C
x
x – a
a
+
−
+
=
2
2
log
+ C
x
x – a
+
, where C = C1 – log |a|
(5) Let x = a sinθ Then dx = a cosθ dθ Therefore,
2
2
dx
a
−x
∫
=
2
2
2
θ θ
θ
cos
sin
a
d
a – a
∫
=
θ = θ + C = sin1
C
–
x
d
a
+
∫
(6) Let x = a tanθ |
1 | 3563-3566 | Therefore,
2
2
dx
x
−a
∫
=
2
2
2
secθ tanθ θ
sec θ
a
d
a
−a
∫
=
1
secθ θ
d =log secθ + tanθ + C
∫
=
2
1
2
log
1
C
x
x –
a
a
+
+
=
2
2
1
log
log
C
x
x – a
a
+
−
+
=
2
2
log
+ C
x
x – a
+
, where C = C1 – log |a|
(5) Let x = a sinθ Then dx = a cosθ dθ Therefore,
2
2
dx
a
−x
∫
=
2
2
2
θ θ
θ
cos
sin
a
d
a – a
∫
=
θ = θ + C = sin1
C
–
x
d
a
+
∫
(6) Let x = a tanθ Then dx = a sec2θ dθ |
1 | 3564-3567 | Then dx = a cosθ dθ Therefore,
2
2
dx
a
−x
∫
=
2
2
2
θ θ
θ
cos
sin
a
d
a – a
∫
=
θ = θ + C = sin1
C
–
x
d
a
+
∫
(6) Let x = a tanθ Then dx = a sec2θ dθ Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
θ
sec
tan
a
d
a
+a
∫
=
1
θ
θ
secθ θ = log (sec
tan )
C
d
+
+
∫
310
MATHEMATICS
=
2
1
2
log
1
C
x
x
a
a
+
+
+
=
2
1
log
log
C
x
x
a
|a|
2
+
+
−
+
=
2
log
C
x
x
a2
+
+
+
, where C = C1 – log |a|
Applying these standard formulae, we now obtain some more formulae which
are useful from applications point of view and can be applied directly to evaluate
other integrals |
1 | 3565-3568 | Therefore,
2
2
dx
a
−x
∫
=
2
2
2
θ θ
θ
cos
sin
a
d
a – a
∫
=
θ = θ + C = sin1
C
–
x
d
a
+
∫
(6) Let x = a tanθ Then dx = a sec2θ dθ Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
θ
sec
tan
a
d
a
+a
∫
=
1
θ
θ
secθ θ = log (sec
tan )
C
d
+
+
∫
310
MATHEMATICS
=
2
1
2
log
1
C
x
x
a
a
+
+
+
=
2
1
log
log
C
x
x
a
|a|
2
+
+
−
+
=
2
log
C
x
x
a2
+
+
+
, where C = C1 – log |a|
Applying these standard formulae, we now obtain some more formulae which
are useful from applications point of view and can be applied directly to evaluate
other integrals (7)
To find the integral
2
dx
ax
bx
c
+
+
∫
, we write
ax 2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
–
a
a
a
a
a
+
+
=
+
+
Now, put
b2
x
t
a
+
= so that dx = dt and writing
2
2
42
c
–b
k
a
a
= ± |
1 | 3566-3569 | Then dx = a sec2θ dθ Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
θ
sec
tan
a
d
a
+a
∫
=
1
θ
θ
secθ θ = log (sec
tan )
C
d
+
+
∫
310
MATHEMATICS
=
2
1
2
log
1
C
x
x
a
a
+
+
+
=
2
1
log
log
C
x
x
a
|a|
2
+
+
−
+
=
2
log
C
x
x
a2
+
+
+
, where C = C1 – log |a|
Applying these standard formulae, we now obtain some more formulae which
are useful from applications point of view and can be applied directly to evaluate
other integrals (7)
To find the integral
2
dx
ax
bx
c
+
+
∫
, we write
ax 2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
–
a
a
a
a
a
+
+
=
+
+
Now, put
b2
x
t
a
+
= so that dx = dt and writing
2
2
42
c
–b
k
a
a
= ± We find the
integral reduced to the form
2
2
1
dt
a
t
∫±k
depending upon the sign of
2
42
c
a–b
a
and hence can be evaluated |
1 | 3567-3570 | Therefore,
2
2
dx
x
+a
∫
=
2
2
2
2
θ
θ
θ
sec
tan
a
d
a
+a
∫
=
1
θ
θ
secθ θ = log (sec
tan )
C
d
+
+
∫
310
MATHEMATICS
=
2
1
2
log
1
C
x
x
a
a
+
+
+
=
2
1
log
log
C
x
x
a
|a|
2
+
+
−
+
=
2
log
C
x
x
a2
+
+
+
, where C = C1 – log |a|
Applying these standard formulae, we now obtain some more formulae which
are useful from applications point of view and can be applied directly to evaluate
other integrals (7)
To find the integral
2
dx
ax
bx
c
+
+
∫
, we write
ax 2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
–
a
a
a
a
a
+
+
=
+
+
Now, put
b2
x
t
a
+
= so that dx = dt and writing
2
2
42
c
–b
k
a
a
= ± We find the
integral reduced to the form
2
2
1
dt
a
t
∫±k
depending upon the sign of
2
42
c
a–b
a
and hence can be evaluated (8)
To find the integral of the type
2
dx
ax
bx
c
+
+
∫
, proceeding as in (7), we
obtain the integral using the standard formulae |
1 | 3568-3571 | (7)
To find the integral
2
dx
ax
bx
c
+
+
∫
, we write
ax 2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
–
a
a
a
a
a
+
+
=
+
+
Now, put
b2
x
t
a
+
= so that dx = dt and writing
2
2
42
c
–b
k
a
a
= ± We find the
integral reduced to the form
2
2
1
dt
a
t
∫±k
depending upon the sign of
2
42
c
a–b
a
and hence can be evaluated (8)
To find the integral of the type
2
dx
ax
bx
c
+
+
∫
, proceeding as in (7), we
obtain the integral using the standard formulae (9)
To find the integral of the type
2
px
q
dx
ax
bx
c
++
+
∫
, where p, q, a, b, c are
constants, we are to find real numbers A, B such that
2
+
= A
(
) + B = A (2
) + B
d
px
q
ax
bx
c
ax
b
dx
+
+
+
To determine A and B, we equate from both sides the coefficients of x and the
constant terms |
1 | 3569-3572 | We find the
integral reduced to the form
2
2
1
dt
a
t
∫±k
depending upon the sign of
2
42
c
a–b
a
and hence can be evaluated (8)
To find the integral of the type
2
dx
ax
bx
c
+
+
∫
, proceeding as in (7), we
obtain the integral using the standard formulae (9)
To find the integral of the type
2
px
q
dx
ax
bx
c
++
+
∫
, where p, q, a, b, c are
constants, we are to find real numbers A, B such that
2
+
= A
(
) + B = A (2
) + B
d
px
q
ax
bx
c
ax
b
dx
+
+
+
To determine A and B, we equate from both sides the coefficients of x and the
constant terms A and B are thus obtained and hence the integral is reduced to
one of the known forms |
1 | 3570-3573 | (8)
To find the integral of the type
2
dx
ax
bx
c
+
+
∫
, proceeding as in (7), we
obtain the integral using the standard formulae (9)
To find the integral of the type
2
px
q
dx
ax
bx
c
++
+
∫
, where p, q, a, b, c are
constants, we are to find real numbers A, B such that
2
+
= A
(
) + B = A (2
) + B
d
px
q
ax
bx
c
ax
b
dx
+
+
+
To determine A and B, we equate from both sides the coefficients of x and the
constant terms A and B are thus obtained and hence the integral is reduced to
one of the known forms INTEGRALS 311
(10) For the evaluation of the integral of the type
2
(
)
px
q dx
ax
bx
c
+
+
+
∫
, we proceed
as in (9) and transform the integral into known standard forms |
1 | 3571-3574 | (9)
To find the integral of the type
2
px
q
dx
ax
bx
c
++
+
∫
, where p, q, a, b, c are
constants, we are to find real numbers A, B such that
2
+
= A
(
) + B = A (2
) + B
d
px
q
ax
bx
c
ax
b
dx
+
+
+
To determine A and B, we equate from both sides the coefficients of x and the
constant terms A and B are thus obtained and hence the integral is reduced to
one of the known forms INTEGRALS 311
(10) For the evaluation of the integral of the type
2
(
)
px
q dx
ax
bx
c
+
+
+
∫
, we proceed
as in (9) and transform the integral into known standard forms Let us illustrate the above methods by some examples |
1 | 3572-3575 | A and B are thus obtained and hence the integral is reduced to
one of the known forms INTEGRALS 311
(10) For the evaluation of the integral of the type
2
(
)
px
q dx
ax
bx
c
+
+
+
∫
, we proceed
as in (9) and transform the integral into known standard forms Let us illustrate the above methods by some examples Example 8 Find the following integrals:
(i)
2
16
dx
∫x −
(ii)
2
2
dx
x
−x
∫
Solution
(i)
We have
2
2
2
16
4
dx
dx
x
x –
=
−
∫
∫
=
4
log
C
8
xx –4
1
+
+
[by 7 |
1 | 3573-3576 | INTEGRALS 311
(10) For the evaluation of the integral of the type
2
(
)
px
q dx
ax
bx
c
+
+
+
∫
, we proceed
as in (9) and transform the integral into known standard forms Let us illustrate the above methods by some examples Example 8 Find the following integrals:
(i)
2
16
dx
∫x −
(ii)
2
2
dx
x
−x
∫
Solution
(i)
We have
2
2
2
16
4
dx
dx
x
x –
=
−
∫
∫
=
4
log
C
8
xx –4
1
+
+
[by 7 4 (1)]
(ii)
(
)
2
2
2
1
1
=
−
∫
∫
dx
dx
x
x
– x –
Put x – 1 = t |
1 | 3574-3577 | Let us illustrate the above methods by some examples Example 8 Find the following integrals:
(i)
2
16
dx
∫x −
(ii)
2
2
dx
x
−x
∫
Solution
(i)
We have
2
2
2
16
4
dx
dx
x
x –
=
−
∫
∫
=
4
log
C
8
xx –4
1
+
+
[by 7 4 (1)]
(ii)
(
)
2
2
2
1
1
=
−
∫
∫
dx
dx
x
x
– x –
Put x – 1 = t Then dx = dt |
1 | 3575-3578 | Example 8 Find the following integrals:
(i)
2
16
dx
∫x −
(ii)
2
2
dx
x
−x
∫
Solution
(i)
We have
2
2
2
16
4
dx
dx
x
x –
=
−
∫
∫
=
4
log
C
8
xx –4
1
+
+
[by 7 4 (1)]
(ii)
(
)
2
2
2
1
1
=
−
∫
∫
dx
dx
x
x
– x –
Put x – 1 = t Then dx = dt Therefore,
2
2
dx
x
−x
∫
=
2
1
dt
– t
∫
=
sin1
( )
C
–
t +
[by 7 |
1 | 3576-3579 | 4 (1)]
(ii)
(
)
2
2
2
1
1
=
−
∫
∫
dx
dx
x
x
– x –
Put x – 1 = t Then dx = dt Therefore,
2
2
dx
x
−x
∫
=
2
1
dt
– t
∫
=
sin1
( )
C
–
t +
[by 7 4 (5)]
=
sin1
( –1)
C
–
x
+
Example 9 Find the following integrals :
(i)
2
6
13
dx
x
−x
+
∫
(ii)
32
13
10
dx
x
x
+
−
∫
(iii)
52
2
dx
x
x
−
∫
Solution
(i)
We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4
So,
6
13
dx
x
2 −x
+
∫
=
(
)
2
2
1
3
2
dx
x –
+
∫
Let
x – 3 = t |
1 | 3577-3580 | Then dx = dt Therefore,
2
2
dx
x
−x
∫
=
2
1
dt
– t
∫
=
sin1
( )
C
–
t +
[by 7 4 (5)]
=
sin1
( –1)
C
–
x
+
Example 9 Find the following integrals :
(i)
2
6
13
dx
x
−x
+
∫
(ii)
32
13
10
dx
x
x
+
−
∫
(iii)
52
2
dx
x
x
−
∫
Solution
(i)
We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4
So,
6
13
dx
x
2 −x
+
∫
=
(
)
2
2
1
3
2
dx
x –
+
∫
Let
x – 3 = t Then dx = dt
Therefore,
6
13
dx
x
2 −x
+
∫
=
1
2
2
1 tan
C
2
2
2
–
dt
t
t
=
+
+
∫
[by 7 |
1 | 3578-3581 | Therefore,
2
2
dx
x
−x
∫
=
2
1
dt
– t
∫
=
sin1
( )
C
–
t +
[by 7 4 (5)]
=
sin1
( –1)
C
–
x
+
Example 9 Find the following integrals :
(i)
2
6
13
dx
x
−x
+
∫
(ii)
32
13
10
dx
x
x
+
−
∫
(iii)
52
2
dx
x
x
−
∫
Solution
(i)
We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4
So,
6
13
dx
x
2 −x
+
∫
=
(
)
2
2
1
3
2
dx
x –
+
∫
Let
x – 3 = t Then dx = dt
Therefore,
6
13
dx
x
2 −x
+
∫
=
1
2
2
1 tan
C
2
2
2
–
dt
t
t
=
+
+
∫
[by 7 4 (3)]
=
1
1
3
tan
C
2
2
–
x –
+
312
MATHEMATICS
(ii)
The given integral is of the form 7 |
1 | 3579-3582 | 4 (5)]
=
sin1
( –1)
C
–
x
+
Example 9 Find the following integrals :
(i)
2
6
13
dx
x
−x
+
∫
(ii)
32
13
10
dx
x
x
+
−
∫
(iii)
52
2
dx
x
x
−
∫
Solution
(i)
We have x2 – 6x + 13 = x2 – 6x + 32 – 32 + 13 = (x – 3)2 + 4
So,
6
13
dx
x
2 −x
+
∫
=
(
)
2
2
1
3
2
dx
x –
+
∫
Let
x – 3 = t Then dx = dt
Therefore,
6
13
dx
x
2 −x
+
∫
=
1
2
2
1 tan
C
2
2
2
–
dt
t
t
=
+
+
∫
[by 7 4 (3)]
=
1
1
3
tan
C
2
2
–
x –
+
312
MATHEMATICS
(ii)
The given integral is of the form 7 4 (7) |
1 | 3580-3583 | Then dx = dt
Therefore,
6
13
dx
x
2 −x
+
∫
=
1
2
2
1 tan
C
2
2
2
–
dt
t
t
=
+
+
∫
[by 7 4 (3)]
=
1
1
3
tan
C
2
2
–
x –
+
312
MATHEMATICS
(ii)
The given integral is of the form 7 4 (7) We write the denominator of the integrand,
32
13
10
x
x –
+
=
2
13
10
3
3
3
x
x
–
+
=
2
2
13
17
3
6
6
x
–
+
(completing the square)
Thus
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
13
17
6
6
dx
x
+
−
∫
Put
13
6
x
t
+
= |
1 | 3581-3584 | 4 (3)]
=
1
1
3
tan
C
2
2
–
x –
+
312
MATHEMATICS
(ii)
The given integral is of the form 7 4 (7) We write the denominator of the integrand,
32
13
10
x
x –
+
=
2
13
10
3
3
3
x
x
–
+
=
2
2
13
17
3
6
6
x
–
+
(completing the square)
Thus
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
13
17
6
6
dx
x
+
−
∫
Put
13
6
x
t
+
= Then dx = dt |
1 | 3582-3585 | 4 (7) We write the denominator of the integrand,
32
13
10
x
x –
+
=
2
13
10
3
3
3
x
x
–
+
=
2
2
13
17
3
6
6
x
–
+
(completing the square)
Thus
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
13
17
6
6
dx
x
+
−
∫
Put
13
6
x
t
+
= Then dx = dt Therefore,
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
17
6
dt
t
−
∫
=
1
17
1
6
log
C
17
17
3
2
6
6
t –
t
+
× ×
+
[by 7 |
1 | 3583-3586 | We write the denominator of the integrand,
32
13
10
x
x –
+
=
2
13
10
3
3
3
x
x
–
+
=
2
2
13
17
3
6
6
x
–
+
(completing the square)
Thus
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
13
17
6
6
dx
x
+
−
∫
Put
13
6
x
t
+
= Then dx = dt Therefore,
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
17
6
dt
t
−
∫
=
1
17
1
6
log
C
17
17
3
2
6
6
t –
t
+
× ×
+
[by 7 4 (i)]
=
1
13
17
1
6
6
log
C
13
17
17
6
6
x
–
x
+
+
+
+
=
1
1
6
4
log
C
17
6
xx30
−
+
+
=
1
1
3
2
1
1
log
C
log
17
5
17
3
xx
−
+
+
+
=
1
3
2
log
C
17
xx5
−
+
+
, where C =
1
1
1
C
17log
3
+
INTEGRALS 313
(iii)
We have
2
2
5
2
5
5
dx
dx
x
x
x
x –
2
=
−
∫
∫
=
2
2
1
5
1
1
5
5
dx
x –
–
∫
(completing the square)
Put
1
5
x –
t
= |
1 | 3584-3587 | Then dx = dt Therefore,
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
17
6
dt
t
−
∫
=
1
17
1
6
log
C
17
17
3
2
6
6
t –
t
+
× ×
+
[by 7 4 (i)]
=
1
13
17
1
6
6
log
C
13
17
17
6
6
x
–
x
+
+
+
+
=
1
1
6
4
log
C
17
6
xx30
−
+
+
=
1
1
3
2
1
1
log
C
log
17
5
17
3
xx
−
+
+
+
=
1
3
2
log
C
17
xx5
−
+
+
, where C =
1
1
1
C
17log
3
+
INTEGRALS 313
(iii)
We have
2
2
5
2
5
5
dx
dx
x
x
x
x –
2
=
−
∫
∫
=
2
2
1
5
1
1
5
5
dx
x –
–
∫
(completing the square)
Put
1
5
x –
t
= Then dx = dt |
1 | 3585-3588 | Therefore,
3
13
10
dx
x
x
2 +
−
∫
=
2
2
31
17
6
dt
t
−
∫
=
1
17
1
6
log
C
17
17
3
2
6
6
t –
t
+
× ×
+
[by 7 4 (i)]
=
1
13
17
1
6
6
log
C
13
17
17
6
6
x
–
x
+
+
+
+
=
1
1
6
4
log
C
17
6
xx30
−
+
+
=
1
1
3
2
1
1
log
C
log
17
5
17
3
xx
−
+
+
+
=
1
3
2
log
C
17
xx5
−
+
+
, where C =
1
1
1
C
17log
3
+
INTEGRALS 313
(iii)
We have
2
2
5
2
5
5
dx
dx
x
x
x
x –
2
=
−
∫
∫
=
2
2
1
5
1
1
5
5
dx
x –
–
∫
(completing the square)
Put
1
5
x –
t
= Then dx = dt Therefore,
5
2
dx
x
x
2 −
∫
=
2
2
1
5
1
5
dt
t –
∫
=
2
2
1
1
log
C
5
5
t
t –
+
+
[by 7 |
1 | 3586-3589 | 4 (i)]
=
1
13
17
1
6
6
log
C
13
17
17
6
6
x
–
x
+
+
+
+
=
1
1
6
4
log
C
17
6
xx30
−
+
+
=
1
1
3
2
1
1
log
C
log
17
5
17
3
xx
−
+
+
+
=
1
3
2
log
C
17
xx5
−
+
+
, where C =
1
1
1
C
17log
3
+
INTEGRALS 313
(iii)
We have
2
2
5
2
5
5
dx
dx
x
x
x
x –
2
=
−
∫
∫
=
2
2
1
5
1
1
5
5
dx
x –
–
∫
(completing the square)
Put
1
5
x –
t
= Then dx = dt Therefore,
5
2
dx
x
x
2 −
∫
=
2
2
1
5
1
5
dt
t –
∫
=
2
2
1
1
log
C
5
5
t
t –
+
+
[by 7 4 (4)]
=
2
1
1
2
log
C
5
5
5
x
x –
x –
+
+
Example 10 Find the following integrals:
(i)
2
2
6
5
x
dx
x
x
2
++
+
∫
(ii)
2
3
5
4
x
dx
x
x
+
−
+
∫
Solution
(i)
Using the formula 7 |
1 | 3587-3590 | Then dx = dt Therefore,
5
2
dx
x
x
2 −
∫
=
2
2
1
5
1
5
dt
t –
∫
=
2
2
1
1
log
C
5
5
t
t –
+
+
[by 7 4 (4)]
=
2
1
1
2
log
C
5
5
5
x
x –
x –
+
+
Example 10 Find the following integrals:
(i)
2
2
6
5
x
dx
x
x
2
++
+
∫
(ii)
2
3
5
4
x
dx
x
x
+
−
+
∫
Solution
(i)
Using the formula 7 4 (9), we express
x + 2 =
(
)
2
A
2
6
5
B
d
x
x
dx
+
+
+
= A (4
6)
B
x +
+
Equating the coefficients of x and the constant terms from both sides, we get
4A = 1 and 6A + B = 2 or A = 1
4 and B = 1
2 |
1 | 3588-3591 | Therefore,
5
2
dx
x
x
2 −
∫
=
2
2
1
5
1
5
dt
t –
∫
=
2
2
1
1
log
C
5
5
t
t –
+
+
[by 7 4 (4)]
=
2
1
1
2
log
C
5
5
5
x
x –
x –
+
+
Example 10 Find the following integrals:
(i)
2
2
6
5
x
dx
x
x
2
++
+
∫
(ii)
2
3
5
4
x
dx
x
x
+
−
+
∫
Solution
(i)
Using the formula 7 4 (9), we express
x + 2 =
(
)
2
A
2
6
5
B
d
x
x
dx
+
+
+
= A (4
6)
B
x +
+
Equating the coefficients of x and the constant terms from both sides, we get
4A = 1 and 6A + B = 2 or A = 1
4 and B = 1
2 Therefore,
2
2
6
5
xx
x
2
++
+
∫
=
1
4
6
1
4
2
2
6
5
2
6
5
x
dx
dx
x
x
x
x
2
2
+
+
+
+
+
+
∫
∫
=
1
2
1
1
I
I
4
2
+
(say) |
1 | 3589-3592 | 4 (4)]
=
2
1
1
2
log
C
5
5
5
x
x –
x –
+
+
Example 10 Find the following integrals:
(i)
2
2
6
5
x
dx
x
x
2
++
+
∫
(ii)
2
3
5
4
x
dx
x
x
+
−
+
∫
Solution
(i)
Using the formula 7 4 (9), we express
x + 2 =
(
)
2
A
2
6
5
B
d
x
x
dx
+
+
+
= A (4
6)
B
x +
+
Equating the coefficients of x and the constant terms from both sides, we get
4A = 1 and 6A + B = 2 or A = 1
4 and B = 1
2 Therefore,
2
2
6
5
xx
x
2
++
+
∫
=
1
4
6
1
4
2
2
6
5
2
6
5
x
dx
dx
x
x
x
x
2
2
+
+
+
+
+
+
∫
∫
=
1
2
1
1
I
I
4
2
+
(say) (1)
314
MATHEMATICS
In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt
Therefore,
I1 =
1
log
C
dt
t
t
=
+
∫
=
2
1
log | 2
6
5|
C
x
x
+
+
+ |
1 | 3590-3593 | 4 (9), we express
x + 2 =
(
)
2
A
2
6
5
B
d
x
x
dx
+
+
+
= A (4
6)
B
x +
+
Equating the coefficients of x and the constant terms from both sides, we get
4A = 1 and 6A + B = 2 or A = 1
4 and B = 1
2 Therefore,
2
2
6
5
xx
x
2
++
+
∫
=
1
4
6
1
4
2
2
6
5
2
6
5
x
dx
dx
x
x
x
x
2
2
+
+
+
+
+
+
∫
∫
=
1
2
1
1
I
I
4
2
+
(say) (1)
314
MATHEMATICS
In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt
Therefore,
I1 =
1
log
C
dt
t
t
=
+
∫
=
2
1
log | 2
6
5|
C
x
x
+
+
+ (2)
and
I2 =
2
2
1
5
2
2
6
5
3
2
dx
dx
x
x
x
x
=
+
+
+
+
∫
∫
=
2
2
21
3
1
2
2
dx
x
+
+
∫
Put
3
2
x
t
+
= , so that dx = dt, we get
I2 =
2
2
21
1
2
dt
t
+
∫
=
1
2
1
tan 2
C
1
2
2
–
t +
×
[by 7 |
1 | 3591-3594 | Therefore,
2
2
6
5
xx
x
2
++
+
∫
=
1
4
6
1
4
2
2
6
5
2
6
5
x
dx
dx
x
x
x
x
2
2
+
+
+
+
+
+
∫
∫
=
1
2
1
1
I
I
4
2
+
(say) (1)
314
MATHEMATICS
In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt
Therefore,
I1 =
1
log
C
dt
t
t
=
+
∫
=
2
1
log | 2
6
5|
C
x
x
+
+
+ (2)
and
I2 =
2
2
1
5
2
2
6
5
3
2
dx
dx
x
x
x
x
=
+
+
+
+
∫
∫
=
2
2
21
3
1
2
2
dx
x
+
+
∫
Put
3
2
x
t
+
= , so that dx = dt, we get
I2 =
2
2
21
1
2
dt
t
+
∫
=
1
2
1
tan 2
C
1
2
2
–
t +
×
[by 7 4 (3)]
=
1
2
3
tan 2
+ C
2
–
x
+
=
(
)
1
2
tan
2
3 + C
–
x + |
1 | 3592-3595 | (1)
314
MATHEMATICS
In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt
Therefore,
I1 =
1
log
C
dt
t
t
=
+
∫
=
2
1
log | 2
6
5|
C
x
x
+
+
+ (2)
and
I2 =
2
2
1
5
2
2
6
5
3
2
dx
dx
x
x
x
x
=
+
+
+
+
∫
∫
=
2
2
21
3
1
2
2
dx
x
+
+
∫
Put
3
2
x
t
+
= , so that dx = dt, we get
I2 =
2
2
21
1
2
dt
t
+
∫
=
1
2
1
tan 2
C
1
2
2
–
t +
×
[by 7 4 (3)]
=
1
2
3
tan 2
+ C
2
–
x
+
=
(
)
1
2
tan
2
3 + C
–
x + (3)
Using (2) and (3) in (1), we get
(
)
2
1
2
1
1
log 2
6
5
tan
2
3
C
4
2
2
6
5
–
x
dx
x
x
x
x
x
2
+
=
+
+
+
+
+
+
+
∫
where,
C =
1
2
C
C
4
2
+
(ii)
This integral is of the form given in 7 |
1 | 3593-3596 | (2)
and
I2 =
2
2
1
5
2
2
6
5
3
2
dx
dx
x
x
x
x
=
+
+
+
+
∫
∫
=
2
2
21
3
1
2
2
dx
x
+
+
∫
Put
3
2
x
t
+
= , so that dx = dt, we get
I2 =
2
2
21
1
2
dt
t
+
∫
=
1
2
1
tan 2
C
1
2
2
–
t +
×
[by 7 4 (3)]
=
1
2
3
tan 2
+ C
2
–
x
+
=
(
)
1
2
tan
2
3 + C
–
x + (3)
Using (2) and (3) in (1), we get
(
)
2
1
2
1
1
log 2
6
5
tan
2
3
C
4
2
2
6
5
–
x
dx
x
x
x
x
x
2
+
=
+
+
+
+
+
+
+
∫
where,
C =
1
2
C
C
4
2
+
(ii)
This integral is of the form given in 7 4 (10) |
1 | 3594-3597 | 4 (3)]
=
1
2
3
tan 2
+ C
2
–
x
+
=
(
)
1
2
tan
2
3 + C
–
x + (3)
Using (2) and (3) in (1), we get
(
)
2
1
2
1
1
log 2
6
5
tan
2
3
C
4
2
2
6
5
–
x
dx
x
x
x
x
x
2
+
=
+
+
+
+
+
+
+
∫
where,
C =
1
2
C
C
4
2
+
(ii)
This integral is of the form given in 7 4 (10) Let us express
x + 3 =
2
A
(5
4
) + B
d
– x – x
dx
= A (– 4 – 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
– 2A = 1 and – 4 A + B = 3, i |
1 | 3595-3598 | (3)
Using (2) and (3) in (1), we get
(
)
2
1
2
1
1
log 2
6
5
tan
2
3
C
4
2
2
6
5
–
x
dx
x
x
x
x
x
2
+
=
+
+
+
+
+
+
+
∫
where,
C =
1
2
C
C
4
2
+
(ii)
This integral is of the form given in 7 4 (10) Let us express
x + 3 =
2
A
(5
4
) + B
d
– x – x
dx
= A (– 4 – 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
– 2A = 1 and – 4 A + B = 3, i e |
1 | 3596-3599 | 4 (10) Let us express
x + 3 =
2
A
(5
4
) + B
d
– x – x
dx
= A (– 4 – 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
– 2A = 1 and – 4 A + B = 3, i e , A =
1
2
–
and B = 1
INTEGRALS 315
Therefore,
2
3
5
4
x
dx
x
x
+
−
−
∫
=
(
)
2
2
4
2
21
5
4
5
4
–
– x dx
dx
–
x
x
x
x
+
−
−
−
−
∫
∫
=
–21
I1 + I2 |
1 | 3597-3600 | Let us express
x + 3 =
2
A
(5
4
) + B
d
– x – x
dx
= A (– 4 – 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
– 2A = 1 and – 4 A + B = 3, i e , A =
1
2
–
and B = 1
INTEGRALS 315
Therefore,
2
3
5
4
x
dx
x
x
+
−
−
∫
=
(
)
2
2
4
2
21
5
4
5
4
–
– x dx
dx
–
x
x
x
x
+
−
−
−
−
∫
∫
=
–21
I1 + I2 (1)
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt |
1 | 3598-3601 | e , A =
1
2
–
and B = 1
INTEGRALS 315
Therefore,
2
3
5
4
x
dx
x
x
+
−
−
∫
=
(
)
2
2
4
2
21
5
4
5
4
–
– x dx
dx
–
x
x
x
x
+
−
−
−
−
∫
∫
=
–21
I1 + I2 (1)
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt Therefore,
I1= (
)
2
4
2
5
4
–
x dx
dt
t
x
x
−
=
−
−
∫
∫
=
1
2
C
t +
=
2
1
2 5 4
– x – x +C |
1 | 3599-3602 | , A =
1
2
–
and B = 1
INTEGRALS 315
Therefore,
2
3
5
4
x
dx
x
x
+
−
−
∫
=
(
)
2
2
4
2
21
5
4
5
4
–
– x dx
dx
–
x
x
x
x
+
−
−
−
−
∫
∫
=
–21
I1 + I2 (1)
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt Therefore,
I1= (
)
2
4
2
5
4
–
x dx
dt
t
x
x
−
=
−
−
∫
∫
=
1
2
C
t +
=
2
1
2 5 4
– x – x +C (2)
Now consider
I2 =
2
2
5
4
9
(
2)
dx
dx
x
x
– x
=
−
−
+
∫
∫
Put x + 2 = t, so that dx = dt |
1 | 3600-3603 | (1)
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt Therefore,
I1= (
)
2
4
2
5
4
–
x dx
dt
t
x
x
−
=
−
−
∫
∫
=
1
2
C
t +
=
2
1
2 5 4
– x – x +C (2)
Now consider
I2 =
2
2
5
4
9
(
2)
dx
dx
x
x
– x
=
−
−
+
∫
∫
Put x + 2 = t, so that dx = dt Therefore,
I2 =
1
2
2
2
sin
3+ C
3
–
dt
t
t
=
−
∫
[by 7 |
1 | 3601-3604 | Therefore,
I1= (
)
2
4
2
5
4
–
x dx
dt
t
x
x
−
=
−
−
∫
∫
=
1
2
C
t +
=
2
1
2 5 4
– x – x +C (2)
Now consider
I2 =
2
2
5
4
9
(
2)
dx
dx
x
x
– x
=
−
−
+
∫
∫
Put x + 2 = t, so that dx = dt Therefore,
I2 =
1
2
2
2
sin
3+ C
3
–
dt
t
t
=
−
∫
[by 7 4 (5)]
=
1
2
2
sin
C
3
– x +
+ |
1 | 3602-3605 | (2)
Now consider
I2 =
2
2
5
4
9
(
2)
dx
dx
x
x
– x
=
−
−
+
∫
∫
Put x + 2 = t, so that dx = dt Therefore,
I2 =
1
2
2
2
sin
3+ C
3
–
dt
t
t
=
−
∫
[by 7 4 (5)]
=
1
2
2
sin
C
3
– x +
+ (3)
Substituting (2) and (3) in (1), we obtain
2
1
2
3
2
5 – 4 –
+ sin
C
3
5
4
–
x
x
–
x
x
– x – x
+
+
=
+
∫
, where
1
2
C
C
C
2
–
=
EXERCISE 7 |
1 | 3603-3606 | Therefore,
I2 =
1
2
2
2
sin
3+ C
3
–
dt
t
t
=
−
∫
[by 7 4 (5)]
=
1
2
2
sin
C
3
– x +
+ (3)
Substituting (2) and (3) in (1), we obtain
2
1
2
3
2
5 – 4 –
+ sin
C
3
5
4
–
x
x
–
x
x
– x – x
+
+
=
+
∫
, where
1
2
C
C
C
2
–
=
EXERCISE 7 4
Integrate the functions in Exercises 1 to 23 |
1 | 3604-3607 | 4 (5)]
=
1
2
2
sin
C
3
– x +
+ (3)
Substituting (2) and (3) in (1), we obtain
2
1
2
3
2
5 – 4 –
+ sin
C
3
5
4
–
x
x
–
x
x
– x – x
+
+
=
+
∫
, where
1
2
C
C
C
2
–
=
EXERCISE 7 4
Integrate the functions in Exercises 1 to 23 1 |
1 | 3605-3608 | (3)
Substituting (2) and (3) in (1), we obtain
2
1
2
3
2
5 – 4 –
+ sin
C
3
5
4
–
x
x
–
x
x
– x – x
+
+
=
+
∫
, where
1
2
C
C
C
2
–
=
EXERCISE 7 4
Integrate the functions in Exercises 1 to 23 1 2
6
3
1
x
x +
2 |
1 | 3606-3609 | 4
Integrate the functions in Exercises 1 to 23 1 2
6
3
1
x
x +
2 2
1
1
4x
+
3 |
1 | 3607-3610 | 1 2
6
3
1
x
x +
2 2
1
1
4x
+
3 (
)
2
1
2
1
– x
+
4 |
1 | 3608-3611 | 2
6
3
1
x
x +
2 2
1
1
4x
+
3 (
)
2
1
2
1
– x
+
4 2
1
9
–25
x
5 |
1 | 3609-3612 | 2
1
1
4x
+
3 (
)
2
1
2
1
– x
+
4 2
1
9
–25
x
5 4
3
1 2
x
x
+
6 |
1 | 3610-3613 | (
)
2
1
2
1
– x
+
4 2
1
9
–25
x
5 4
3
1 2
x
x
+
6 2
6
1
x
x
−
7 |
1 | 3611-3614 | 2
1
9
–25
x
5 4
3
1 2
x
x
+
6 2
6
1
x
x
−
7 2
1
1
x –
x –
8 |
1 | 3612-3615 | 4
3
1 2
x
x
+
6 2
6
1
x
x
−
7 2
1
1
x –
x –
8 2
6
6
x
x
a
+
9 |
1 | 3613-3616 | 2
6
1
x
x
−
7 2
1
1
x –
x –
8 2
6
6
x
x
a
+
9 2
2
sec
tan
4
x
x +
316
MATHEMATICS
10 |
1 | 3614-3617 | 2
1
1
x –
x –
8 2
6
6
x
x
a
+
9 2
2
sec
tan
4
x
x +
316
MATHEMATICS
10 2
1
2
2
x
+x
+
11 |
1 | 3615-3618 | 2
6
6
x
x
a
+
9 2
2
sec
tan
4
x
x +
316
MATHEMATICS
10 2
1
2
2
x
+x
+
11 2
1
9
6
5
x
+x
+
12 |
1 | 3616-3619 | 2
2
sec
tan
4
x
x +
316
MATHEMATICS
10 2
1
2
2
x
+x
+
11 2
1
9
6
5
x
+x
+
12 2
1
7
–6
x – x
13 |
1 | 3617-3620 | 2
1
2
2
x
+x
+
11 2
1
9
6
5
x
+x
+
12 2
1
7
–6
x – x
13 (
)(
)
1
1
2
x –
x –
14 |
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