Standard integration 371
Problem 9. Determine
(a)
7 sec 2 4t dt (b) 3
cosec 2 2θ dθ.
(a) From Table 37.1(iv),
7 sec
2 4t dt = (7)
1
4
tan 4t + c
=
7
4
tan 4t + c
(b) From Table 37.1(v),
3
cosec
2 2θ dθ = (3)
−
1
2
cot 2θ + c
= −
3
2
cot 2θ + c
Problem 10. Determine
(a)
5 e
3x dx (b)
2
3 e 4t dt.
(a) From Table 37.1(viii),
5 e
3x dx = (5)
1
3
e
3x
+ c =
5
3
e
3x
+ c
(b)
2
3 e 4t dt =
2
3
e
−4t dt =
2
3
−
1
4
e
−4t
+ c
=−
1
6
e
−4t
+ c = −
1
6e 4t + c
Problem 11. Determine
(a)
3
5x
dx (b)
2m 2 + 1
m
dm.
(a)
3
5x
dx =
3
5
1
x
dx =
3
5
ln x +c
(from Table 37.1(ix))
(b)
2m 2 + 1
m
dm =
2m 2
m
+
1
m
dm
=
2m +
1
m
dm
=
2m 2
2
+ ln m + c
= m
2
+ ln m + c
Now try the following exercise
Exercise 145 Further problems on standard
integrals
In Problems 1 to 12, determine the indefinite
integrals.
1. (a)
4 dx (b)
7x dx
(a) 4x + c (b)
7x 2
2
+ c
2. (a)
2
5
x
2 dx (b)
5
6
x
3 dx
(a)
2
15
x 3 + c (b)
5
24
x 4 + c
3. (a)
3x 2 − 5x
x
dx (b)
(2 + θ)
2 dθ
⎡
⎢
⎢
⎣
(a)
3x 2
2
− 5x + c
(b) 4θ + 2θ 2 +
θ 3
3
+ c
⎤
⎥
⎥
⎦
4. (a)
4
3x 2 dx (b)
3
4x 4 dx
(a)
−4
3x
+ c (b)
−1
4x 3 + c
5. (a) 2
x 3 dx (b)
1
4
4
x 5 dx
(a)
4
5
√
x 5 + c (b)
1
9
4
√
x 9 + c
6. (a)
−5
√
t 3
dt (b)
3
7
5
√
x 4
dx
(a)
10
√
t
+ c (b)
15
7
5
√
x + c