370 INTEGRAL CALCULUS
(a) From Table 37.1(iv),
∫
7 sec^24 tdt=(7)
(
1
4)
tan 4t+c=7
4tan 4t+c(b) From Table 37.1(v),
3∫
cosec^22 θdθ=(3)(
−1
2)
cot 2θ+c=−3
2cot 2θ+cProblem 10. Determine(a)∫
5e^3 xdx (b)∫
2
3e^4 tdt.(a) From Table 37.1(viii),
∫
5e^3 xdx=(5)
(
1
3)
e^3 x+c=5
3e^3 x+c(b)
∫
2
3e^4 tdt=∫
2
3e−^4 tdt=(
2
3)(
−1
4)
e−^4 t+c=−1
6e−^4 t+c=−1
6e^4 t+cProblem 11. Determine(a)∫
3
5 xdx (b)∫(
2 m^2 + 1
m)
dm.(a)
∫
3
5 xdx=∫(
3
5)(
1
x)
dx=3
5lnx+c(from Table 37.1(ix))(b)
∫(
2 m^2 + 1
m)
dm=∫(
2 m^2
m+1
m)
dm=∫(
2 m+1
m)
dm=2 m^2
2+lnm+c=m^2 +lnm+cNow try the following exercise.Exercise 146 Further problems on standard
integralsIn Problems 1 to 12, determine the indefinite
integrals.- (a)
∫
4dx (b)∫
7 xdx[
(a) 4x+c (b)7 x^2
2+c]- (a)
∫
2
5x^2 dx (b)∫
5
6x^3 dx
[
(a)2
15x^3 +c (b)5
24x^4 +c]- (a)
∫(
3 x^2 − 5 x
x)
dx (b)∫
(2+θ)^2 dθ⎡⎢
⎢
⎣(a)3 x^2
2− 5 x+c(b) 4θ+ 2 θ^2 +θ^3
3+c⎤⎥
⎥
⎦- (a)
∫
4
3 x^2dx (b)∫
3
4 x^4dx[
(a)− 4
3 x+c (b)− 1
4 x^3+c]- (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]- (a)
∫
− 5
√
t^3dt (b)∫
37√ 5
x^4dx[
(a)10
√
t+c (b)15
7√ (^5) x+c
]
- (a)
∫
3 cos 2xdx (b)∫
7 sin 3θdθ⎡⎢
⎢
⎣(a)3
2sin 2x+c(b)−7
3cos 3θ+c⎤⎥
⎥
⎦