370 Higher Engineering Mathematics
(b) Rearranging∫
( 1 −t)^2 dtgives:
∫
( 1 − 2 t+t^2 )dt=t−2 t^1 +^1
1 + 1+t^2 +^1
2 + 1+c=t−2 t^2
2+t^3
3+c=t−t^2 +1
3t^3 +cThis problem shows that functions often have to be
rearranged into the standard form of∫
axndxbefore
it is possible to integrate them.Problem 4. Determine∫
3
x^2dx.∫
3
x^2dx=∫
3 x−^2 dx. Using the standard integral,
∫
axndxwhena=3andn=−2gives:∫
3 x−^2 dx=3 x−^2 +^1
− 2 + 1+c=3 x−^1
− 1+c=− 3 x−^1 +c=− 3
x+cProblem 5. Determine∫
3√
xdx.For fractional powers it is necessary to appreciate
√nam=amn∫
3√
xdx=∫
3 x1(^2) dx=
3 x
1
2 +^1
1
2
1
+c
3 x
3
2
3
2
+c= 2 x
3
(^2) +c= 2
√
x^3 +c
Problem 6. Determine
∫
− 5
9
√ 4
t^3
dt.
∫
− 5
9
√ 4
t^3
dt=
∫
− 5
9 t
3
4
dt=
∫(
−
5
9
)
t−
43
dt
(
−
5
9
)
t
−
3
4
- 1
−
3
4
1
+c
(
−
5
9
)
t
1
4
1
4
+c=
(
−
5
9
)(
4
1
)
t
1
(^4) +c
=−
20
9
√ (^4) t+c
Problem 7. Determine
∫
( 1 +θ)^2
√
θ
dθ.
∫
( 1 +θ)^2
√
θ
dθ=
∫
( 1 + 2 θ+θ^2 )
√
θ
dθ
∫(
1
θ
1
2
2 θ
θ
1
2
θ^2
θ
1
2
)
dθ
∫(
θ
− 1
(^2) + 2 θ^1 −
(
(^12)
)
+θ
2 −
(
(^12)
))
dθ
∫(
θ
− 21
2 θ
(^12)
+θ
32 )
dθ
θ
(− 1
2
)
- 1
−^12 + 1
2 θ
( 1
2
)
- 1
1
2 +^1
θ
( 3
2
)
1
3
2 +^1
+c
θ
1
2
1
2
2 θ
3
2
3
2
θ
5
2
5
2
+c
= 2 θ
1
(^2) +
4
3
θ
3
(^2) +
2
5
θ
5
(^2) +c
= 2
√
θ+
4
3
√
θ^3 +
2
5
√
θ^5 +c
Problem 8. Determine
(a)
∫
4cos3xdx (b)
∫
5sin2θdθ.
(a) From Table 37.1(ii),
∫
4cos3xdx=( 4 )
(
1
3
)
sin3x+c
4
3
sin3x+c
(b) From Table 37.1(iii),
∫
5sin2θdθ=( 5 )
(
−
1
2
)
cos2θ+c
=−
5
2
cos2θ+c