420 INTEGRAL CALCULUS

From Problem 1,

∫

xcosxdx=xsinx+cosx

Hence

∫

x^2 sinxdx

=−x^2 cosx+ 2 {xsinx+cosx}+c

=−x^2 cosx+ 2 xsinx+2 cosx+c

= (2−x^2 )cosx+ 2 xsinx+c

In general, if the algebraic term of a product is of
powern, then the integration by parts formula is
appliedntimes.

Now try the following exercise.

Exercise 168 Further problems on integra-

tion by parts

Determine the integrals in Problems 1 to 5 using

integration by parts.

1.

∫

xe^2 xdx

[[

e^2 x

2

(

x−

1

2

)]

+c

]

2.

∫

4 x

e^3 x

dx

[

−

4

3

e−^3 x

(

x+

1

3

)

+c

]

3.

∫

xsinxdx [−xcosx+sinx+c]

4.

∫

5 θcos 2θdθ

[

5

2

(

θsin 2θ+^12 cos 2θ

)

+c

]

5.

∫

3 t^2 e^2 tdt

[ 3

2 e

2 t(t (^2) −t+ 1
2
)
+c
]
Evaluate the integrals in Problems 6 to 9, correct
to 4 significant figures.
6.
∫ 2
0
2 xexdx [16.78]
7.
∫π
4
0
xsin 2xdx [0.2500]
8.
∫π
2
0
t^2 costdt [0.4674]
9.
∫ 2
1
3 x^2 e
x
(^2) dx [15.78]
43.3 Further worked problems on
integration by parts
Problem 6. Find
∫
xlnxdx.
The logarithmic function is chosen as the ‘upart’.
Thus whenu=lnx, then
du
dx

1
x
, i.e. du=
dx
x
Letting dv=xdxgivesv=
∫
xdx=
x^2
2
Substituting into
∫
udv=uv−
∫
vdugives:
∫
xlnxdx=(lnx)
(
x^2
2
)
−
∫ (
x^2
2
)
dx
x

x^2
2
lnx−
1
2
∫
xdx

x^2
2
lnx−
1
2
(
x^2
2
)
+c
Hence
∫
xlnxdx=
x^2
2
(
lnx−
1
2
)
+cor
x^2
4
(2 lnx−1)+c
Problem 7. Determine
∫
lnxdx.
∫
lnxdxis the same as
∫
(1) lnxdx
Letu=lnx, from which,
du
dx

1
x
, i.e. du=
dx
x
and let dv=1dx, from which,v=
∫
1dx=x
Substituting into
∫
udv=uv−
∫
vdugives:
∫
lnxdx=(lnx)(x)−
∫
x
dx
x
=xlnx−
∫
dx=xlnx−x+c
Hence
∫
lnxdx=x(lnx−1)+c
Problem 8. Evaluate
∫ 9
1
√
xlnxdx, correct to
3 significant figures.