422 Higher Engineering Mathematics
The integral,
x cos x dx, is not a ‘standard integral’
and it can only be determined by using the integration
by parts formula again.
From Problem 1,
x cos x dx = x sin x + cos x
Hence
x
2 sin x dx
= −x
2 cos x + 2{x sin x + cos x} + c
= −x
2 cos x + 2x sin x + 2 cos x + c
= (2 −x
2 )cos x +2x sin x +c
In general, if the algebraic term of a product is of power
n, then the integration by parts formula is applied n
times.
Now try the following exercise
Exercise 167 Further problems on
integration by parts
Determine the integrals in Problems 1 to 5 using
integration by parts.
1.
xe
2x dx
e 2x
2
x −
1
2
+ c
2.
4x
e 3x dx
−
4
3
e
−3x
x +
1
3
+ c
3.
x sin x dx
[−x cos x + sin x + c]
4.
5θ cos 2θ dθ
5
2
θ sin 2θ +
1
2 cos 2θ
+ c
5.
3t
2 e
2t dt
3
2 e
2t
t
2
− t +
1
2
+ c
Evaluate the integrals in Problems 6 to 9, correct
to 4 significant figures.
6.
2
0
2xe
x dx
[16.78]
7.
π
4
0
x sin 2x dx
[0.2500]
8.
π
2
0
t
2 cos t dt
[0.4674]
9.
2
1
3x
2 e
x
2 dx
[15.78]
43.3 Further worked problems on
integration by parts
Problem 6. Find
x ln x dx.
The logarithmic function is chosen as the ‘u part’.
Thus when u = ln x, then
du
dx
=
1
x
, i.e. du =
dx
x
Letting dv = x dx gives v =
x dx =
x 2
2
Substituting into
u dv = uv −
v du gives:
x ln x dx = (ln x)
x 2
2
−
x 2
2
dx
x
=
x 2
2
ln x −
1
2
x dx
=
x 2
2
ln x −
1
2
x 2
2
+ c
Hence
x ln x dx =
x
2
2
lnx −
1
2
+ c or
x 2
4
(2 ln x −1)+ c
Problem 7. Determine
ln x dx.
ln x dx is the same as
(1) ln x dx
Let u = ln x, from which,
du
dx
=
1
x
, i.e. du =
dx
x
and let dv = 1dx, from which, v =
1 dx = x
Substituting into
u dv = uv −
v du gives:
ln x dx = (ln x)(x) −
x
dx
x
= x ln x −
dx = x ln x − x + c
Hence
ln x dx = x(ln x −1)+ c