Higher Engineering Mathematics, Sixth Edition

422 Higher Engineering Mathematics

The integral,

∫

xcosxdx, is not a ‘standard integral’

and it can only be determined by using the integration

by parts formula again.

From Problem 1,

∫

xcosxdx=xsinx+cosx

Hence

∫

x^2 sinxdx

=−x^2 cosx+ 2 {xsinx+cosx}+c

=−x^2 cosx+ 2 xsinx+2cosx+c

=(2−x^2 )cosx+ 2 xsinx+c

In general, if the algebraic term of a product is of power

n, then the integration by parts formula is appliedn

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^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 θcos2θdθ

[

5

2

(

θsin2θ+^12 cos2θ

)

+c

]

5.

∫

3 t^2 e^2 tdt

[

3

2 e

2 t

(

t^2 −t+^12

)

+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

xsin2xdx [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

(2lnx−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