Higher Engineering Mathematics, Sixth Edition

424 Higher Engineering Mathematics

i.e.

(

1 +

a^2

b^2

)∫

eaxcosbxdx

=

1

b

eaxsinbx+

a

b^2

eaxcosbx

i.e.

(

b^2 +a^2

b^2

)∫

eaxcosbxdx

=

eax

b^2

(bsinbx+acosbx)

Hence

∫

eaxcosbxdx

=

(

b^2

b^2 +a^2

)(

eax

b^2

)

(bsinbx+acosbx)

=

eax

a^2 +b^2

(bsinbx+acosbx)+c

Using a similar method to above, that is, integrating by

parts twice, the following result may be proved:

∫

eaxsinbxdx

=

eax

a^2 +b^2

(asinbx−bcosbx)+c (2)

Problem 10. Evaluate

∫ π

4

0

etsin2tdt, correct to

4 decimal places.

Comparing

∫

etsin2tdtwith

∫

eaxsinbxdxshows that

x=t,a=1andb=2.

Hence, substituting into equation (2) gives:

∫ π

4

0

etsin2tdt

=

[

et

12 + 22

(1sin2t−2cos2t)

]π

4

0

=

[

e

π

4

5

(

sin2

(π

4

)

−2cos2

(π

4

))

]

−

[

e^0

5

(sin0−2cos0)

]

=

[

e

π

4

5

( 1 − 0 )

]

−

[

1

5

( 0 − 2 )

]

=

e

π

4

5

+

2

5

=0.8387,correct to 4 decimal places.

Now try the following exercise

Exercise 168 Further problems on

integration by parts

Determine the integrals in Problems 1 to 5 using

integration by parts.

1.

∫

2 x^2 lnxdx

[

2

3

x^3

(

lnx−

1

3

)

+c

]

2.

∫

2ln3xdx [2x(ln3x− 1 )+c]

3.

∫

x^2 sin3xdx

[

cos3x

27

( 2 − 9 x^2 )+

2

9

xsin3x+c

]

4.

∫

2e^5 xcos2xdx

[

2

29

e^5 x(2sin2x+5cos2x)+c

]

5.

∫

2 θsec^2 θdθ [2[θtanθ−ln(secθ)]+c]

Evaluate the integrals in Problems 6 to 9, correct

to 4 significant figures.

6.

∫ 2

1

xlnxdx [0.6363]

7.

∫ 1

0

2e^3 xsin2xdx [11.31]

8.

∫ π

2

0

etcos3tdt [− 1 .543]

9.

∫ 4

1

√

x^3 lnxdx [12.78]