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]