424 Higher Engineering Mathematics
i.e.
1 +
a 2
b 2
e
ax cos bx dx
=
1
b
e
ax sin bx +
a
b 2 e
ax cos bx
i.e.
b 2 + a 2
b 2
e
ax cos bx dx
=
e ax
b 2 (b sin bx + a cos bx)
Hence
e
ax cos bx dx
=
b 2
b 2 + a 2
e ax
b 2
(b sin bx + a cos bx)
=
e ax
a 2 + b 2 (b sin bx + a cos bx) + c
Using a similar method to above, that is, integrating by
parts twice, the following result may be proved:
e
ax sin bx dx
=
e ax
a 2 + b 2 (a sin bx − b cos bx)+ c
(2)
Problem 10. Evaluate
π
4
0
e
t sin 2t dt , correct to
4 decimal places.
Comparing
e t sin 2t dt with
e ax sin bx dx shows that
x = t , a = 1 and b = 2.
Hence, substituting into equation (2) gives:
π
4
0
e
t sin 2t dt
=
e t
1 2 + 2 2 (1 sin 2t − 2 cos 2t )
π
4
0
=
e
π
4
5
sin 2
π
4
− 2 cos2
π
4
−
e 0
5
(sin 0 − 2 cos 0)
=
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.
2x
2 ln x dx
2
3
x
3
ln x −
1
3
+ c
2.
2 ln 3x dx
[2x(ln 3x − 1) + c]
3.
x
2 sin 3x dx
cos 3x
27
(2 − 9x
2
) +
2
9
x sin 3x + c
4.
2e
5x cos 2x dx
2
29
e
5x
(2 sin 2x + 5 cos 2x) + c
5.
2θ sec
2
θ dθ [2[θ tan θ − ln(sec θ)] + c]
Evaluate the integrals in Problems 6 to 9, correct
to 4 significant figures.
6.
2
1
x ln x dx
[0.6363]
7.
1
0
2e
3x sin 2x dx
[11.31]
8.
π
2
0
e
t cos 3t dt
[−1.543]
9.
4
1
x 3 ln x dx
[12.78]