6.3 The method of substitution 171
Integral 1 in the table is evaluated by means of the substitutionx 1 = 1 a 1 sin 1 θ. Then
dx 1 = 1 a 1 cos 1 θ 1 dθand
Therefore
Similarly, integral 2 is evaluated by means of the substitutionx 1 = 1 a 1 cosh 1 u. Integral 4 ,
evaluated by means of the substitution x 1 = 1 a 1 tan 1 θ, is used in Section 6.6 for the
integration of rational functions. Integral 5 can be evaluated either by means of
the substitution x 1 = 1 a 1 tanh 1 uor by expressing the integrand in terms of partial
fractions to give the logarithmic form (see Section 6.6). Alternatively, all the integrals
in Table 6.3 are readily obtained by integrating the standard derivatives listed in
Tables 4.5 and 4.6.
Such substitutions are useful when the integrand contains the square root of a
quadratic function.
EXAMPLE 6.8Evaluate.
Letx 1 = 1 a 1 sin 1 θ. Thendx 1 = 1 a 1 cos 1 θ 1 dθ, , and
This is integral 1 in Table 6.1. Therefore
Nowsin 1 θ 1 = 1 x 2 a, , andθ 1 = 1 sin
− 11 (x 2 a). Therefore,
0 Exercises 29–32
Z axdx a
x
a
xa x C
22 2 1 221
2
1
2
−=
+−+
−sin
cosθ=−axa
22ZZaxdx
a
d
a
C
22222
12
2
−= +( cos )θθ= +( sin cos )θ θθ +
ZZaxdxa d
22 2 2−=cos θθ
axa
22−=cosθ
Z axdx
22−
ZZ Z
dx
ax
ad
a
dC
x
a
221−
===+=
−cos
cos
sin
θθ
θ
θθ
+C
ax aa a a
22 222 2−= −sin θθθ= − 1 sin =cos.