Integration using trigonometric and hyperbolic substitutions 407
2x − 3
(x 2 − 9)
dx =
2x
(x 2 − 9)
dx
−
3
(x 2 − 9)
dx
The first integral is determined using the algebraic substitution u =(x 2 − 9), and the second integral is of the
form
1
(x 2 − a 2 )
dx (see Problem 24)
Hence
2x
(x 2 − 9)
dx −
3
(x 2 − 9)
dx
= 2
(x 2 − 9)− 3 cosh −1 x
3
+ c
Problem 26.
(x 2 − a 2 ) dx.
Let x = a cosh θ then
dx
dθ
= a sinh θ and
dx = a sinh θ dθ
Hence
(x 2 − a 2 ) dx
=
(a 2 cosh
2
θ − a 2 )(a sinh θ dθ)
=
[a 2 (cosh
2
θ − 1)] (a sinh θ dθ)
=
(a 2 sinh 2 θ)(a sinh θ dθ)
= a 2
sinh
2
θ dθ = a
2
cosh 2θ − 1
2
dθ
since cosh 2θ = 1 + 2 sinh 2 θ
from Table 5.1, page 45,
=
a 2
2
sinh 2θ
2
− θ
+ c
=
a 2
2
[sinh θ cosh θ − θ] + c,
since sinh 2θ = 2 sinhθ cosh θ
Since x = a cosh θ then cosh θ =
x
a
and
θ = cosh −1 x
a
Also, since cosh 2 θ − sinh 2 θ = 1, then
sinh θ =
(cosh 2 θ − 1)
=
x
a
2 − 1
=
(x 2 − a 2 )
a
Hence
(x 2 − a 2 ) dx
=
a 2
2
(x 2 − a 2 )
a
x
a
− cosh
−1 x
a
+ c
=
x
2
(x 2 − a 2 ) −
a 2
2
cosh −1 x
a
+ c
Problem 27. Evaluate
3
2
(x 2 − 4) dx.
3
2
(x 2 − 4) dx =
x
2
(x 2 − 4) −
4
2
cosh −1 x
2
3
2
from Problem 26, when a = 2,
=
3
5
√
5 − 2 cosh
−1 3
2
− (0 − 2 cosh −1 1)
Since cosh −1 x
a
= ln
x +
(x 2 − a 2 )
a
then
cosh −1 3
2
= ln
3 +
(3 2 − 2 2 )
2
= ln 2.6180 = 0.9624
Similarly, cosh −1 1 = 0
Hence
3
2
(x 2 − 4) dx
=
3
2
√
5 − 2(0.9624)
− [0]
= 1.429, correct to 4 significant figures.
Now try the following exercise
Exercise 161 Further problems on
integration using the cosh θ substitution
1. Find
1
(t 2 − 16)
dt .
cosh −1 x
4
+ c