406 Higher Engineering Mathematics
=
a 2
2
θ +
sinh 2θ
2
+ c
=
a 2
2
[θ + sinh θ cosh θ] + c,
since sinh 2θ = 2 sinhθ cosh θ
Since x = a sinh θ, then sinhθ =
x
a
and θ = sinh
−1 x
a
Also since cosh 2 θ − sinh 2 θ = 1
then cosh θ =
(1 + sinh 2 θ)
=
1 +
x
a
2
=
a
2
+ x
2
a 2
=
(a 2 + x 2 )
a
Hence
(x 2 + a 2 ) dx
=
a 2
2
sinh −1 x
a
+
x
a
(x 2 + a 2 )
a
+ c
=
a 2
2
sinh −1 x
a
+
x
2
(x 2 + a 2 ) + c
Now try the following exercise
Exercise 160 Further problems on
integration using the sinh θ substitution
1. Find
2
(x 2 + 16)
dx.
2 sinh
−1 x
4
+ c
2. Find
3
(9 + 5x 2 )
dx.
3
√
5
sinh −1
√
5
3
x + c
3. Find
(x 2 + 9) dx.
9
2
sinh −1 x
3
+
x
2
(x 2 + 9) + c
4. Find
(4t 2 + 25) dt .
25
4
sinh −1 2t
5
+
t
2
(4t 2 + 25) + c
5. Evaluate
3
0
4
(t 2 + 9)
dt .
[3.525]
6. Evaluate
1
0
(16 + 9θ 2 ) dθ.
[4.348]
40.8 Worked problems on integration
using the cosh θ substitution
Problem 24. Determine
1
(x 2 − a 2 )
dx.
Let x = a cosh θ then
dx
dθ
= a sinh θ and
dx = a sinh θ dθ
Hence
1
(x 2 − a 2 )
dx
=
1
(a 2 cosh 2 θ − a 2 )
(a sinh θ dθ)
=
a sinh θ dθ
[a 2 (cosh 2 θ − 1)]
=
a sinh θ dθ
(a 2 sinh
2
θ)
,
since cosh 2 θ − sinh 2 θ = 1
=
a sinh θ dθ
a sinh θ
=
dθ = θ + c
= cosh
−1 x
a
+ c, since x = a cosh θ
It is shown on page 339 that
cosh
−1 x
a
= ln
x +
(x 2 − a 2 )
a
which provides as alternative solution to
1
(x 2 − a 2 )
dx
Problem 25. Determine
2x − 3
(x 2 − 9)
dx.