INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS 405
H
=
∫ √
(a^2 cosh^2 θ)(acoshθdθ),
since cosh^2 θ−sinh^2 θ= 1
=
∫
(acoshθ)(acoshθ)dθ=a^2
∫
cosh^2 θdθ
=a^2
∫ (
1 +cosh 2θ
2
)
dθ
=
a^2
2
(
θ+
sinh 2θ
2
)
+c
=
a^2
2
[θ+sinhθcoshθ]+c,
since sinh 2θ=2 sinhθcoshθ
Sincex=asinhθ, then sinhθ=
x
a
andθ=sinh−^1
x
a
Also since cosh^2 θ−sinh^2 θ= 1
then coshθ=
√
(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 161 Further problems on integra-
tion using the sinhθsubstitution
- Find
∫
2
√
(x^2 +16)
dx
[
2 sinh−^1
x
4
+c
]
- Find
∫
3
√
(9+ 5 x^2 )
dx
[
3
√
5
sinh−^1
√
5
3
x+c
]
- Find
∫ √
(x^2 +9) dx
[
9
2
sinh−^1
x
3
+
x
2
√
(x^2 +9)+c
]
- Find
∫ √
(4t^2 +25) dt
[
25
4
sinh−^1
2 t
5
+
t
2
√
(4t^2 +25)+c
]
- Evaluate
∫ 3
0
4
√
(t^2 +9)
dt [3.525]
- Evaluate
∫ 1
0
√
(16+ 9 θ^2 )dθ [4.348]
40.8 Worked problems on integration
using the coshθsubstitution
Problem 24. Determine
∫
1
√
(x^2 −a^2 )
dx.
Letx=acoshθthen
dx
dθ
=asinhθand
dx=asinhθdθ
Hence
∫
1
√
(x^2 −a^2 )
dx
=
∫
1
√
(a^2 cosh^2 θ−a^2 )
(asinhθdθ)
=
∫
asinhθdθ
√
[a^2 ( cosh^2 θ−1)]
=
∫
asinhθdθ
√
(a^2 sinh^2 θ)
,
since cosh^2 θ−sinh^2 θ= 1
=
∫
asinhθdθ
asinhθ
=
∫
dθ=θ+c
=cosh−^1
x
a
+c, sincex=acoshθ
It is shown on page 337 that
cosh−^1
x
a
=ln
{
x+
√
(x^2 −a^2 )
a
}