Integration using trigonometric and hyperbolic substitutions 405
Hence
∫
1
√
(x^2 +a^2 )
dx
=
∫
1
√
(a^2 sinh^2 θ+a^2 )
(acoshθdθ)
=
∫
acoshθdθ
√
(a^2 cosh^2 θ)
,
since cosh^2 θ−sinh^2 θ= 1
=
∫
acoshθ
acoshθ
dθ=
∫
dθ=θ+c
=sinh−^1
x
a
+c,sincex=asinhθ
Itisshownonpage339that
sinh−^1
x
a
=ln
{
x+
√
(x^2 +a^2 )
a
}
,
which provides an alternative solution to
∫
1
√
(x^2 +a^2 )
dx
Problem 21. Evaluate
∫ 2
0
1
√
(x^2 + 4 )
dx, correct
to 4 decimal places.
∫ 2
0
1
√
(x^2 + 4 )
dx=
[
sinh−^1
x
2
] 2
0
or
[
ln
{
x+
√
(x^2 + 4 )
2
}] 2
0
from Problem 20, wherea= 2
Using the logarithmic form,
∫ 2
0
1
√
(x^2 + 4 )
dx
=
[
ln
(
2 +
√
8
2
)
−ln
(
0 +
√
4
2
)]
=ln2. 4142 −ln1= 0. 8814 ,
correct to 4 decimal places.
Problem 22. Evaluate
∫ 2
1
2
x^2
√
( 1 +x^2 )
dx,
correct to 3 significant figures.
Since the integral contains a term of the form√
(a^2 +x^2 ),thenletx=sinhθ, from which
dx
dθ
=coshθand dx=coshθdθ
Hence
∫
2
x^2
√
( 1 +x^2 )
dx
=
∫
2 (coshθdθ)
sinh^2 θ
√
( 1 +sinh^2 θ)
= 2
∫
coshθdθ
sinh^2 θcoshθ
,
since cosh^2 θ−sinh^2 θ= 1
= 2
∫
dθ
sinh^2 θ
= 2
∫
cosech^2 θdθ
=−2cothθ+c
cothθ=
coshθ
sinhθ
=
√
( 1 +sinh^2 θ)
sinhθ
=
√
( 1 +x^2 )
x
Hence
∫ 2
1
2
x^2
√
1 +x^2 )
dx
=−[2cothθ]^21 =− 2
[√
( 1 +x^2 )
x
] 2
1
=− 2
[√
5
2
−
√
2
1
]
= 0. 592 ,
correct to 3 significant figures
Problem 23. Find
∫√
(x^2 +a^2 )dx.
Letx=asinhθthen
dx
dθ
=acoshθand
dx=acoshθdθ
Hence
∫√
(x^2 +a^2 )dx
=
∫√
(a^2 sinh^2 θ+a^2 )(acoshθdθ)
=
∫√
[a^2 (sinh^2 θ+ 1 )](acoshθdθ)
=
∫ √
(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