integration

404 INTEGRAL CALCULUS
Let x = a sinh θ, then dx = a cosh θ and
dθ
dx = a cosh θ dθ
∫
1
Hence √
(x 2 + a 2 ) dx
∫
=
∫
=
1
√ (a cosh θ dθ)
(a 2 sinh 2 θ + a 2 )
a cosh θ dθ
√
(a 2 cosh 2 θ) ,
since cosh 2 θ − sinh 2 θ = 1
∫ ∫
a cosh θ
=
a cosh θ dθ = dθ = θ + c
= sinh −1 x + c, since x = a sinh θ
a
It is shown on page 337 that
{
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
1
√ dx, cor-
(x 2 + 4)
rect to 4 decimal places.
∫ 2
0
∫ 2
0
1
[
√
(x 2 + 4) dx = sinh
2] −1 x 2 or
0
[ {
x + √ }] 2
(x
ln
2 + 4)
2
0
from Problem 20, where a = 2
Using the logarithmic form,
∫ 2
0
=
1
√
(x 2 + 4) dx
[
ln
(
2 + √ ) (
8 0 + √ )]
4
− ln
2
2
= ln 2.4142 − ln 1 = 0.8814,
correct to 4 decimal places
Problem 22. Evaluate
∫ 2
correct to 3 significant figures.
1
2
x 2√ (1 + x 2 ) dx,
Since
√
the integral contains a term of the form
(a 2 + x 2 ), then let x = sinh θ, from which
dx
= cosh θ and dx = cosh θ dθ
dθ
∫
2
Hence
x 2√ (1 + x 2 ) dx
∫
2(cosh θ dθ)
=
sinh 2 θ √ (1 + sinh 2 θ)
∫
cosh θ dθ
= 2
sinh 2 θ cosh θ ,
since cosh 2 θ − sinh 2 θ = 1
∫
∫
dθ
= 2
sinh 2 θ = 2 cosech 2 θ dθ
=−2 coth θ + c
coth θ = cosh θ
√ √
(1 + sinh 2
sinh θ = θ) (1 + x
=
2 )
sinh θ
x
Hence
∫ 2
1
2
x 2√ 1 + x 2 ) dx
] 2
1
[√
(1 + x
=−[2 coth θ] 2 1 =−2 2 )
x
[√ √ ]
5 2
=−2
2 − = 0.592,
1
correct to 3 significant figures
Problem 23. Find
∫ √(x 2 + a 2 )dx.
Let x = a sinh θ then dx = a cosh θ and
dθ
dx = a cosh θ dθ
∫ √(x
Hence 2 + a 2 )dx
∫ √
= (a 2 sinh 2 θ + a 2 )(a cosh θ dθ)
∫ √
= [a 2 (sinh 2 θ + 1)](a cosh θ dθ)