Integration using trigonometric and hyperbolic substitutions 403
=
∫
acosθdθ
√
(a^2 cos^2 θ)
,sincesin^2 θ+cos^2 θ= 1
=
∫
acosθdθ
acosθ
=
∫
dθ=θ+c
Sincex=asinθ,thensinθ=
x
a
andθ=sin−^1
x
a
.
Hence
∫
1
√
(a^2 −x^2 )
dx=sin−^1
x
a
+c
Problem 14. Evaluate
∫ 3
0
1
√
( 9 −x^2 )
dx.
From Problem 13,
∫ 3
0
1
√
( 9 −x^2 )
dx
=
[
sin−^1
x
3
] 3
0
, sincea= 3
=(sin−^11 −sin−^10 )=
π
2
or 1. 5708
Problem 15. Find
∫ √
(a^2 −x^2 )dx.
Letx=asinθthen
dx
dθ
=acosθand dx=acosθdθ.
Hence
∫ √
(a^2 −x^2 )dx
=
∫ √
(a^2 −a^2 sin^2 θ)(acosθdθ)
=
∫ √
[a^2 ( 1 −sin^2 θ)](acosθdθ)
=
∫ √
(a^2 cos^2 θ)(acosθdθ)
=
∫
(acosθ)(acosθdθ)
=a^2
∫
cos^2 θdθ=a^2
∫ (
1 +cos2θ
2
)
dθ
(since cos 2θ=2cos^2 θ− 1 )
=
a^2
2
(
θ+
sin2θ
2
)
+c
=
a^2
2
(
θ+
2sinθcosθ
2
)
+c
since from Chapter 17, sin2θ=2sinθcosθ
=
a^2
2
[θ+sinθcosθ]+c
Sincex=asinθ,thensinθ=
x
a
andθ=sin−^1
x
a
Also, cos^2 θ+sin^2 θ=1, from which,
cosθ=
√
( 1 −sin^2 θ)=
√[
1 −
(x
a
) 2 ]
=
√(
a^2 −x^2
a^2
)
=
√
(a^2 −x^2 )
a
Thus
∫ √
(a^2 −x^2 )dx=
a^2
2
[θ+sinθcosθ]
=
a^2
2
[
sin−^1
x
a
+
(x
a
)√(a (^2) −x (^2) )
a
]
+c
a^2
2
sin−^1
x
a
x
2
√
(a^2 −x^2 )+c
Problem 16. Evaluate
∫ 4
0
√
( 16 −x^2 )dx.
From Problem 15,
∫ 4
0
√
( 16 −x^2 )dx
[
16
2
sin−^1
x
4
x
2
√
( 16 −x^2 )
] 4
0
[
8sin−^11 + 2
√
( 0 )
]
−[8 sin−^10 +0]
=8sin−^11 = 8
(π
2
)
= 4 πor 12. 57
Now try the following exercise
Exercise 158 Further problemson
integration using the sineθsubstitution
- Determine
∫
5
√
( 4 −t^2 )
dt.
[
5sin−^1
x
2
+c
]
- Determine
∫
3
√
( 9 −x^2 )
dx.
[
3sin−^1
x
3
+c
]
- Determine
∫ √
( 4 −x^2 )dx.
[
2sin−^1
x
2
+
x
2
√
( 4 −x^2 )+c
]