Integration using trigonometric and hyperbolic substitutions 403
=
a cos θ dθ
(a 2 cos 2 θ)
, since sin
2
θ + cos 2 θ = 1
=
a cos θ dθ
a cos θ
=
dθ = θ + c
Since x = a sin θ, then sinθ =
x
a
and θ = 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
, since a = 3
= (sin
−1 1 − sin
−1 0) =
π
2
or 1.5708
Problem 15. Find
(a 2 − x 2 ) dx.
Let x = a sin θ then
dx
dθ
= a cos θ and dx = a cos θ dθ.
Hence
(a 2 − x 2 ) dx
=
(a 2 − a 2 sin 2 θ)(a cos θ dθ)
=
[a 2 (1 − sin 2 θ)] (a cos θ dθ)
=
(a 2 cos 2 θ)(a cos θ dθ)
=
(a cos θ)(a cos θ dθ)
= a 2
cos
2
θ dθ = a
2
1 + cos 2θ
2
dθ
(since cos 2θ = 2 cos 2 θ − 1)
=
a 2
2
θ +
sin 2θ
2
+ c
=
a 2
2
θ +
2 sinθ cos θ
2
+ c
since from Chapter 17, sin 2θ = 2 sinθ cos θ
=
a 2
2
[θ + sin θ cos θ] + c
Since x = a sin θ, then sinθ =
x
a
and θ = sin −1 x
a
Also, cos 2 θ + sin 2 θ = 1, from which,
cos θ =
(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 + sin θ cos θ]
=
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
=
8 sin
−1 1 + 2
(0)
− [8 sin
−1 0 + 0]
= 8 sin
−1 1 = 8
π
2
= 4π or 12.57
Now try the following exercise
Exercise 158 Further problems on
integration using the sine θ substitution
1. Determine
5
(4 − t 2 )
dt .
5 sin −1 x
2
+ c
2. Determine
3
(9 − x 2 )
dx.
3 sin −1 x
3
+ c
3. Determine
(4 − x 2 ) dx.
2 sin −1 x
2
+
x
2
(4 − x 2 ) + c