402 INTEGRAL CALCULUS

Sincex=asinθ, then sinθ=

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−^1 0)=

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 +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 sinθcosθ

2

)

+c

since from Chapter 18, sin 2θ=2 sinθcosθ

=

a^2

2

[θ+sinθcosθ]+c

Sincex=asinθ, then sinθ=

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

  
  

  [
  8 sin−^11 + 2
  √
  (0)
  ]
  −[8 sin−^10 +0]
  =8 sin−^11 = 8
  (π
  2
  )
  = 4 πor 12. 57
  Now try the following exercise.
  Exercise 159 Further problems on integra-
  tion using the sineθsubstitution

  
  

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

]