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

  
  

1. Determine

∫

5

√

( 4 −t^2 )

dt.

[

5sin−^1

x

2

+c

]

2. Determine

∫

3

√

( 9 −x^2 )

dx.

[

3sin−^1

x

3

+c

]

3. Determine

∫ √

( 4 −x^2 )dx.

[

2sin−^1

x

2

+

x

2

√

( 4 −x^2 )+c

]