Higher Engineering Mathematics

INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS 403

H

4. Determine

∫ √

(16− 9 t^2 )dt

[

8

3

sin−^1

3 t

4

+

t

2

√

(16− 9 t^2 )+c

]

5. Evaluate

∫ 4

0

1

√

(16−x^2 )

dx

[π

2

or 1. 571

]

6. Evaluate

∫ 1

0

√

(9− 4 x^2 )dx [2.760]

40.6 Worked problems on integration

using tanθsubstitution

Problem 17. Determine

∫

1

(a^2 +x^2 )

dx.

Letx=atanθthen

dx

dθ

=asec^2 θand dx=asec^2 θdθ.

Hence

∫

1

(a^2 +x^2 )

dx

=

∫

1

(a^2 +a^2 tan^2 θ)

(asec^2 θdθ)

=

∫

asec^2 θdθ

a^2 (1+tan^2 θ)

=

∫

asec^2 θdθ

a^2 sec^2 θ

, since 1+tan^2 θ=sec^2 θ

=

∫

1

a

dθ=

1

a

(θ)+c

Sincex=atanθ,θ=tan−^1

x

a

Hence

∫

1

(a^2 +x^2 )

dx=

1

a

tan−^1

x

a

+c.

Problem 18. Evaluate

∫ 2

0

1

(4+x^2 )

dx.

From Problem 17,

∫ 2

0

1

(4+x^2 )

dx

=

1

2

[

tan−^1

x

2

] 2

0

sincea= 2

=

1

2

(tan−^11 −tan−^1 0)=

1

2

(π

4

− 0

)

=

π

8

or 0. 3927

Problem 19. Evaluate

∫ 1

0

5

(3+ 2 x^2 )

dx, cor-

rect to 4 decimal places.

∫ 1

0

5

(3+ 2 x^2 )

dx=

∫ 1

0

5

2[(3/2)+x^2 ]

dx

=

5

2

∫ 1

0

1

[

√

(3/2)]^2 +x^2

dx

=

5

2

[

1

√

(3/2)

tan−^1

x

√

(3/2)

] 1

0

=

5

2

√(

2

3

)[

tan−^1

√(

2

3

)

−tan−^10

]

=(2.0412)[0. 6847 −0]

= 1. 3976 , correct to 4 decimal places

Now try the following exercise.

Exercise 160 Further problems on integra-

tion using the tanθsubstitution

1. Determine

∫

3

4 +t^2

dt

[

3

2

tan−^1

t

2

+c

]

2. Determine

∫

5

16 + 9 θ^2

dθ

[

5

12

tan−^1

3 θ

4

+c

]

3. Evaluate

∫ 1

0

3

1 +t^2

dt [2.356]

4. Evaluate

∫ 3

0

5

4 +x^2

dx [2.457]

40.7 Worked problems on integration

using the sinhθsubstitution

Problem 20. Determine

∫

1

√

(x^2 +a^2 )

dx.