404 Higher Engineering Mathematics
4. Determine
(16 − 9t 2 ) dt .
8
3
sin −1 3t
4
+
t
2
(16 − 9t 2 ) + c
5. Evaluate
4
0
1
(16 − x 2 )
dx.
π
2
or 1.571
6. Evaluate
1
0
(9 − 4x 2 ) dx.
[2.760]
40.6 Worked problems on integration
using tan θ substitution
Problem 17. Determine
1
(a 2 + x 2 )
dx.
Let x = a tan θ then
dx
dθ
= a sec
2
θ and dx = a sec
2
θ dθ.
Hence
1
(a 2 + x 2 )
dx
=
1
(a 2 + a 2 tan 2 θ)
(a sec
2
θ dθ)
=
a sec 2 θ dθ
a 2 (1 + tan 2 θ)
=
a sec 2 θ dθ
a 2 sec 2 θ
, since 1+tan 2 θ = sec 2 θ
=
1
a
dθ =
1
a
(θ) + c
Since x = a tan θ, 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
since a = 2
=
1
2
(tan −1 1 −tan −1 0) =
1
2
π
4
− 0
=
π
8
or 0.3927
Problem 19. Evaluate
1
0
5
(3 + 2x 2 )
dx, correct
to 4 decimal places.
1
0
5
(3 + 2x 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 −1 0
= (2.0412)[0.6847 − 0]
= 1.3976, correct to 4 decimal places.
Now try the following exercise
Exercise 159 Further problems on
integration using the tan θ substitution
1. Determine
3
4 + t 2 dt .
3
2
tan −1 t
2
+ c
2. Determine
5
16 + 9θ 2 dθ.
5
12
tan −1 3θ
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 sinh θ substitution
Problem 20. Determine
1
(x 2 + a 2 )
dx.
Let x = a sinh θ, then
dx
dθ
= a cosh θ and
dx = a cosh θ dθ