PRELIMINARY CALCULUS
Since
dt
dx
=
1
2
sec^2
x
2
=
1
2
(
1+tan^2
x
2
)
=
1+t^2
2
,
the required relationship is
dx=
2
1+t^2
dt. (2.34)
Evaluate the integral
I=
∫
2
1+3cosx
dx.
Rewriting cosxin terms oftand using (2.34) yields
I=
∫
2
1+3
[
(1−t^2 )(1 +t^2 )−^1
]
(
2
1+t^2
)
dt
=
∫
2(1 +t^2 )
1+t^2 +3(1−t^2 )
(
2
1+t^2
)
dt
=
∫
2
2 −t^2
dt=
∫
2
(
√
2 −t)(
√
2+t)
dt
=
∫
1
√
2
(
1
√
2 −t
+
1
√
2+t
)
dt
=−
1
√
2
ln(
√
2 −t)+
1
√
2
ln(
√
2+t)+c
=
1
√
2
ln
[√
2+tan(x/2)
√
2 −tan (x/2)
]
+c.
Integrals of a similar form to (2.33), but involving sin 2x,cos2x, tan 2x,sin^2 x,
cos^2 xor tan^2 xinstead of cosxand sinx, should be evaluated by using the
substitutiont=tanx.Inthiscase
sinx=
t
√
1+t^2
, cosx=
1
√
1+t^2
and dx=
dt
1+t^2
. (2.35)
A final example of the evaluation of integrals using substitution is the method
of completing the square (cf. subsection 1.7.3).