5)
The example is to integrate
|
dx
4 + sin(x)
|
If instead this were
|
dx
1 + sin(x)
|
it would be easy. You could just multiply numerator and denominator
by
1 - sin(x), apply the trig
identity,
cos2(x) = 1 - sin2(x),
and then apply simple substitution. I urge you to carry that one to completion
on your own as an exercise.
But when the constant in the denominator is anything other than 1, the
difference of squares trick doesn't yield anything useful. Instead the immediate
goal becomes getting the integrand into an expression in tangents so that we can
use our tangent substitution trick on it. Your first impulse toward that end
is to try using the identity
tan(x)
sin(x) = ___________
√tan2(x) + 1
|
When we try that, we can do some simplification and apply the tangent substitution,
but we still end up with something pretty horrendous (although probably doable
if you had enough time). Instead of attacking something horrendous,
the mental leap here is to realize that in order
to do the tangent substitution trick, you don't necessarily need the integrand to be
a function of
tan(x). You just need it to be a function
of tangent of
something.
The first step I'm going to do here won't seem obvious at all, but remember
it as something to try if you ever get into a jam trying to integrate something
that's made out of trig functions. The substitution is
x
u =
2
|
and
|
2 du = dx eq. 11.8-53a
|
This is the first key we need to unlock the door to this integral.
Using the
double angle formula
for sine, the integral becomes
2
|
|
du
=
4 + sin(2u)
|
2
|
|
du
= eq. 11.8-53b
4 + 2 sin(u)cos(u)
|
|
du
2 + sin(u)cos(u)
|
|
Recall that
|
1
sec2(u) = = tan2(u) + 1
cos2(u)
|
and
|
sin(u)
tan(u) =
cos(u)
|
So if we multiply numerator and denominator by
sec2(u),
we get
|
du
=
2 + sin(u)cos(u)
|
|
sec2(u) du
= eq. 11.8-53c
2 sec2(u) + tan(u)
|
|
tan2(u) + 1
du
2 tan2(u) + 2 + tan(u)
|
which is the second key we need to unlock the door to this integral.
With both keys in place, the form we now have easily admits our tangent substitution trick.
v = tan(u)
|
and
|
dv
= du eq. 11.8-54a
v2 + 1
|
Making that substitution yields (after taking the cancellation of
v2 + 1 and rearranging the
terms in the denominator)
|
tan2(u) + 1
du =
2 tan2(u) + 2 + tan(u)
|
|
dv
eq. 11.8-54b
2v2 + v + 2
|
We have to
complete the square now.
But this one has a new wrinkle to it. The
v2 coefficient
is not
1. When you have a denominator of
av2 + bv + c, you complete
the square by letting
w = v + b/(2a).
In this case that means
1
w = v +
4
|
and
|
2w2 = 2v2 + v +
|
1
8
|
|
hence
|
2w2 +
|
15
=
8
|
2v2 + v + 2
|
and
|
dw = dv eq. 11.8-55a
|
Making the substitution then multiplying numerator and denominator
by
8/15 we get
|
dv
=
2v2 + v + 2
|
|
dw
=
2w2 + 15/8
|
8
15
|
|
dw
__ eq. 11.8-55b
(4w/√15)2 + 1
|
which paves the way to a trig substitution. The trig substitution is
4w
tan(t) = __
√15
|
and
|
__
√15
4
|
(tan2(t) + 1) dt = dw eq. 11.8-56a
|
yielding
8
15
|
|
dw
__ =
(4w/√15)2 + 1
|
2
__
√15
|
|
2
dt = __ t + C eq. 11.8-56b
√15
|
Now we substitute back
t to
w to
v to
u to
x.
|
dx
=
4 + sin(x)
|
2
__ t + C =
√15
|
2
__ arctan
√15
|
|
4w
__
√15
|
|
+ C = eq. 11.8-56c
|
2
__ arctan
√15
|
|
4v + 1
__
√15
|
|
+ C =
|
2
__ arctan
√15
|
|
4 tan(u) + 1
__
√15
|
|
+ C =
|
2
__ arctan
√15
|
|
4 tan(x/2) + 1
__
√15
|
|
+ C
|
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