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errors can cancel each other out∗
pahio†
2013-03-22 2:59:42
If one uses the change of variable
tan x := t,
dx =
dt
,
1+t2
cos2 x =
1
1+t2
(1)
for finding the value of the definite integral
3π
4
Z
I :=
π
4
dx
,
2 cos2 x + 1
the following calculation looks appropriate and faultless:
Z
I =
1
−1
−1
1 5π π
dt
1 .
t
2π
√
√
√
=
=
arctan
−
= √
3+t2
6
3 1
3
3 6
3 3
(2)
The result is quite right. Unfortunately, the calculation contains two errors,
the effects of which cancel each other out.
The crucial error in (2) is using the substitution (1) when tan x is discontinuous in the point x = π2 on the interval [ π4 , 3π
4 ] of integration. The error
−1
√
is however canceled out by the second error using the value 5π
6 for arctan 3 ,
when the right value were − π6 (the values of arctan lie only between − π2 and π2 ;
see cyclometric functions). The value 5π
6 belongs to a different branch of the
inverse tangent function than π6 ; parts of two distinct branches cannot together
form the antiderivative which must be continuous.
What were a right way to calculate I? The universal trigonometric substitution produces an awkward integrand
2+2t2
3−2t2 +3t4
∗ hErrorsCanCancelEachOtherOuti
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1
√
√
and limits 2−1 and 1− 2, therefore it is unusable. It is now better to change
the interval of integration, using the properties of trigonometric functions.
Since the (graph of) cosine squared is symmetric about the line x = π2 , we
could integrate only over [ π4 , π2 ] and multiply the integral by 2 (cf. integral of
even and odd functions):
π
2
Z
dx
.
2 cos2 x + 1
I = 2
π
4
We can also get rid of the inconvenient upper limit
sine in virtue of the complement formula
π
2
by changing over to the
π
cos( −x) = sin x,
2
getting
π
4
Z
dx
.
2 sin2 x + 1
I = 2
0
2
Then (1) is usable, and because sin x =
Z
I = 2
0
1
2t2
1+t2
dt
= 2
+ 1 (1+t2 )
Z
0
1
t2
1+t2 ,
we obtain
1
√
2π
dt
2 .
= √
arctan t 3 = √ .
3t2 +1
3 0
3 3
2