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单词 ErrorsCanCancelEachOtherOut
释义

errors can cancel each other out


If one uses the http://planetmath.org/ChangeOfVariableInDefiniteIntegralchange of variable

tanx:=t,dx=dt1+t2,cos2x=11+t2(1)

for finding the value of the definite integral

I:=π43π4dx2cos2x+1,

the following calculation looks appropriate and faultless:

I=1-1dt3+t2=13/1-1arctant3=13(5π6-π6)=2π33(2)

The result is quite .  Unfortunately, the calculation two errors, the effects of which cancel each other out.

The crucial error in (2) is using the substitution (1) when tanx is discontinuousMathworldPlanetmath in the point x=π2  on the interval  [π4,3π4]  of integration.  The error is however canceled out by the second error using the value 5π6 for arctan-13, when the right value were -π6 (the values of arctan lie only between -π2 and π2; see cyclometric functions).  The value 5π6 belongs to a different branch of the inverse tangent functionMathworldPlanetmath than π6; parts of two distinct branches cannot together form the antiderivative which must be continuousMathworldPlanetmath.

What were a right way to calculate I?  The universal trigonometric substitutionPlanetmathPlanetmath produces an awkward integrand

2+2t23-2t2+3t4

and 2-1 and 1-2,  therefore it is unusable.  It is now better to change the interval of integration, using the properties of trigonometric functionsDlmfMathworldPlanetmath.

Since the (graph of) cosine squared is symmetric about the line  x=π2,  we could integrate only over [π4,π2] and multiply the integralDlmfPlanetmath by 2 (cf. integral of even and odd functions):

I= 2π4π2dx2cos2x+1.

We can also get rid of the inconvenient upper limit π2 by changing over to the sine in virtue of the complement formula

cos(π2-x)=sinx,

getting

I= 20π4dx2sin2x+1.

Then (1) is usable, and because  sin2x=t21+t2,  we obtain

I= 201dt(2t21+t2+1)(1+t2)= 201dt3t2+1=23/01arctant3=2π33.
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更新时间:2025/5/4 22:48:07