# errors can cancel each other out

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

 $\displaystyle\tan{x}\;:=\;t,\quad dx\;=\;\frac{dt}{1\!+\!t^{2}},\quad\cos^{2}x% \;=\;\frac{1}{1\!+\!t^{2}}$ (1)

for finding the value of the definite integral

 $I\;:=\;\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{dx}{2\cos^{2}x+1},$

the following calculation looks appropriate and faultless:

 $\displaystyle I\;=\;\int_{1}^{-1}\!\!\frac{dt}{3\!+\!t^{2}}\;=\;\frac{1}{\sqrt% {3}}\!\!\operatornamewithlimits{\Big{/}}_{\!\!\!1}^{\,\;\quad-1}\!\arctan\frac% {t}{\sqrt{3}}\;=\;\frac{1}{\sqrt{3}}\!\left(\frac{5\pi}{6}-\frac{\pi}{6}\right% )\;=\;\frac{2\pi}{3\sqrt{3}}$ (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 $\tan{x}$ is discontinuous  in the point  $x=\frac{\pi}{2}$  on the interval$[\frac{\pi}{4},\,\frac{3\pi}{4}]$  of integration.  The error is however canceled out by the second error using the value $\frac{5\pi}{6}$ for $\arctan\frac{-1}{\sqrt{3}}$, when the right value were $-\frac{\pi}{6}$ (the values of arctan lie only between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$; see cyclometric functions).  The value $\frac{5\pi}{6}$ belongs to a different branch of the inverse tangent function  than $\frac{\pi}{6}$; 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

 $\frac{2\!+\!2t^{2}}{3\!-\!2t^{2}\!+\!3t^{4}}$

and $\sqrt{2}-1$ and $1-\sqrt{2}$,  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  $x=\frac{\pi}{2}$,  we could integrate only over  $[\frac{\pi}{4},\,\frac{\pi}{2}]$ and multiply the integral  by 2 (cf. integral of even and odd functions):

 $I\;=\;2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{dx}{2\cos^{2}x+1}.$

We can also get rid of the inconvenient upper limit $\frac{\pi}{2}$ by changing over to the sine in virtue of the complement formula

 $\cos(\frac{\pi}{2}\!-\!x)\;=\;\sin{x},$

getting

 $I\;=\;2\int_{0}^{\frac{\pi}{4}}\frac{dx}{2\sin^{2}x+1}.$

Then (1) is usable, and because  $\sin^{2}x=\frac{t^{2}}{1\!+\!t^{2}}$,  we obtain

 $I\;=\;2\int_{0}^{1}\frac{dt}{\left(\frac{2t^{2}}{1\!+\!t^{2}}+1\right)(1\!+\!t% ^{2})}\;=\;2\int_{0}^{1}\frac{dt}{3t^{2}\!+\!1}\;=\;\frac{2}{\sqrt{3}}\!\!% \operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,\;\quad 1}\!\arctan{t\sqrt{3}}\;% =\;\frac{2\pi}{3\sqrt{3}}.$
Title errors can cancel each other out ErrorsCanCancelEachOtherOut 2013-03-22 18:59:39 2013-03-22 18:59:39 pahio (2872) pahio (2872) 13 pahio (2872) Example msc 00A35 msc 26A06 msc 97D70 UniversalTrigonometricSubstitution SubstitutionNotation IntegrationOfRationalFunctionOfSineAndCosine