Английская Википедия:Bioche's rules
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Bioche's rules, formulated by the French mathematician Шаблон:Ill (1859–1949), are rules to aid in the computation of certain indefinite integrals in which the integrand contains sines and cosines.
In the following, <math>f(t)</math> is a rational expression in <math>\sin t</math> and <math>\cos t</math>. In order to calculate <math>\int f(t)\,dt</math>, consider the integrand <math>\omega(t)=f(t)\,dt</math>. We consider the behavior of this entire integrand, including the <math display="inline> dt</math>, under translation and reflections of the t axis. The translations and reflections are ones that correspond to the symmetries and periodicities of the basic trigonometric functions.
Bioche's rules state that:
- If <math>\omega(-t)=\omega(t)</math>, a good change of variables is <math>u=\cos t</math>.
- If <math>\omega(\pi-t)=\omega(t)</math>, a good change of variables is <math>u=\sin t</math>.
- If <math>\omega(\pi+t)=\omega(t)</math>, a good change of variables is <math>u=\tan t</math>.
- If two of the preceding relations both hold, a good change of variables is <math>u=\cos 2t</math>.
- In all other cases, use <math>u=\tan(t/2)</math>.
Because rules 1 and 2 involve flipping the t axis, they flip the sign of dt, and therefore the behavior of ω under these transformations differs from that of ƒ by a sign. Although the rules could be stated in terms of ƒ, stating them in terms of ω has a mnemonic advantage, which is that we choose the change of variables u(t) that has the same symmetry as ω.
These rules can be, in fact, stated as a theorem: one shows[1] that the proposed change of variable reduces (if the rule applies and if f is actually of the form <math>f(t) = \frac{P(\sin t, \cos t)}{Q(\sin t, \cos t)}</math>) to the integration of a rational function in a new variable, which can be calculated by partial fraction decomposition.
Case of polynomials
To calculate the integral <math>\int\sin^p(t)\cos^q(t)dt</math>, Bioche's rules apply as well.
- If p and q are odd, one uses <math>u = \cos(2t)</math>;
- If p is odd and q even, one uses <math>u = \cos(t)</math>;
- If p is even and q odd, one uses <math>u = \sin(t)</math>;
- If not, one is reduced to lineariz.
Another version for hyperbolic functions
Suppose one is calculating <math>\int g(\cosh t, \sinh t)dt</math>.
If Bioche's rules suggest calculating <math>\int g(\cos t, \sin t)dt</math> by <math>u = \cos(t)</math> (respectively, <math>\sin t, \tan t, \cos(2t), \tan(t/2)</math>), in the case of hyperbolic sine and cosine, a good change of variable is <math>u = \cosh(t)</math> (respectively, <math>\sinh(t), \tanh(t), \cosh(2t), \tanh(t/2)</math>). In every case, the change of variable <math>u = e^t</math> allows one to reduce to a rational function, this last change of variable being most interesting in the fourth case (<math>u = \tanh(t/2)</math>).
Examples
Example 1
As a trivial example, consider
- <math>\int \sin t \,dt.</math>
Then <math>f(t)=\sin t</math> is an odd function, but under a reflection of the t axis about the origin, ω stays the same. That is, ω acts like an even function. This is the same as the symmetry of the cosine, which is an even function, so the mnemonic tells us to use the substitution <math>u=\cos t</math> (rule 1). Under this substitution, the integral becomes <math>-\int du</math>. The integrand involving transcendental functions has been reduced to one involving a rational function (a constant). The result is <math>-u+c=-\cos t+c</math>, which is of course elementary and could have been done without Bioche's rules.
Example 2
The integrand in
- <math>\int \frac{dt}{\sin t}</math>
has the same symmetries as the one in example 1, so we use the same substitution <math>u=\cos t</math>. So
- <math>\frac{dt}{\sin t} = - \frac{du}{\sin^2 t} = - \frac{du}{\ 1-\cos^2 t}. </math>
This transforms the integral into
- <math>\int - \frac{du}{1 - u^2},</math>
which can be integrated using partial fractions, since <math>\frac {1}{1-u^2} = \frac {1}{2} \left( \frac{1}{1+u}+\frac{1}{1-u}\right)</math>. The result is that
- <math>\int \frac{dt}{\sin t}=-\frac{1}{2}\ln\frac{1+\cos t}{1-\cos t}+c.</math>
Example 3
Consider
- <math>\int \frac{dt}{1+\beta\cos t},</math>
where <math>\beta^2<1</math>. Although the function f is even, the integrand as a whole ω is odd, so it does not fall under rule 1. It also lacks the symmetries described in rules 2 and 3, so we fall back to the last-resort substitution <math>u=\tan(t/2)</math>.
Using <math>\cos t=\frac{1-\tan^2(t/2)}{1+\tan^2(t/2)}</math> and a second substitution <math>v=\sqrt{\frac{1-\beta}{1+\beta}}u</math> leads to the result
- <math>\int \frac{\mathrm{d}t}{1+\beta\cos t} = \frac{2}{\sqrt{1-\beta^2}}\arctan\left[\sqrt{\frac{1-\beta}{1+\beta}}\tan \frac{t}{2}\right] + c.</math>
References
Шаблон:WikiversityШаблон:Reflist
- Zwillinger, Handbook of Integration, p. 108
- Stewart, How to Integrate It: A practical guide to finding elementary integrals, pp. 190−197.
