Bioche's_rules Knowpia

Bioche's rules, formulated by the French mathematician Charles Bioche [fr] (1859–1949), are rules to aid in the computation of certain indefinite integrals in which the integrand contains sines and cosines.

In the following, $f(t)$ is a rational expression in $\sin t$ and $\cos t$ . In order to calculate $\int f(t)\,dt$ , consider the integrand $\omega (t)=f(t)\,dt$ . We consider the behavior of this entire integrand, including the ${\textstyle dt}$ , 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 $\omega (-t)=\omega (t)$ , a good change of variables is $u=\cos t$ .
If $\omega (\pi -t)=\omega (t)$ , a good change of variables is $u=\sin t$ .
If $\omega (\pi +t)=\omega (t)$ , a good change of variables is $u=\tan t$ .
If two of the preceding relations both hold, a good change of variables is $u=\cos 2t$ .
In all other cases, use $u=\tan(t/2)$ .

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 $f(t)={\frac {P(\sin t,\cos t)}{Q(\sin t,\cos t)}}$ ) to the integration of a rational function in a new variable, which can be calculated by partial fraction decomposition.

Case of polynomials edit

To calculate the integral $\int \sin ^{p}(t)\cos ^{q}(t)dt$ , Bioche's rules apply as well.

If p and q are odd, one uses $u=\cos(2t)$ ;
If p is odd and q even, one uses $u=\cos(t)$ ;
If p is even and q odd, one uses $u=\sin(t)$ ;
If not, one is reduced to lineariz.

Another version for hyperbolic functions edit

Suppose one is calculating $\int g(\cosh t,\sinh t)dt$ .

If Bioche's rules suggest calculating $\int g(\cos t,\sin t)dt$ by $u=\cos(t)$ (respectively, $\sin t,\tan t,\cos(2t),\tan(t/2)$ ), in the case of hyperbolic sine and cosine, a good change of variable is $u=\cosh(t)$ (respectively, $\sinh(t),\tanh(t),\cosh(2t),\tanh(t/2)$ ). In every case, the change of variable $u=e^{t}$ allows one to reduce to a rational function, this last change of variable being most interesting in the fourth case ( $u=\tanh(t/2)$ ).

Examples edit

Example 1 edit

As a trivial example, consider

\int \sin t\,dt.

Then $f(t)=\sin t$ 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 $u=\cos t$ (rule 1). Under this substitution, the integral becomes $-\int du$ . The integrand involving transcendental functions has been reduced to one involving a rational function (a constant). The result is $-u+c=-\cos t+c$ , which is of course elementary and could have been done without Bioche's rules.

Example 2 edit

The integrand in

\int {\frac {dt}{\sin t}}

has the same symmetries as the one in example 1, so we use the same substitution $u=\cos t$ . So

{\frac {dt}{\sin t}}=-{\frac {du}{\sin ^{2}t}}=-{\frac {du}{\ 1-\cos ^{2}t}}.

This transforms the integral into

\int -{\frac {du}{1-u^{2}}},

which can be integrated using partial fractions, since ${\frac {1}{1-u^{2}}}={\frac {1}{2}}\left({\frac {1}{1+u}}+{\frac {1}{1-u}}\right)$ . The result is that

\int {\frac {dt}{\sin t}}=-{\frac {1}{2}}\ln {\frac {1+\cos t}{1-\cos t}}+c.

Example 3 edit

Consider

\int {\frac {dt}{1+\beta \cos t}},

where $\beta ^{2}<1$ . 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 $u=\tan(t/2)$ .

Using $\cos t={\frac {1-\tan ^{2}(t/2)}{1+\tan ^{2}(t/2)}}$ and a second substitution $v={\sqrt {\frac {1-\beta }{1+\beta }}}u$ leads to the result

\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.

References edit

^ Vidiani, L.G. (October 1976). "Règles de Bioche" [Bioche's rules] (PDF). Revue de mathématiques et de sciences physiques (in French): 1–2. Archived from the original on 18 July 2022. Retrieved 10 June 2022.{{cite journal}}: CS1 maint: bot: original URL status unknown (link)

Zwillinger, Handbook of Integration, p. 108
Stewart, How to Integrate It: A practical guide to finding elementary integrals, pp. 190−197.

[1] Vidiani, L.G. (October 1976). "Règles de Bioche" [Bioche's rules] (PDF). Revue de mathématiques et de sciences physiques (in French): 1–2. Archived from the original on 18 July 2022. Retrieved 10 June 2022.{{cite journal}}: CS1 maint: bot: original URL status unknown (link)

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