Bioche's rules

Bioche's rules, formulated by the French mathematician  [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, ${\displaystyle f(t)}$ is a rational expression in ${\displaystyle \sin t}$ and ${\displaystyle \cos t}$. In order to calculate ${\displaystyle \int f(t)\,dt}$, consider the integrand ${\displaystyle \omega (t)=f(t)\,dt}$. 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:

1. If ${\displaystyle \omega (-t)=\omega (t)}$, a good change of variables is ${\displaystyle u=\cos t}$.
2. If ${\displaystyle \omega (\pi -t)=\omega (t)}$, a good change of variables is ${\displaystyle u=\sin t}$.
3. If ${\displaystyle \omega (\pi +t)=\omega (t)}$, a good change of variables is ${\displaystyle u=\tan t}$.
4. If two of the preceding relations both hold, a good change of variables is ${\displaystyle u=\cos 2t}$.
5. In all other cases, use ${\displaystyle u=\tan(t/2)}$.

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.

Examples[]

Example 1[]

As a trivial example, consider

${\displaystyle \int \sin t\,dt.}$

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

Example 2[]

The integrand in

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

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

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

This transforms the integral into

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

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

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

Example 3[]

Consider

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

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

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

${\displaystyle \int {\frac {\mathrm {d} t}{1+\beta \cos t}}={\frac {2}{\sqrt {1-\beta ^{2}}}}\arctan \left[{\frac {(1-\beta )\sin t}{{\sqrt {1-\beta ^{2}}}(1+\cos t)}}\right]+c.}$

References[]

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

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