Bioche's rules

Bioche's rules, formulated by the French mathematician Template: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, $f(t)$ 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 $dt$, 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 $\omega (-t)=\omega (t)$, a good change of variables is $u=\cos t$.
2. If $\omega (\pi -t)=\omega (t)$, a good change of variables is $u=\sin t$.
3. If $\omega (\pi +t)=\omega (t)$, a good change of variables is $u=\tan t$.
4. If two of the preceding relations both hold, a good change of variables is $u=\cos 2t$.
5. 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 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

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

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

Example 1

As a trivial example, consider

$\int \sin tdt.$

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

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 $\frac{1}{1-u^2}=\frac{1}{2}(\frac{1}{1+u}+\frac{1}{1-u})$. The result is that

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

Example 3

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

$\int \frac{\mathrm{d}t}{1+\beta \cos t}=\frac{2}{\sqrt{1-{\beta }^2}}\arctan [\sqrt{\frac{1-\beta }{1+\beta }}\tan \frac{t}{2}]+c.$

References

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