Bioche's rules

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

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

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.

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

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

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 ${\displaystyle v={\sqrt {\frac {1-\beta }{1+\beta }}}u}$ leads to the result

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

Wikiversity has learning resources about Bioche's rules

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