The Weierstrass Substitution

November 13, 2009

tags: anti-differentiation tricks, calculus, indefinite integrals, trigonometric identities, Weierstrass substitution

Back in an earlier post we considered a rational parameterization of the unit circle. We saw there that for $\theta = 2\tan^{-1}(x)$ we have $\displaystyle{ \cos \theta = \frac{1-x^2}{1+x^2} }$ and $\displaystyle{ \sin \theta = \frac{2x}{1+x^2}}$ . A moment’s reflection reveals that this substitution would transform any rational function of $\sin \theta$ and $\cos \theta$ into a rational function of $x$ . This is the Weierstrass Substitution. Its main application is to the anti-differentiation of rational functions of $\sin \theta$ and $\cos \theta$ . We would have

$\displaystyle{ \int\! R(\sin \theta, \cos \theta)\,d\theta = \int\! R\left(\frac{2x}{1+x^2}, \frac{1-x^2}{1+x^2}\right)\,\frac{2}{1+x^2}\,dx }$

where we calculated $d\theta = \frac{2}{1+x^2}\,dx$ from $\theta = 2\tan^{-1} x$ .

Here are a pair of examples.

Consider the integral $\displaystyle{ \int\!\frac{d\theta}{1-\sin \theta + \cos \theta} }$ . We apply the Weierstrass substitution to find

$\displaystyle{ \int\!\frac{d\theta}{1-\sin \theta + \cos \theta} = \int\!\frac{\frac{2\,dx}{1+x^2}}{1 - \frac{2x}{1+x^2} + \frac{1-x^2}{1+x^2}} }$

$\displaystyle{ = \int\! \frac{2\,dx}{1+x^2 - 2x + 1-x^2} }$

$\displaystyle{ = \int\! \frac{2\,dx}{2 - 2x} }$

$\displaystyle{ = \int\! \frac{dx}{1 - x} }$

$\displaystyle{ = -\ln|1-x| + C }$

$\displaystyle{ = -\ln|1-\tan(\theta/2)| + C }$

We considered the integral $\int\! \sec \theta\,d\theta$ in an earlier post. Let’s do it again with the help of the Weierstrass substitution.

$\displaystyle{ \int\! \sec \theta\,d\theta = \int\! \frac{1+x^2}{1-x^2} \cdot \frac{2}{1+x^2}\,dx }$

$\displaystyle{ = \int\! \frac{2}{1-x^2}\,dx }$

$\displaystyle{ = \int\! \left( \frac{1}{1+x} + \frac{1}{1-x}\right)\,dx }$

$\displaystyle{ = \ln|1+x| - \ln|1-x| + C }$

$\displaystyle{ = \ln\left|\frac{1+x}{1-x}\right| + C }$

$\displaystyle{ = \ln\left|\frac{1+\tan(\theta/2)}{1-\tan(\theta/2)}\right| + C }$

This answer has a very different form from the ones given in our earlier post. We can reconcile this answer with the older ones through algebra and trigonometric identities.

We have

$\displaystyle{ \frac{1+\tan(\theta/2)}{1-\tan(\theta/2)} = \frac{1+\tan(\theta/2)}{1-\tan(\theta/2)} \cdot \frac{\cos(\theta/2)}{\cos(\theta/2)} }$

$\displaystyle{ = \frac{\cos(\theta/2) + \sin(\theta/2)}{\cos(\theta/2) - \sin(\theta/2)} = \frac{\cos(\theta/2) + \sin(\theta/2)}{\cos(\theta/2) - \sin(\theta/2)} \cdot \frac{\cos(\theta/2) + \sin(\theta/2)}{\cos(\theta/2) + \sin(\theta/2)} }$

$\displaystyle{ = \frac{\cos^2(\theta/2) + \sin^2(\theta/2) + 2\sin(\theta/2)\cos(\theta/2)}{\cos^2(\theta/2) - \sin^2(\theta/2)}}$

$\displaystyle{ = \frac{1 + \sin \theta}{\cos \theta} }$

$\displaystyle{ = \sec \theta + \tan \theta }$

as expected. (In the penultimate equality we used the usual $\cos^2 A + \sin^2 A = 1$ , $\sin 2A = 2\sin A\cos A$ , and $\cos 2A = \cos^2 A - \sin^2 A$ identities.)

Notes Mathematical and Otherwise

The Weierstrass Substitution

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