# The Weierstrass Substitution

A standard way to calculate \(\int{\frac{dx}{1+\text{sin}x}}\) is via a substitution \(u=\text{tan}(x/2)\). As I'll show in a moment, this substitution leads to

\( \begin{align} \text{sin}x&=\frac{2u}{1+u^2} \\ \text{cos}x&=\frac{1-u^2}{1+u^2} \\ \text{tan}x&=\frac{2u}{1-u^2} \\ dx&=\frac{2du}{1+u^2} \end{align} \)

Using these,

\( \begin{align} \int{\frac{dx}{1+\text{sin}x}}&=\int{\frac{1}{1+2u/(1+u^{2})}\frac{2}{1+u^2}du} \\ &=\int{\frac{2du}{1+2u+u^2}} \\ &=\int{\frac{2du}{(1+u)^2}} \\ &=-\frac{2}{1+u}+C \\ &=-\frac{2}{1+\text{tan}(x/2)}+C. \end{align} \)

The method is known as the *Weierstrass substitution*. It applies to trigonometric integrals that include a mixture of constants and trigonometric function. H. Anton, though, warns the student that *the substitution can lead to cumbersome partial fractions decompositions and consequently should be used only in the absence of finding a simpler method*.

For another example,

\( \begin{align} \int{\frac{dx}{\text{sin}x+\text{tan}x}}&=\int{\frac{1}{\frac{2u}{1+u^2}+\frac{2u}{1-u^2}}\frac{2}{1+u^2}du} \\ &=\int{\frac{2(1-u^{2})}{2u}du} \\ &=\int{(\frac{1}{u}-u)du} \\ &=\text{ln}|u|-\frac{u^2}{2} + C \\ &=\text{ln}|\text{tan}(x/2)|-\frac{\text{tan}^2(x/2)}{2} + C. \end{align} \)

Now, let's return to the substitution formulas. It's not difficult to derive them using trigonometric identities. But here is a proof without words due to Sidney Kung:

The diagram is suggestive of

\(\text{sin}\theta=\frac{AC}{AB}=\frac{2u}{1+u^2}\) and \(\text{cos}\theta=\frac{BC}{AB}=\frac{1-u^2}{1+u^2}\).

Finally, it must be clear that, since \(\text{tan}x\) is undefined for \(\frac{\pi}{2}+k\pi\), \(k\) any integer, the substitution is only meaningful when restricted to intervals that do not contain those values, e.g., for \(-\pi\lt x\lt\pi\).

## References

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