Sine half-angle substitution

 If you are a student of integral calculus, it is highly likely that you have come across or will come across the famous Weierstrass substitution, which is very useful for converting rational expressions involving trigonometric functions into ordinary rational expressions involving $t$, where $t=\tan{\frac12{x}}$. The general transformation formula is as follows:

$$\int f(\sin{x},\cos{x})\,dx =  \int f\left(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}\right)\frac{2dt}{1+t^2}.$$

In this note, we introduce another substitution as a companion to the Weierstrass substitution that can transform certain rational expressions of trigonometric functions into simpler ordinary rational expressions than the Weierstrass substitution. For instance, it would be useful when a common linear factor of $\sin{x}$ appears in the numerator or denominator of the integrand. The general transformation formula is given by:

$$\int f(\sin{x},\cos{x})\,dx =  \int f(2\sqrt{s-s^2}, 1-2s)\frac{ds}{\sqrt{s-s^2}}.$$

Here $s=\sin^2{\frac12x}$. 

Derivation

Using the double-angle formulas and the Pythagorean identity, one gets
$$\sin{x}=2\sin{\frac12x}\cos{\frac12x}=2\sin{\frac12x}\sqrt{1-\sin^2{\frac12x}}=2\sqrt{\sin^2{\frac12x}-\sin^4{\frac12x}}=2\sqrt{s-s^2},$$
$$\cos{x}=1-2\sin^2{\frac12x}=1-2s.$$
Finally, since $s=\sin^2{\frac12x}$, differentiation rules imply
$$ds=\sin{\frac12x}\cos{\frac12x}\,dx=\frac{\sin{x}}{2}\,dx,$$
and thus,
$$dx=\frac{ds}{\sqrt{s-s^2}}.$$

Example 1

By applying the sine half-angle substitution and simplifying,

$$\int \frac{\sin{x}}{\sin^2{x}+2\cos{x}}\,dx=\int \frac{1}{1-2s^2}\,ds.$$

Using standard integrals,

$$\int \frac{1}{1-2s^2}\,ds=\frac{\tanh^{-1}{(\sqrt{2}s)}}{\sqrt{2}}+C=\frac{\tanh^{-1}{(\sqrt{2}\sin^2{\frac12x})}}{\sqrt{2}}+C.$$

The advantage of this substitution is evident when comparing it to the solution provided in this integral calculator (which solution, by the way, is equivalent to the one given here) or when using the Weierstrass substitution.

Example 2 

By applying the sine half-angle substitution,

$$\int \frac{1}{\sin^3{x}}\,dx=\frac18\int \frac{1}{(s-s^2)^2}\,ds.$$

Using partial fraction decomposition, 

$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac18\int \left(\frac{1}{s}+\frac{1}{s^2}-\frac{2}{s-1}+\frac{1}{(s-1)^2}\right)\,ds=\frac14\ln{\left|\frac{s}{1-s}\right|}+\frac18\left(\frac{1}{1-s}-\frac{1}{s}\right)+C.$$

Substituting $s$ by $\sin^2{\frac12x}$, we have

$$\frac18\int \frac{1}{(s-s^2)^2}\,ds=\frac14\ln{\left|\tan^2{\frac12x}\right|}+\frac18\left(\frac{1}{\cos^2{\frac12x}}-\frac{1}{\sin^2{\frac12x}}\right)+C=\frac12\left(\ln{|\tan{\frac12x}|}-\frac{\cos{x}}{\sin^2{x}}\right)+C.$$

A solution using integration by parts is given at Integrals For You.