The substitution: Let t = tan$\frac{x}{2}$. Then:

• sin(x) = 2$\frac{t}{1+t^2}$
• cos(x) = $\frac{1-t^2}{(1+t^2)}$
• tan(x) = 2$\frac{t}{1-t^2}$
• dx = 2$\frac{dt}{1+t^2}$

When to use: Integrals of rational functions of sin(x) and cos(x), especially when other methods fail. The substitution converts any such integral into a rational function of t.

Example: integral $\frac{dx}{1+sin x}$ = integral [$\frac{2}{1+t^2}$] / [1+2$\frac{t}{1+t^2}$] dt = integral 2$\frac{dt}{(1+t}$^2) = -$\frac{2}{1+t}$ = -$\frac{2}{1+tan(x/2}$) + C.

Drawback: Often produces complicated expressions. Try simpler substitutions first (like t = tan x for integral of sec x, or multiplying by conjugates).

Alternative: For integrals of type integral R(sin x, cos x) dx where R(-sin x, -cos x) = R(sin x, cos x), try t = tan x instead.