The Weierstrass substitutionEdexcel A-Level Further Maths: Revision notes

The Weierstrass substitution (the tangent half-angle substitution) is $t=\tan\frac{x}{2}$. It turns an integral of trigonometric functions into an integral of a rational function of $t$. The three results needed are $\sin x=\frac{2t}{1+t^2},\quad\cos x=\frac{1-t^2}{1+t^2},\quad\mathrm{d}x=\frac{2\,\mathrm{d}t}{1+t^2}.$ The last comes from $\frac{\mathrm{d}t}{\mathrm{d}x}=\frac12\sec^2\frac{x}{2}=\frac12(1+t^2)$. Use it for integrals whose denominators involve $\sin x$ and $\cos x$ in a combination such as $1+\sin x$, $2+\cos x$ or $\operatorname{cosec}x$.

Find $\int\operatorname{cosec}x\,\mathrm{d}x$. With $\operatorname{cosec}x=\frac{1+t^2}{2t}$ and $\mathrm{d}x=\frac{2\,\mathrm{d}t}{1+t^2}$, $\int\frac{1+t^2}{2t}\cdot\frac{2}{1+t^2}\,\mathrm{d}t=\int\frac{1}{t}\,\mathrm{d}t=\ln|t|+c=\ln\left|\tan\frac{x}{2}\right|+c.$ The $(1+t^2)$ factors cancel completely. The answer must be returned in terms of $x$ for an indefinite integral.

For a definite integral, change the limits using $t=\tan\frac{x}{2}$: $x=0\to t=0$, $x=\frac{\pi}{3}\to t=\frac{1}{\sqrt3}$, $x=\frac{\pi}{2}\to t=1$. Example: $\int_0^{\pi/2}\frac{1}{1+\sin x}\,\mathrm{d}x$. Since $1+\sin x=\frac{(1+t)^2}{1+t^2}$, the integrand becomes $\frac{2}{(1+t)^2}$ and $\int_0^1\frac{2}{(1+t)^2}\,\mathrm{d}t=\left[-\frac{2}{1+t}\right]_0^1=1.$ Another: $\int\frac{1}{1+\cos x}\,\mathrm{d}x$ becomes $\int\mathrm{d}t=\tan\frac{x}{2}+c$.

Evaluate $\int_{\pi/3}^{\pi/2}\frac{1}{1+\sin x-\cos x}\,\mathrm{d}x$. First $1+\sin x-\cos x=\frac{(1+t^2)+2t-(1-t^2)}{1+t^2}=\frac{2t(t+1)}{1+t^2}$, so the integrand becomes $\frac{1}{t(t+1)}$. Partial fractions: $\frac{1}{t(t+1)}=\frac1t-\frac{1}{t+1}$. Then $\int_{1/\sqrt3}^{1}\left(\frac1t-\frac{1}{t+1}\right)\mathrm{d}t=\left[\ln\frac{t}{t+1}\right]_{1/\sqrt3}^{1}=\ln\frac12-\ln\frac{1}{1+\sqrt3}=\ln\frac{1+\sqrt3}{2}.$

If the substituted integrand has a quadratic denominator of the form $a^2+t^2$, use $\int\frac{1}{a^2+t^2}\,\mathrm{d}t=\frac1a\arctan\frac ta$. Example: $\int_0^{\pi/2}\frac{1}{2+\cos x}\,\mathrm{d}x$. Here $2+\cos x=\frac{3+t^2}{1+t^2}$, so the integrand is $\frac{2}{3+t^2}$ and the integral is $\left[\frac{2}{\sqrt3}\arctan\frac{t}{\sqrt3}\right]_0^1=\frac{\pi\sqrt3}{9}$. Check: $\arctan\frac{1}{\sqrt3}=\frac{\pi}{6}$.