FP1 2.5 - The Weierstrass substitution | Edexcel A-Level Further Mathematics

Turn rational expressions in sine and cosine into rational functions of $t=\tan(x/2)$ , then integrate using familiar algebra. You will handle indefinite and definite integrals, retain original exclusions and split intervals when the half-angle tangent jumps.

Sine, cosine and the differential become rational

How can $\displaystyle\int_0^{2\pi}\frac{\mathrm{d}x}{2+\cos x}$ become a rational integral, when both endpoints give $t=\tan(x/2)=0$ ? Those matching values look as though they should give zero. The missing detail is a jump halfway through the interval. Track that jump and the exact answer is $2\pi/\sqrt3$ .

You already know

From FP1 1.1, $t=\tan(x/2)$ gives $\sin x=2t/(1+t^2)$ and $\cos x=(1-t^2)/(1+t^2)$ , with odd multiples of $\pi$ omitted by finite $t$ . From A-level Mathematics, change the differential and the bounds when substituting. Core Pure supplies partial fractions, logarithmic integrals and $\int\mathrm{d}t/(a^2+t^2)=a^{-1}\arctan(t/a)+C$ , for $a>0$ .

The extra ingredient is the differential. Differentiate the half-angle tangent:

\frac{\mathrm{d}t}{\mathrm{d}x} =\frac12\sec^2(x/2)=\frac{1+t^2}{2} \quad\Longrightarrow\quad \mathrm{d}x=\frac{2\,\mathrm{d}t}{1+t^2}

Combining this with the two t-formulae replaces every sine, cosine and differential by rational expressions in the same variable. This is the Weierstrass substitution.

The Weierstrass substitution

On an interval where $t=\tan(x/2)$ is defined,

\sin x=\frac{2t}{1+t^2},\qquad \cos x=\frac{1-t^2}{1+t^2},\qquad \mathrm{d}x=\frac{2\,\mathrm{d}t}{1+t^2}

The substitution is continuous and strictly increasing on each interval

(2k-1)\pi<x<(2k+1)\pi,\qquad k\in\mathbb Z

Within one such interval, $x=2\arctan t+2k\pi$ . Preserve every exclusion from the original integrand as well.

Not in the formula booklet: learn it.

The denominator $1+t^2$ is positive for real $t$ , so it introduces no real poles. However, denominators in the original integrand may vanish, and the substitution itself still misses $x=(2k+1)\pi$ . Algebraic cancellation cannot erase those restrictions.

Key idea

Change the whole integrand into one variable: sine, cosine and the differential must all become expressions in $t$ . Keep the original domain restrictions as well as the substitution's omitted points.

This is why the method is so useful: a fraction built from sine and cosine becomes a fraction of polynomials, so partial fractions and completing the square become available. A simpler substitution may still be shorter when one derivative is already present; the half-angle substitution is particularly helpful when neither sine nor cosine alone will do.

Watch out

Wrong: change the trigonometric functions but keep $\mathrm{d}x$ . Right: replace the differential too, and simplify only after every part of the integrand is expressed in $t$ .

Check the differential before using the method.

Integrate the resulting quadratic fraction

When the denominator is a constant plus a cosine term, the common factor $1+t^2$ often cancels neatly with the differential. What remains is an inverse-tangent integral.

Return an indefinite answer to $x$ . A primitive obtained on one substitution interval has an arbitrary constant on that interval; do not assume a principal inverse tangent automatically joins correctly across an omitted point.

Worked example 1: a positive quadratic denominator

Question. Use $t=\tan(x/2)$ to find $\displaystyle\int\frac{\mathrm{d}x}{2+\cos x}$ .

Plan. Substitute cosine and the differential, cancel the common denominator and use the inverse-tangent integral.

\begin{aligned} \int\frac{\mathrm{d}x}{2+\cos x} &=\int\frac{2\,\mathrm{d}t}{2(1+t^2)+(1-t^2)} && \text{combine the substitution factors} \\ &=\int\frac{2\,\mathrm{d}t}{3+t^2} && \text{collect terms} \\ &=\frac2{\sqrt3}\arctan\left(\frac{t}{\sqrt3}\right)+C && \text{inverse-tangent integral} \end{aligned}

Why this step? The factor $1+t^2$ from writing the denominator as a single fraction cancels the one in $\mathrm{d}x$ . It should not survive as an extra numerator.

Answer.

\frac2{\sqrt3}\arctan\left(\frac{\tan(x/2)}{\sqrt3}\right)+C

on each interval avoiding an odd multiple of $\pi$ .

Check. Differentiating gives $(1+t^2)/(3+t^2)$ , which equals $1/(2+\cos x)$ . The original denominator is always at least $1$ ; its apparent break at an odd multiple of $\pi$ comes from this expression for the primitive.

Change only the cosine sign. The resulting quadratic is still positive, but its coefficients swap.

Worked example 2: changing the cosine sign

Question. Use $t=\tan(x/2)$ to find $\displaystyle\int\frac{\mathrm{d}x}{2-\cos x}$ .

Plan. Use the same substitution, retaining the minus sign when clearing the denominator.

\begin{aligned} \int\frac{\mathrm{d}x}{2-\cos x} &=\int\frac{2\,\mathrm{d}t}{2(1+t^2)-(1-t^2)} && \text{substitute} \\ &=\int\frac{2\,\mathrm{d}t}{1+3t^2} && \text{collect terms} \\ &=\frac2{\sqrt3}\arctan(\sqrt3\,t)+C && \text{scale the variable} \end{aligned}

Why this step? In $\arctan(\sqrt3\,t)$ , differentiation brings out a factor $\sqrt3$ , so the outside coefficient must be $2/\sqrt3$ .

Answer. $\displaystyle\frac2{\sqrt3}\arctan\bigl(\sqrt3\,\tan(x/2)\bigr)+C$ , on each substitution interval.

Check. Differentiation gives $(1+t^2)/(1+3t^2)=1/(2-\cos x)$ .

The standard inverse-tangent integral is in the formula booklet. These particular trigonometric primitives are derived results and are not printed there.

Watch out

Wrong: interpret the inverse tangent's jump as an infinite value of the original integrand. Right: check the original denominator. A substitution can have a break where the original function is smooth.

Change the constant term while retaining the minus sign.

Use partial fractions and keep excluded points

Logarithms appear when the rational denominator has real linear factors. The simple cosecant case illustrates the mechanism:

\begin{aligned} \int\operatorname{cosec}x\,\mathrm{d}x &=\int\frac{1+t^2}{2t}\frac{2\,\mathrm{d}t}{1+t^2} \\ &=\int\frac{\mathrm{d}t}{t} =\ln|t|+C =\ln|\tan(x/2)|+C \end{aligned}

The cosecant primitive, including this half-angle logarithmic form, is in the formula booklet; here the substitution derives it. The original condition $\sin x\ne0$ remains. Next, rational expressions involving cosecant give both quadratic and repeated linear factors.

For an irreducible quadratic, start with a linear numerator in the partial-fraction setup. For a repeated linear factor, include each power. Finding that one coefficient is zero is a conclusion from the algebra.

Worked example 3: a repeated factor after substitution

Question. Use the Weierstrass substitution to find $\displaystyle\int\frac{\mathrm{d}x}{1+\operatorname{cosec}x}$ , retaining the original exclusions.

Plan. First express the integrand as $\sin x/(1+\sin x)$ , then decompose the rational function.

\begin{aligned} \int\frac{\mathrm{d}x}{1+\operatorname{cosec}x} &=\int\frac{4t\,\mathrm{d}t}{(1+t^2)(1+t)^2} && \sin x\ne0 \\ \frac{4t}{(1+t^2)(1+t)^2} &=\frac{At+B}{1+t^2}+\frac{C}{1+t}+\frac{D}{(1+t)^2} && \text{complete setup} \\ 4t&=(At+B)(1+t)^2+C(1+t)(1+t^2)+D(1+t^2) && \text{clear denominators} \end{aligned}

Coefficient comparison gives $A=C=0$ , $B=2$ , $D=-2$ . Therefore

\begin{aligned} I&=\int\left(\frac2{1+t^2}-\frac2{(1+t)^2}\right)\mathrm{d}t && \text{partial fractions} \\ &=2\arctan t+\frac2{1+t}+C && \text{integrate each term} \\ &=x+\frac2{1+\tan(x/2)}+C && \text{absorb the branch constant} \end{aligned}

Why this step? Replacing $1/(1+\operatorname{cosec}x)$ by $\sin x/(1+\sin x)$ requires $\sin x\ne0$ . The simplified fraction has a value at some formerly excluded points, but the original integral still excludes them.

Answer. $\displaystyle x+\frac2{1+\tan(x/2)}+C$ , on each interval where the original expression and the substitution are defined. Retain $\sin x\ne0$ and $\sin x\ne-1$ ; in finite $t$ , these exclude $t=0,-1$ .

Check. The decomposition numerator is $2(1+t)^2-2(1+t^2)=4t$ . Differentiating the answer gives $2t/(1+t)^2=\sin x/(1+\sin x)$ , on the original domain.

Change only the sign before cosecant. The repeated factor changes from $1+t$ to $t-1$ .

Worked example 4: the changed repeated factor

Question. Use the Weierstrass substitution to find $\displaystyle\int\frac{\mathrm{d}x}{1-\operatorname{cosec}x}$ , retaining the original exclusions.

Plan. Use $\sin x/(\sin x-1)$ and the same full partial-fraction structure with the new linear factor.

\begin{aligned} I&=\int\frac{-4t\,\mathrm{d}t}{(1+t^2)(t-1)^2} && \text{substitute and simplify} \\ &=\int\left(\frac2{1+t^2}-\frac2{(t-1)^2}\right)\mathrm{d}t && \text{compare coefficients} \\ &=2\arctan t+\frac2{t-1}+C && \text{integrate} \\ &=x+\frac2{\tan(x/2)-1}+C && \text{return to }x \end{aligned}

Why this step? The original denominator is $1-\operatorname{cosec}x$ , so multiplying top and bottom by sine produces $\sin x-1$ , not $1-\sin x$ . That sign controls the rational numerator.

Answer. $\displaystyle x+\frac2{\tan(x/2)-1}+C$ , on each valid interval. Keep $\sin x\ne0$ and $\sin x\ne1$ ; in finite $t$ , exclude $t=0,1$ .

Check. Combining the partial fractions gives numerator $2(t-1)^2-2(1+t^2)=-4t$ . Differentiation of the final answer gives $-2t/(t-1)^2=\sin x/(\sin x-1)$ .

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FP1 2.5 - The Weierstrass substitution

Sine, cosine and the differential become rational

Integrate the resulting quadratic fraction

Use partial fractions and keep excluded points

Continue this lesson.

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