The Weierstrass Substitution (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

Author

Mark Curtis

Last updated

6 January 2026

The Weierstrass substitution

What is the Weierstrass substitution?

The Weierstrass substitution refers to using the tangent half-angle substitution $t = \tan (\frac{θ}{2})$ when performing integration by substitution
- i.e. integration using the $t$ -substitution
It is helpful to know the t-formulae, namely
- $\sin θ = \frac{2 t}{1 + t^{2}}$
- $\cos θ = \frac{1 - t^{2}}{1 + t^{2}}$
- $\tan θ = \frac{2 t}{1 - t^{2}}$
- and hence
  - $cosec θ = \frac{1 + t^{2}}{2 t}$
  - $\sec θ = \frac{1 + t^{2}}{1 - t^{2}}$
  - $\cot θ = \frac{1 - t^{2}}{2 t}$

Examiner Tips and Tricks

Exam questions don't say to "use a Weierstrass substitution" but they do give the $t$ -substitution in the question.

How do I use the Weierstrass substitution for indefinite integration?

To use the Weierstrass substitution $t = \tan (\frac{θ}{2})$ for indefinite integrals:
STEP 1
Rewrite the $θ$ expression in terms of $t$
- Use the t-formulae to help
STEP 2
Convert $d θ$ into $d t$
- Differentiate the substitution $t = \tan (\frac{θ}{2})$
  - $\frac{d t}{d θ} = \frac{1}{2} \sec^{2} (\frac{θ}{2})$
- Use the reciprocal trig identity $1 + \tan^{2} A \equiv \sec^{2} A$
  - so $\frac{d t}{d θ} = \frac{1}{2} (1 + \tan^{2} (\frac{θ}{2})) = \frac{1}{2} (1 + t^{2})$
- Find $d θ$ in terms of $d t$
  - $\frac{d θ}{d t} = \frac{2}{1 + t^{2}}$
  - so $d θ = \frac{2}{1 + t^{2}} d t$
STEP 3
Integrate and add a constant of integration
- using any of the known methods in the course
- e.g.
  - $\int \frac{f' (t)}{f (t)} d t = \ln | f (t) | + c$
  - Integration by partial fractions
  - Integration by trigonometric or hyperbolic substitutions
STEP 4
Rewrite the $t$ expression back in terms of $θ$

Worked Example

Use the substitution $t = \tan (\frac{θ}{2})$ to show that

$\int \sec θ d θ = \ln |\begin{array}{rcl} \frac{1 + \tan (\frac{θ}{2})}{1 - \tan (\frac{θ}{2})} \end{array}| + c$

Answer:

Write $\sec θ$ in terms of the cosine t-formula, $\cos θ = \frac{1 - t^{2}}{1 + t^{2}}$

$\begin{array}{rcl} \sec θ & = & \frac{1}{\cos θ} \\ \sec θ & = & \frac{1 + t^{2}}{1 - t^{2}} \end{array}$

Change $d θ$ into $d t$ by differentiating $t = \tan (\frac{θ}{2})$

$\frac{d t}{d θ} = \frac{1}{2} \sec^{2} (\frac{θ}{2})$

Use $1 + \tan^{2} A \equiv \sec^{2} A$ to write it in terms of $t$

$\begin{array}{rcl} \frac{d t}{d θ} & = & \frac{1}{2} (1 + \tan^{2} (\frac{θ}{2})) \\ \frac{d t}{d θ} & = & \frac{1}{2} (1 + t^{2}) \end{array}$

Find $d θ$ in terms of $d t$

$\begin{array}{rcl} \frac{d θ}{d t} & = & \frac{2}{1 + t^{2}} \\ d θ & = & \frac{2}{1 + t^{2}} d t \end{array}$

Substitute $\sec θ$ and $d θ$ into the integral

$\begin{array}{rcl} \int \sec θ d θ & = & \int \frac{1 + t^{2}}{1 - t^{2}} \times \frac{2}{1 + t^{2}} d t \\ = & \int \frac{2}{1 - t^{2}} d t \end{array}$

Write $\frac{2}{1 - t^{2}}$ in partial fractions

$\begin{array}{rcl} \frac{2}{1 - t^{2}} & = & \frac{A}{1 + t} + \frac{B}{1 - t} \\ 2 & = & A (1 - t) + B (1 + t) \end{array}$

Find $A$ and $B$

$\begin{array}{rcl} t & = & 1 \Rightarrow 2 = 2 B \Rightarrow B = 1 \\ t & = & - 1 \Rightarrow 2 = 2 A \Rightarrow A = 1 \end{array}$

Substitute the partial fractions into the integral

$\begin{array}{rcl} \int \sec θ d θ & = & \int \frac{2}{1 - t^{2}} d t \\ = & \int (\frac{1}{1 + t} + \frac{1}{1 - t}) d t \end{array}$

Integrate each partial fraction

$\begin{array}{rcl} \int \sec θ d θ & = & \ln | 1 + t | - \ln | 1 - t | + c \end{array}$

Use log laws to combine the two log terms

$\begin{array}{rcl} \int \sec θ d θ & = & \ln |\frac{1 + t}{1 - t}| + c \end{array}$

Write the answer back in terms of $θ$ using $t = \tan (\frac{θ}{2})$

$\begin{array}{rcl} \int \sec θ d θ & = & \ln |\frac{1 + \tan (\frac{θ}{2})}{1 - \tan (\frac{θ}{2})}| + c \end{array}$

How do I use the Weierstrass substitution for definite integration?

To use the Weierstrass substitution $t = \tan (\frac{θ}{2})$ for definite integrals:
STEP 1
Rewrite the $θ$ expression in terms of $t$
- Use the t-formulae to help
STEP 2
Convert $d θ$ into $d t$
- Differentiate the substitution $t = \tan (\frac{θ}{2})$
  - $\frac{d t}{d θ} = \frac{1}{2} \sec^{2} (\frac{θ}{2})$
- Use the reciprocal trig identity $1 + \tan^{2} A \equiv \sec^{2} A$
  - so $\frac{d t}{d θ} = \frac{1}{2} (1 + \tan^{2} (\frac{θ}{2})) = \frac{1}{2} (1 + t^{2})$
- Find $d θ$ in terms of $d t$
  - $\frac{d θ}{d t} = \frac{2}{1 + t^{2}}$
  - so $d θ = \frac{2}{1 + t^{2}} d t$
STEP 3
Change the limits
STEP 4
Integrate
- using any of the known methods in the course
- e.g.
  - $\int_{a}^{b} \frac{f' (t)}{f (t)} d t = {[\ln | f (t) |]}_{a}^{b}$
  - Integration by partial fractions
  - Integration by trigonometric or hyperbolic substitutions
STEP 5
Substitute in the new limits

Worked Example

Use the substitution $t = \tan (\frac{θ}{2})$ to determine the exact value of

$\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{2 + 2 \cos θ + \sin θ} d θ$

giving your answer in the form $\ln (p + q \sqrt{3})$ , where $p$ and $q$ are constants to be found.

Answer:

Use the t-formulae $\cos θ = \frac{1 - t^{2}}{1 + t^{2}}$ and $\sin θ = \frac{2 t}{1 + t^{2}}$ to rewrite the expression inside the integral

$\begin{array}{rcl} \frac{1}{2 + 2 \cos θ + \sin θ} & = & \frac{1}{2 + \frac{2 (1 - t^{2})}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \end{array}$

Multiply top and bottom by $(1 + t^{2})$ to simplify the result

$\begin{array}{rcl} \frac{1}{2 + 2 \cos θ + \sin θ} & = & \frac{1}{2 + \frac{2 (1 - t^{2})}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \times \frac{1 + t^{2}}{1 + t^{2}} \\ = & \frac{1 + t^{2}}{2 (1 + t^{2}) + 2 (1 - t^{2}) + 2 t} \\ = & \frac{1 + t^{2}}{2 + 2 t^{2} + 2 - 2 t^{2} + 2 t} \\ = & \frac{1 + t^{2}}{2 (t + 2)} \end{array}$

Change $d θ$ into $d t$ by differentiating $t = \tan (\frac{θ}{2})$

$\frac{d t}{d θ} = \frac{1}{2} \sec^{2} (\frac{θ}{2})$

Use $1 + \tan^{2} A \equiv \sec^{2} A$ to write it in terms of $t$

$\begin{array}{rcl} \frac{d t}{d θ} & = & \frac{1}{2} (1 + \tan^{2} (\frac{θ}{2})) \\ \frac{d t}{d θ} & = & \frac{1}{2} (1 + t^{2}) \end{array}$

Find $d θ$ in terms of $d t$

$\begin{array}{rcl} \frac{d θ}{d t} & = & \frac{2}{1 + t^{2}} \\ d θ & = & \frac{2}{1 + t^{2}} d t \end{array}$

Change the limits

$θ = \frac{π}{3} \Rightarrow t = \tan (\frac{\frac{π}{3}}{2}) = \tan (\frac{π}{6}) = \frac{\sqrt{3}}{3} θ = \frac{π}{2} \Rightarrow t = \tan (\frac{\frac{π}{2}}{2}) = \tan (\frac{π}{4}) = 1$

Substitute the expression, $d θ$ and new limits into the integral and simplify

$\begin{array}{rcl} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{2 + 2 \cos θ + \sin θ} d θ & = & \int_{\frac{\sqrt{3}}{3}}^{1} \frac{1 + t^{2}}{2 (t + 2)} \times \frac{2}{1 + t^{2}} d t \\ = & \int_{\frac{\sqrt{3}}{3}}^{1} \frac{1}{t + 2} d t \end{array}$

Integrate and substitute in the limits

$\begin{array}{rcl} \int_{\frac{\sqrt{3}}{3}}^{1} \frac{1}{t + 2} d t & = & {[\ln | t + 2 |]}_{\frac{\sqrt{3}}{3}}^{1} \\ = & \ln 3 - \ln (\frac{\sqrt{3}}{3} + 2) \end{array}$

Use log laws and rationalising the denominator (or your calculator) to write out the final answer in the form $\ln (p + q \sqrt{3})$

$\begin{array}{rcl} \int_{\frac{\sqrt{3}}{3}}^{1} \frac{1}{t + 2} d t & = & \ln (\frac{3}{\frac{\sqrt{3}}{3} + 2}) \\ = & \ln (\frac{9}{\sqrt{3} + 6}) \\ = & \ln (\frac{9}{\sqrt{3} + 6} \times \frac{\sqrt{3} - 6}{\sqrt{3} - 6}) \\ = & \ln (\frac{9 \sqrt{3} - 54}{- 33}) \\ = & \ln (\frac{18 - 3 \sqrt{3}}{11}) \end{array}$

Write as $\ln (p + q \sqrt{3})$

$\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{2 + 2 \cos θ + \sin θ} d θ = \ln (\frac{18}{11} - \frac{3}{11} \sqrt{3})$

How do I use the Weierstrass substitution for improper integrals?

The Weierstrass substitution $t = \tan (\frac{θ}{2})$ is undefined at
- $θ = (2 k + 1) π$
  - where $k \in ℤ$
- i.e. odd multiples of $π$
Any definite integrals that contain an odd multiple of $π$ between the lower and upper limit are improper integrals
- They must be split into two separate integrals
- e.g. by inserting the integration limits:
  - $a \to π^{-}$ from below $π$
  - $b \to π^{+}$ from above $π$
- which, from the tan graph, gives the $t$ -limits:
  - $t = \tan (\frac{π^{-}}{2}) = \infty$
  - $t = \tan (\frac{π^{+}}{2}) = - \infty$
- See the worked example below

Worked Example

Use the substitution $t = \tan (\frac{θ}{2})$ to determine the exact value of

$\int_{0}^{\frac{3 π}{2}} \frac{1}{2 + \cos θ} d θ$

Answer:

The substitution $t = \tan (\frac{θ}{2})$ is undefined at $θ = π$ which lies between the two limits

$0 < π < \frac{3 π}{2}$

Split the integral into two separate integrals, either side of $θ = π$

$\int_{0}^{\frac{3 π}{2}} \frac{1}{2 + \cos θ} d θ = \lim_{a \to π^{-}} \int_{0}^{a} \frac{1}{2 + \cos θ} d θ + \lim_{b \to π^{+}} \int_{b}^{\frac{3 π}{2}} \frac{1}{2 + \cos θ} d θ$

Use the t-formula $\cos θ = \frac{1 - t^{2}}{1 + t^{2}}$ to rewrite the expression inside the integral

$\begin{array}{rcl} \frac{1}{2 + \cos θ} & = & \frac{1}{2 + \frac{1 - t^{2}}{1 + t^{2}}} \end{array}$

Multiply top and bottom by $(1 + t^{2})$ to simplify the result

$\begin{array}{rcl} \frac{1}{2 + \cos θ} & = & \frac{1}{2 + \frac{1 - t^{2}}{1 + t^{2}}} \times \frac{1 + t^{2}}{1 + t^{2}} \\ = & \frac{1 + t^{2}}{2 (1 + t^{2}) + (1 - t^{2})} \\ = & \frac{1 + t^{2}}{2 + 2 t^{2} + 1 - t^{2}} \\ = & \frac{1 + t^{2}}{t^{2} + 3} \end{array}$

Change $d θ$ into $d t$ by differentiating $t = \tan (\frac{θ}{2})$

$\frac{d t}{d θ} = \frac{1}{2} \sec^{2} (\frac{θ}{2})$

Use $1 + \tan^{2} A \equiv \sec^{2} A$ to write it in terms of $t$

$\begin{array}{rcl} \frac{d t}{d θ} & = & \frac{1}{2} (1 + \tan^{2} (\frac{θ}{2})) \\ \frac{d t}{d θ} & = & \frac{1}{2} (1 + t^{2}) \end{array}$

Find $d θ$ in terms of $d t$

$\begin{array}{rcl} \frac{d θ}{d t} & = & \frac{2}{1 + t^{2}} \\ d θ & = & \frac{2}{1 + t^{2}} d t \end{array}$

Change the four limits

$θ = 0 \Rightarrow t = \tan (\frac{0}{2}) = \tan (0) = 0 θ = a \to π^{-} \Rightarrow \lim_{a \to π^{-}} \tan (\frac{a}{2}) = \infty θ = b \to π^{+} \Rightarrow \lim_{b \to π^{+}} \tan (\frac{b}{2}) = - \infty θ = \frac{3 π}{2} \Rightarrow t = \tan (\frac{\frac{3 π}{2}}{2}) = \tan (\frac{3 π}{4}) = - 1$

Substitute the expression, $d θ$ and new limits into the integrals and simplify

$\begin{array}{rcl} \int_{0}^{\frac{3 π}{2}} \frac{1}{2 + \cos θ} d θ & = & \int_{0}^{\infty} \frac{1 + t^{2}}{t^{2} + 3} \times \frac{2}{1 + t^{2}} d t + \int_{- \infty}^{- 1} \frac{1 + t^{2}}{t^{2} + 3} \times \frac{2}{1 + t^{2}} d t \\ = & 2 \int_{0}^{\infty} \frac{1}{t^{2} + 3} d t + 2 \int_{- \infty}^{- 1} \frac{1}{t^{2} + 3} d t \end{array}$

Integrate $\frac{1}{t^{2} + 3}$ , e.g. using that $\frac{1}{a^{2} + x^{2}}$ integrates to $\frac{1}{a} \arctan (\frac{x}{a})$ (from the formulae booklet)

$\int \frac{1}{t^{2} + 3} d t = \frac{1}{\sqrt{3}} \arctan (\frac{t}{\sqrt{3}}) + c$

Substitute this into the working above

$\begin{array}{rcl} \int_{0}^{\frac{3 π}{2}} \frac{1}{2 + \cos θ} d θ & = & \frac{2}{\sqrt{3}} {[\arctan (\frac{t}{\sqrt{3}})]}_{0}^{\infty} + \frac{2}{\sqrt{3}} {[\arctan (\frac{t}{\sqrt{3}})]}_{- \infty}^{- 1} \end{array}$

Find the value of the first integral (using the idea that $\arctan (\infty) = \frac{π}{2}$ )

$\begin{array}{rcl} \frac{2}{\sqrt{3}} {[\arctan (\frac{t}{\sqrt{3}})]}_{0}^{\infty} & = & \frac{2}{\sqrt{3}} (\lim_{t \to \infty} \arctan (\frac{t}{\sqrt{3}}) - \arctan (0)) \\ = & \frac{2}{\sqrt{3}} (\frac{π}{2} - 0) \\ = & \frac{π \sqrt{3}}{3} \end{array}$

Find the value of the second integral (using the idea that $\arctan (- \infty) = - \frac{π}{2}$ )

$\begin{array}{rcl} \frac{2}{\sqrt{3}} {[\arctan (\frac{t}{\sqrt{3}})]}_{- \infty}^{- 1} & = & \frac{2}{\sqrt{3}} (\arctan (\frac{- 1}{\sqrt{3}}) - \lim_{t \to - \infty} \arctan (\frac{t}{\sqrt{3}})) \\ = & \frac{2}{\sqrt{3}} (- \frac{π}{6} - (- \frac{π}{2})) \\ = & \frac{2 π \sqrt{3}}{9} \end{array}$

Add the answers together to give the final answer

$\begin{array}{rcl} \int_{0}^{\frac{3 π}{2}} \frac{1}{2 + \cos θ} d θ & = & \frac{π \sqrt{3}}{3} + \frac{2 π \sqrt{3}}{9} \end{array}$

$\begin{array}{rcl} \int_{0}^{\frac{3 π}{2}} \frac{1}{2 + \cos θ} d θ & = & \frac{5 π \sqrt{3}}{9} \end{array}$

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The Weierstrass Substitution (Edexcel A Level Further Maths): Revision Note

The Weierstrass substitution

What is the Weierstrass substitution?

How do I use the Weierstrass substitution for indefinite integration?

How do I use the Weierstrass substitution for definite integration?

How do I use the Weierstrass substitution for improper integrals?

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Author: Mark Curtis