Edexcel IAL - Further Pure Mathematics 3- 4.4 Substitution for Integrals with Quadratic Surds- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 3- 4.4 Substitution for Integrals with Quadratic Surds -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 3- 4.4 Substitution for Integrals with Quadratic Surds -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.4 Substitution for Integrals with Quadratic Surds
Use of Substitution for Integrals Involving Quadratic Surds
Integrals involving quadratic surds, such as \( \sqrt{ax^2+bx+c} \), often cannot be evaluated directly. At IAL level, these integrals are handled using algebraic substitution or, where appropriate, trigonometric or hyperbolic substitution. In more complicated cases, the required substitution will be stated in the question.
General Principle
The key idea is to reduce the quadratic surd to a simpler form by choosing a substitution that completes the square or removes the square root.
- Complete the square if necessary
- Choose a substitution that simplifies the surd
- Rewrite the integral entirely in terms of the new variable
Common Quadratic Surd Forms
| Surd Form | Typical Substitution |
|---|---|
| \( \sqrt{a^2+x^2} \) | \( x = a\sinh t \) |
| \( \sqrt{a^2-x^2} \) | \( x = a\sin\theta \) |
| \( \sqrt{x^2-a^2} \) | \( x = a\cosh t \) |
| \( \sqrt{ax+b} \) | \( ax+b = u^2 \) |
At IAL level, once a substitution is chosen, careful algebra and simplification are essential.
Example :
Evaluate
\( \displaystyle \int \sqrt{x^2 + 9}\,dx \).
▶️ Answer/Explanation
Use the substitution \( x = 3\sinh t \).
Then \( dx = 3\cosh t\,dt \) and \( \sqrt{x^2+9} = 3\cosh t \).
\( \displaystyle \int \sqrt{x^2+9}\,dx = \int 9\cosh^2 t\,dt \)
Using \( \cosh^2 t = \dfrac{1+\cosh 2t}{2} \):
\( = \dfrac{9}{2}\int (1+\cosh 2t)\,dt \)
\( = \dfrac{9}{2}\left(t + \dfrac{1}{2}\sinh 2t\right) + C \)
Returning to \( x \):
\( t = \operatorname{arsinh}\!\left(\dfrac{x}{3}\right) \)
Conclusion: The integral is evaluated using a hyperbolic substitution.
Example :
Evaluate
\( \displaystyle \int \dfrac{x}{\sqrt{4x^2+1}}\,dx \).
▶️ Answer/Explanation
Use the substitution
\( u = 4x^2 + 1 \)
Then \( du = 8x\,dx \).
\( \displaystyle \int \dfrac{x}{\sqrt{4x^2+1}}\,dx = \dfrac{1}{8}\int u^{-1/2}\,du \)
\( = \dfrac{1}{4}\sqrt{u} + C \)
Substituting back:
\( = \dfrac{1}{4}\sqrt{4x^2+1} + C \)
Conclusion: A simple algebraic substitution removes the surd.
Example :
Evaluate, given that
\( x = 2\sec\theta \),
the integral
\( \displaystyle \int \dfrac{1}{x\sqrt{x^2-4}}\,dx \).
▶️ Answer/Explanation
The substitution is given:
\( x = 2\sec\theta \Rightarrow dx = 2\sec\theta\tan\theta\,d\theta \)
\( \sqrt{x^2-4} = 2\tan\theta \)
Substituting:
\( \displaystyle \int \dfrac{1}{x\sqrt{x^2-4}}\,dx = \int \dfrac{2\sec\theta\tan\theta}{(2\sec\theta)(2\tan\theta)}\,d\theta \)
\( = \dfrac{1}{2}\int d\theta = \dfrac{\theta}{2} + C \)
Returning to \( x \):
\( \theta = \sec^{-1}\!\left(\dfrac{x}{2}\right) \)
Conclusion: The integral is evaluated using the given trigonometric substitution.