Edexcel IAL - Further Pure Mathematics 3- 4.5 Reduction Formulae- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 3- 4.5 Reduction Formulae -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 3- 4.5 Reduction Formulae -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.5 Reduction Formulae
Reduction Formulae
A reduction formula is a recurrence relation that expresses an integral involving a higher power or order in terms of a similar integral of lower order. Reduction formulae are extremely useful for evaluating families of integrals systematically.
Why Reduction Formulae Are Useful
- They avoid repeated integration by parts
- They allow evaluation of definite integrals recursively
- They are especially useful for integrals involving powers of trigonometric functions
Reduction Formula for \( \displaystyle I_n = \int_0^{\pi/2} \sin^n x \, dx \)
Let
\( I_n = \int_0^{\pi/2} \sin^n x \, dx \)
Using integration by parts with
\( u = \sin^{n-1}x \), \( dv = \sin x\,dx \)
after simplification, we obtain the standard reduction formula
\( I_n = \dfrac{n-1}{n} I_{n-2}, \quad n \geq 2 \)
This expresses \( I_n \) in terms of \( I_{n-2} \).
Base Values
\( I_0 = \dfrac{\pi}{2} \)
\( I_1 = 1 \)
Reduction Formula for \( \displaystyle I_n = \int \dfrac{\sin(nx)}{\sin x} \, dx \)
Using the trigonometric identity
\( \sin (n+2)x = 2\sin (n+1)x \cos x – \sin nx \)
and rearranging, we obtain the reduction formula
\( I_{n+2} = \dfrac{2\sin (n+1)x}{n+1} + I_n, \quad n > 0 \)
This allows higher-order integrals to be reduced step by step.
Example :
Use the reduction formula to find
\( \displaystyle \int_0^{\pi/2} \sin^4 x \, dx \).
▶️ Answer/Explanation
Using the reduction formula:
\( I_4 = \dfrac{3}{4} I_2 \)
\( I_2 = \dfrac{1}{2} I_0 \)
\( I_0 = \dfrac{\pi}{2} \)
Therefore
\( I_4 = \dfrac{3}{4} \cdot \dfrac{1}{2} \cdot \dfrac{\pi}{2} = \dfrac{3\pi}{16} \)
Conclusion: \( \displaystyle \int_0^{\pi/2} \sin^4 x\,dx = \dfrac{3\pi}{16} \).
Example :
Evaluate
\( \displaystyle \int_0^{\pi/2} \sin^5 x \, dx \).
▶️ Answer/Explanation
Using the reduction formula:
\( I_5 = \dfrac{4}{5} I_3 \)
\( I_3 = \dfrac{2}{3} I_1 \)
\( I_1 = 1 \)
Therefore
\( I_5 = \dfrac{4}{5} \cdot \dfrac{2}{3} = \dfrac{8}{15} \)
Conclusion: \( \displaystyle \int_0^{\pi/2} \sin^5 x\,dx = \dfrac{8}{15} \).
Example :
Given that
\( I_n = \displaystyle \int \dfrac{\sin(nx)}{\sin x} \, dx \),
find \( I_3 \).
▶️ Answer/Explanation
Using the reduction formula:
\( I_{n+2} = \dfrac{2\sin (n+1)x}{n+1} + I_n \)
Let \( n = 1 \):
\( I_3 = \dfrac{2\sin 2x}{2} + I_1 \)
But
\( I_1 = \int \dfrac{\sin x}{\sin x} dx = \int 1\,dx = x \)
Therefore
\( I_3 = \sin 2x + x + C \)
Conclusion: \( I_3 = x + \sin 2x + C \).