- Reduction Formulae

If we needed to determine \(\displaystyle \int x^6 e^x\:dx\), we would quickly realise that there would be quite a few instances where INTEGRATION BY PARTS would be used. This would be quite tedious, especially if the power was higher. To make these types of integrations a bit simpler we use what is called a REDUCTION FORMULA.


DERIVING A REDUCTION FORMULA
LESSON 1
Establish a reduction formula that could be used to find \(\displaystyle \int x^n e^x\:dx\).
SOLUTION
We will be using \(I_n\) to represent the integral for which we are developing the REDUCTION FORMULA.
\(\displaystyle I_n=\int x^n e^x\:dx\)
  • Integrate using Integration by Parts
\(u=x^n \to du=nx^{n-1}\)
\(\displaystyle {dv\over dx}=e^x \to v=e^x\)
\(\displaystyle I_n=x^n e^x-\int nx^{n-1}e^x\:dx\)
\(\displaystyle I_n=x^n e^x-n\int x^{n-1}e^x\:dx\)
Since
\(\displaystyle I_n=\int x^ne^x\:dx\)
\(\displaystyle ⟹I_{n-1}=\int x^{n-1}e^x\:dx\)
\(\displaystyle ⟹I_n=x^n e^x-nI_{n-1}\)

USING A REDUCTION FORMULA
LESSON 1
Given that \(\displaystyle I_n=\int x^n e^x\:dx\), use the reduction formula \(I_n=x^n e^x-nI_{n-1}\) to determine \(I_4\).
SOLUTION
If
\(\displaystyle I_n=\int x^n e^x\:dx\)
Then
\(\displaystyle I_4=\int x^4e^x\:dx\)
Therefore \(n=4\)
  • Substitute \(n=4\) into form for \(I_n\)
\(I_4=x^4 e^x-4I_3\)
\(⟹x^4 e^x-4[x^3 e^x-3I_2]\)
\(⟹x^4 e^x-4x^3 e^x+12I_2\)
  • Apply the reduction formula to determine \(I_2\)
\(⟹x^4 e^x-4x^3 e^x+12[x^2 e^x-2I_1]\)
\(⟹x^4 e^x-4x^3 e^x+12x^2 e^x-24I_1\)
  • Apply the reduction formula to determine \(I_1\)
\(⟹x^4 e^x-4x^3 e^x+12x^2 e^x-24[xe^x-I_0]\)
\(⟹x^4 e^x-4x^3 e^x+12x^2 e^x-24xe^x+24I_0\)
  • If we substitute \(n=0\) into the reduction formula we get \(I_0=x^0 e^x-0I_(0-1)=e^x\).
Alternately, we can substitute zero into \(\displaystyle \int x^ne^x \:dx\) to determine \(I_0\). The latter is more common.
\(\displaystyle ⟹x^4 e^x-4x^3 e^x+12x^2 e^x-24xe^x+24\left[\int e^x \:dx\right]\)
\(⟹x^4 e^x-4x^3 e^x+12x^2 e^x-24xe^x+24[e^x+c]\)
\(⟹x^4 e^x-4x^3 e^x+12x^2 e^x-24xe^x+24e^x+A\)

LESSON 2
If \(\displaystyle I_n≡\int \cos^n⁡x\:dx\) show that \(\displaystyle I_n={1\over n}\sin ⁡x \cos^{n-1}⁡x+{n-1\over n} I_{n-2}\).
Hence, find \(\displaystyle \int \cos^5⁡x\:dx\).
SOLUTION
  • Write the integral in the product form
\(I_n=\int \cos^1⁡x \cos^{n-1}⁡x\:dx\)
  • Integrate by parts or an appropriate method
\(u=\cos^{n-1}⁡x\:du=(n-1) \cos^{n-2}⁡x (-\sin ⁡x)\)
\(\displaystyle {dv\over dx}=\cos ⁡x \to v=\sin ⁡x\)
\(\displaystyle I_n=\sin ⁡x \cos^{n-1}⁡x+\int (n-1) \cos^{n-2}⁡x \sin^2⁡x \:dx\)
  • Simplify
\(\displaystyle I_n=\sin ⁡x \cos^{n-1}⁡x+\int (n-1) \cos^{n-2}⁡x (1-\cos^2⁡x)\:dx\)
\(\displaystyle I_n=\sin ⁡x \cos^{n-1}⁡x+(n-1) \int \cos^{n-2}⁡x (1-\cos^2⁡x)\:dx\)
\(\displaystyle I_n=\sin ⁡x \cos^{n-1}⁡x+(n-1) \int \cos^{n-2}⁡x\:dx-(n-1) \int \cos^n⁡x\:dx\)
\(I_n=\sin ⁡x \cos^{n-1}⁡x+(n-1) I_{n-2}-(n-1)I_n\)
\(I_n+nI_n-I_n=\sin ⁡x \cos^{n-1}⁡x+(n-1) I_{n-2}\)
\(nI_n=\sin ⁡x \cos^{n-1}⁡x+(n-1) I_{n-2}\)
\(\displaystyle I_n={1\over n} \sin ⁡x \cos^{n-1}⁡x+{n-1\over n}I_{n-2}\)

  • Apply the derived formula to the rest of the question
\(n=5\)
\(\displaystyle I_5={1\over5}\sin ⁡x \cos^4⁡x+{4\over5}I_3\)
\(n=3\)
\(\displaystyle I_3={1\over3}\sin ⁡x \cos^2⁡x+{2\over3}I_1\)
When you have reduced your integral to its lowest form, go back to the original integral & plug in the final value of \(n\).
\(n=1\)
\(\displaystyle I_1=\int \cos ⁡x\:dx=\sin ⁡x+c\)
\(\displaystyle I_5={1\over5}\sin ⁡x \cos^4⁡x+{4\over5}\left({1\over3} \sin ⁡x \cos^2⁡x+{2\over3}\sin ⁡x\right)+c\)

REDUCTION FORMULA WITH LIMITS
LESSON 1
(a) By using the substitution \(x=4 \sin \theta\), find the exact value of
\[\displaystyle \int_0^2 {1\over\sqrt{16-x^2}}\:dx\]
(b) Show that
\[{d\over dx} [x^{n-1}\sqrt{16-x^2}]={16(n-1) x^{n-2}\over \sqrt{16-x^2}}-{nx^n \over \sqrt{16-x^2}}\]
(c) Deduce, or prove otherwise, that if
\[I_n=\int_0^2 {x^n\over \sqrt{16-x^2}}\:dx\) for \(n≥2\),
then
\[nI_n=16(n-1) I_{n-2}-2^n \sqrt{3}\]
(d) Hence find \(I_2\).
SOLUTION
Part (a)
Let
\(x=4 \sin⁡ \theta\)
\(\displaystyle {dx\over dθ}=4 \cos⁡θ\)
\(dx=4 \cos⁡θ \:dθ\)
When
\(x=2\)
\(2=4 \sin⁡θ\)
\(\displaystyle {1\over2}=\sin⁡θ\)
\(\displaystyle θ={π\over6}\)
When
\(x=0\)
\(0=4 \sin⁡θ\)
\(θ=0\)
  • Rewrite integral using substitutions
\(\displaystyle \int_0^2 {1\over\sqrt{16-x^2}}\:dx=\int_0^{π\over6}{1\over \sqrt{16-(4 \sin⁡θ)^2}} 4 \cos⁡θ \:dθ\)
\(\displaystyle =\int_0^{π\over6} {4\cos⁡θ\over \sqrt{16-16 \sin^2⁡θ}}\:dθ\)
\(\displaystyle =\int_0^{π\over6} {4 \cos⁡θ\over \sqrt{16(1-\sin^2⁡θ}}\:dθ\)
\(\displaystyle =\int_0^{π\over6} {\cos⁡θ\over \sqrt{1-\sin^2⁡θ}}\:dθ\)
\(\displaystyle =\int_0^{π\over6} {\cos⁡θ\over \cos⁡θ}\:dθ\)
\(\displaystyle =\int_0^{{π\over6}} 1\:dθ\)
\(\displaystyle =\left[θ\right]_0^{{\pi\over6}}\)
\(\displaystyle ={π\over6}\)

Part (b)
  • Differentiate using the product rule
\(\displaystyle {d\over dx} [x^{n-1} \sqrt{16-x^2}]=x^{n-1}\left[{1\over2} (16-x^2 )^{{1\over2}} (-2x)\right]+\sqrt{16-x^2} [(n-1) x^{n-2}]\)
\(\displaystyle ⟹-{(2x^{n-1}) x^1\over 2\sqrt{16-x^2}}+(n-1) \sqrt{16-x^2} (x^{n-2})\)
\(\displaystyle ⟹-{x^n\over \sqrt{16-x^2}}+{{(n-1)(16-x^2)x^{n-2}}\over \sqrt{16-x^2}}\)
\(\displaystyle ⟹-{x^n\over \sqrt{16-x^2}}+{(n-1)(16x^{n-2}-x^n) \over \sqrt{16-x^2}}\)
\(\displaystyle ⟹-{x^n\over \sqrt{16-x^2}}+{16(n-1) x^{n-2}\over \sqrt{16-x^2}}-{(n-1) x^n\over \sqrt{16-x^2}}\)
\(\displaystyle ⟹{16(n-1) x^{n-2}\over \sqrt{16-x^2}}-\left[{x^n\over \sqrt{16-x^2}}+{(n-1)x^n\over \sqrt{16-x^2}} \right]\)
\(\displaystyle ⟹{16(n-1)x^{n-2}\over \sqrt{16-x^2}}-{nx^n\over \sqrt{16-x^2}}\)
\(\displaystyle {d\over dx}[x^{n-1} \sqrt{16-x^2}]={16(n-1) x^{n-2}\over \sqrt{16-x^2}}-{nx^n\over \sqrt{16-x^2}}\:\:\:\:\: (1)\)

Part (c)
Rearrange (1)
\(\displaystyle n\int_0^2 {x^n\over \sqrt{16-x^2}}\:dx=16(n-1) \int_0^2 {x^{n-2}\over \sqrt{16-x^2}}\:dx-\int_0^2 {d\over dx} [x^{n-1}\sqrt{16-x^2}]\:dx\)
\(nI_n=16(n-1) I_{n-2}-\left[x^{n-1} \sqrt{16-x^2}\right]_0^2\)
\(nI_n=16(n-1) I_{n-2}-2^{n-1}\sqrt{12}\)
\(\displaystyle nI_n=16(n-1) I_{n-2}-2^n\times{1\over2}\times \sqrt{4}\sqrt{3}\)
\(nI_n=16(n-1) I_{n-2}-2^n \sqrt{3}\)

Part (d)
Determine \(I_2\) using the derived reduction formula
\(2I_2=16I_0-2^2 \sqrt{3}\)
\(2I_2=16I_0-4\sqrt{3}\)
\(\displaystyle I_0=\int_0^2 {x^0\over \sqrt{16-x^2}}\:dx=\int_0^2 {1\over \sqrt{16-x^2}}\:dx=\left[\sin^{-1}\left({x\over4}\right)\right]_0^2={π\over6}\)
\(\displaystyle 2I_2={8π\over3}-4\sqrt{3}\)
\(\displaystyle I_2={4π\over3}-2\sqrt{3}\)