## - 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
$$2I_2=16I_0-2^2 \sqrt{3}$$
$$2I_2=16I_0-4\sqrt{3}$$
$$\displaystyle I_0=\int_0^2 {x^0\over \sqrt{16-x^2}}\:dx=\int_0^2 {1\over \sqrt{16-x^2}}\:dx=\left[\sin^{-1}\left({x\over4}\right)\right]_0^2={π\over6}$$
$$\displaystyle 2I_2={8π\over3}-4\sqrt{3}$$
$$\displaystyle I_2={4π\over3}-2\sqrt{3}$$