2.4 - Reduction formulae

2.4 - Reduction formulae

A high power inside an integral need not mean a long expansion. You will derive relations between neighbouring integrals, choose the starting values they need, and use those relations to find exact definite integrals.

Deriving a recurrence

How could you integrate sin⁡6x\sin^6x without expanding it into multiple angles? First make an easier integral appear inside the calculation.

For example, write sin⁡4x=sin⁡3xsin⁡x\sin^4x=\sin^3x\sin x. Integration by parts differentiates the power and integrates the final sine. The identity cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x then produces both the original integral and one with power two less.

Reduction formula

A reduction formula is a recurrence relating an integral with an integer index to integrals with smaller indices. Together with suitable starting values, it evaluates a whole family of integrals.

The limits belong to the definition of the family. They also determine the boundary term, so retain them throughout the derivation.

Worked example 1: reducing a sine power

Question. For integers n≥0n\ge0, let In=∫0π/2sin⁡nx dxI_n=\int_0^{\pi/2}\sin^n x\,\mathrm{d}x. Show that nIn=(n−1)In−2nI_n=(n-1)I_{n-2} for n≥2n\ge2.

Plan. Use parts with u=sin⁡n−1xu=\sin^{n-1}x and dv=sin⁡x dx\mathrm{d}v=\sin x\,\mathrm{d}x, then replace cos⁡2x\cos^2x.

Here v=−cos⁡xv=-\cos x and u′=(n−1)sin⁡n−2xcos⁡xu'=(n-1)\sin^{n-2}x\cos x. Thus

In=[−sin⁡n−1xcos⁡x]0π/2+(n−1)∫0π/2sin⁡n−2xcos⁡2x dx=0+(n−1)(In−2−In)\begin{aligned} I_n&=[-\sin^{n-1}x\cos x]_0^{\pi/2} +(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,\mathrm{d}x \\ &=0+(n-1)(I_{n-2}-I_n) \end{aligned}

The boundary term vanishes: cos⁡(π/2)=0\cos(\pi/2)=0, and sin⁡n−10=0\sin^{n-1}0=0 because n≥2n\ge2. Collecting the two copies of InI_n gives the result.

Answer. nIn=(n−1)In−2nI_n=(n-1)I_{n-2}, for n≥2n\ge2.

Check. At n=2n=2, the identity sin⁡2x=(1−cos⁡2x)/2\sin^2x=(1-\cos2x)/2 gives I2=π/4I_2=\pi/4. Hence 2I2=π/2=I02I_2=\pi/2=I_0, independently of parts.

Change only the upper limit. The method survives, but its endpoint contribution changes.

Worked example 2: keeping a nonzero boundary term

Question. Let Jn=∫0π/4sin⁡nx dxJ_n=\int_0^{\pi/4}\sin^n x\,\mathrm{d}x. Derive a reduction formula valid for n≥2n\ge2.

Plan. Repeat the same parts calculation with the new upper endpoint.

Jn=[−sin⁡n−1xcos⁡x]0π/4+(n−1)(Jn−2−Jn)=−2−n/2+(n−1)(Jn−2−Jn)\begin{aligned} J_n&=[-\sin^{n-1}x\cos x]_0^{\pi/4}+(n-1)(J_{n-2}-J_n) \\ &=-2^{-n/2}+(n-1)(J_{n-2}-J_n) \end{aligned}

The product at π/4\pi/4 contains nn factors of 1/21/\sqrt2. Moving the final term to the left gives

nJn=(n−1)Jn−2−2−n/2.nJ_n=(n-1)J_{n-2}-2^{-n/2}.

Answer. Jn=n−1nJn−2−2−n/2nJ_n=\frac{n-1}{n}J_{n-2}-\frac{2^{-n/2}}n, for n≥2n\ge2.

Check. Directly integrating (1−cos⁡2x)/2(1-\cos2x)/2 gives J2=π/8−1/4J_2=\pi/8-1/4. Therefore 2J2=π/4−1/2=J0−2−12J_2=\pi/4-1/2=J_0-2^{-1}, as required.

A supplied derivative hint can organise the same argument. For example,

ddx(sin⁡n−1xcos⁡x)=(n−1)sin⁡n−2x−nsin⁡nx.\frac{\mathrm{d}}{\mathrm{d}x}(\sin^{n-1}x\cos x) =(n-1)\sin^{n-2}x-n\sin^n x.

Integrate both sides between the stated endpoints to recover the recurrence. This is the product rule read backwards. Integration by parts is supplied in MF19; these family-specific reduction formulae are not.

Watch out

Dropping [uv]ab[uv]_a^b before substituting the endpoints loses a term in many recurrences. A boundary term is zero only when its evaluated values show that it is.

Use a different endpoint in the same derivation.

Initial integrals

A recurrence is a route down to an integral you can already evaluate. Before using it, follow the index backwards to identify the necessary starting value.

For a step of two, even and odd indices form separate chains. The even chain ends at 00; the odd chain ends at 11. Neither chain needs a negative index, and a recurrence proved only for n≥2n\ge2 cannot be substituted into at n=1n=1.

Worked example 3: two starting values

Question. For In=∫0π/2sin⁡nx dxI_n=\int_0^{\pi/2}\sin^n x\,\mathrm{d}x, find the initial values needed to evaluate I8I_8 and I7I_7 using the step-two recurrence.

Plan. Trace each chain, then integrate its final member directly.

The chains are 8,6,4,2,08,6,4,2,0 and 7,5,3,17,5,3,1. Hence

I0=∫0π/21 dx=π2I1=[−cos⁡x]0π/2=1\begin{aligned} I_0&=\int_0^{\pi/2}1\,\mathrm{d}x=\frac\pi2 \\ I_1&=[-\cos x]_0^{\pi/2}=1 \end{aligned}

Answer. I8I_8 needs I0=π/2I_0=\pi/2; I7I_7 needs I1=1I_1=1.

Check. The first integral is the area of a rectangle of height 11 and width π/2\pi/2. For the second, substituting u=cos⁡xu=\cos x gives −∫10du=1-\int_1^0\mathrm{d}u=1.

Now keep the integrand but shorten the interval. Both starting values must be recalculated, just as the boundary term was.

Continue this lesson.

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