8.06a - Reduction formulae | OCR A-Level Further Mathematics A

A reduction formula turns a difficult integral into simpler members of the same family. You will establish these formulae, keep their boundary terms, and use the right starting value to obtain exact answers without repeatedly finding a new antiderivative.

Reducing an integral to a starting value

Can you find the exact value of $\displaystyle\int_0^{\pi/2}\sin^8x\,dx$ without expanding the eighth power?

The curves $y=\sin^2x$ , $y=\sin^4x$ and $y=\sin^8x$ enclose successively smaller areas on this interval. Although the powers grow, the integrals can be linked by a very short rule. Each link removes two powers, until an ordinary sine integral or the area of a rectangle remains.

[DIAGRAM: asset_name: Sine powers and two reduction chains; asset_slug: h245_8_06_reduction_chains; caption: Even powers reduce to the integral of 1, while odd powers reduce to the integral of sine; description: Left panel shows sine squared, sine to the fourth and sine to the eighth from zero to pi over two, with the eighth-power area shaded. Right panel shows the chain I8, I6, I4, I2, I0 and a separate I7, I5, I3, I1 chain. Arrows point upwards from each simpler integral to the next, labelled with the multiplication factor. The starting values at the bottoms are I0 equals pi over two and I1 equals one.]
Even powers reduce to the integral of 1, while odd powers reduce to the integral of sine

You already know

Integration by parts gives $\int u\,dv=uv-\int v\,du$ , and its formula is in the booklet. For definite integrals, evaluate the product at both limits. The identities $\sin^2x+\cos^2x=1$ and the standard derivatives are assumed A-level Maths knowledge.

A family needs a precise definition: the same limits and integrand pattern, with an integer index controlling the power.

Reduction formula

A reduction formula relates an integral $I_n$ to one or more integrals of the same family with smaller indices. Repeated use ends at known starting values. For example, a link from $I_n$ to $I_{n-2}$ needs separate starting values for the even and odd chains. A relation containing only $I_n$ , with no simpler integral, is not a reduction.

For the sine family, the following rule will be proved later in the lesson. Here the task is to use it accurately.

A sine-power reduction

Define $\displaystyle I_n=\int_0^{\pi/2}\sin^n x\,dx$ for integers $n\ge0$ . Then

I_n=\frac{n-1}{n}I_{n-2}\quad(n\ge2),\qquad I_0=\frac\pi2,\quad I_1=1

The reduction is not printed; a question may give it or ask you to establish it. Learn how to use the index and starting values.

Not in the formula booklet: learn it.

Write the index used at each stage. The rule changes its coefficient every time the index falls.

Worked example 1: an even chain

Question. Given the sine-power reduction above, determine the exact value of $I_6$ .

Plan. Reduce the index by two at each stage, ending at $I_0$ .

\begin{aligned} I_6&=\frac56I_4 && n=6 \\ &=\frac56\times\frac34I_2 && n=4 \\ &=\frac56\times\frac34\times\frac12I_0 && n=2 \\ &=\frac{5\pi}{32} && I_0=\pi/2 \end{aligned}

Why this step? An even index stays even when two is subtracted, so $I_1$ cannot be the starting value for this chain.

Answer. $\displaystyle I_6=\frac{5\pi}{32}$ .

Check. $I_6=(5/6)I_4<I_4$ , as expected because $0\le\sin x\le1$ . Numerically the answer is about $0.491$ , smaller than $I_4=3\pi/16$ .

Change only the parity of the requested power. The recurrence is unchanged, but the chain now ends at $I_1$ .

Worked example 2: an odd chain

Question. Given the same reduction formula, determine the exact value of $I_7$ .

Plan. Follow the odd indices down to $I_1$ , then multiply the rational factors.

\begin{aligned} I_7&=\frac67I_5 && n=7 \\ &=\frac67\times\frac45I_3 && n=5 \\ &=\frac67\times\frac45\times\frac23I_1 && n=3 \\ &=\frac{16}{35} && I_1=1 \end{aligned}

Why this step? $I_1=\int_0^{\pi/2}\sin x\,dx=[-\cos x]_0^{\pi/2}=1$ ; inserting $\pi/2$ here would use the wrong base integral.

Answer. $\displaystyle I_7=\frac{16}{35}$ .

Check. $I_7\approx0.457<I_6\approx0.491$ , agreeing with the ordering of the integrands.

Every odd-power integral in this family is rational, while every even-power integral is a rational multiple of $\pi$ . The same rational reduction factors lead to two different kinds of answer because their starting values differ.

The smallest admissible index matters as much as the starting value.

Watch out

Wrong: applying the rule with $n=1$ to obtain $I_1=0I_{-1}=0$ . The rule was stated only for $n\ge2$ ; $I_{-1}$ would not even converge on these limits. Right: evaluate $I_1$ directly and stop there.

One more even step answers the opening question.

Establishing a one-step reduction

Integration by parts becomes a reduction method when differentiating one factor lowers the index and integrating the other keeps its form.

Consider $\displaystyle J_n=\int_0^1x^ne^x\,dx$ . Differentiating $x^n$ gives $nx^{n-1}$ , and integrating $e^x$ leaves $e^x$ . The remaining integral therefore belongs to the same family. The index $n$ is a constant while you integrate with respect to $x$ .

The integration-by-parts rule is in the booklet; no reduction formula for this family is printed. Establish the relation from the definition, rather than quoting a memorised formula for a different family.

Worked example 3: lowering a polynomial power

Question. For integers $n\ge0$ , let $\displaystyle J_n=\int_0^1x^ne^x\,dx$ . Show that $J_n=e-nJ_{n-1}$ for $n\ge1$ . Hence determine $J_2$ .

Plan. Take $u=x^n$ and $dv=e^x\,dx$ , retain the boundary product, then start the recursion at $J_0$ .

\begin{aligned} u=x^n&\Rightarrow du=nx^{n-1}\,dx && \text{lower the power} \\ dv=e^x\,dx&\Rightarrow v=e^x && \text{integrate the other factor} \\ J_n&=[x^ne^x]_0^1-n\int_0^1x^{n-1}e^x\,dx && \text{parts} \\ &=e-nJ_{n-1} && n\ge1\text{ gives a zero lower term} \end{aligned}

Why this step? The remaining power is exactly $n-1$ , so its integral is $J_{n-1}$ , including the same limits. It is not $J_n$ .

\begin{aligned} J_0&=[e^x]_0^1=e-1 && \text{evaluate directly} \\ J_1&=e-J_0=1 && n=1 \\ J_2&=e-2J_1=e-2 && n=2 \end{aligned}

Answer. $J_n=e-nJ_{n-1}$ for $n\ge1$ , and $J_2=e-2$ .

Check. The integrand for $J_2$ is positive and less than the integrand for $J_1$ on $(0,1)$ . Indeed, $0<e-2\approx0.718<1$ .

Change only the exponential from $e^x$ to $e^{2x}$ . Its antiderivative introduces a factor of $1/2$ in both terms of the recurrence.

Worked example 4: an inner exponential scale

Question. Define $\displaystyle H_n=\int_0^1x^ne^{2x}\,dx$ for integers $n\ge0$ . Establish a reduction formula for $H_n$ when $n\ge1$ , and hence determine $H_2$ .

Plan. Differentiate $x^n$ , integrate $e^{2x}$ , and retain the resulting half throughout.

\begin{aligned} u=x^n,\quad v&=\frac12e^{2x} && \text{chosen parts} \\ H_n&=\left[\frac12x^ne^{2x}\right]_0^1-\frac n2\int_0^1x^{n-1}e^{2x}\,dx && \text{parts} \\ &=\frac{e^2}{2}-\frac n2H_{n-1} && n\ge1 \\ H_0&=\frac{e^2-1}{2} && \text{direct integration} \\ H_1&=\frac{e^2}{2}-\frac12H_0=\frac{e^2+1}{4} && n=1 \\ H_2&=\frac{e^2}{2}-H_1=\frac{e^2-1}{4} && n=2 \end{aligned}

Why this step? Integrating $e^{2x}$ changes the product term and the remaining integral by the same factor. Changing just one would break integration by parts.

Answer. $H_n=e^2/2-(n/2)H_{n-1}$ , for $n\ge1$ , and $H_2=(e^2-1)/4$ .

Check. $H_2\approx1.60>J_2\approx0.718$ , since $e^{2x}>e^x$ for $0<x<1$ .

A boundary product is a value to calculate, not a piece of notation to discard.

Watch out

Wrong: $J_n=-nJ_{n-1}$ , omitting $[x^ne^x]_0^1$ . This would make positive integrals alternate in sign. Right: the product is $e$ for $n\ge1$ , so $J_n=e-nJ_{n-1}$ . At $n=0$ , evaluate the integral directly instead of using this boundary argument.

Show the choice of parts and the evaluation of the product when establishing a given result. Those steps make the recurrence a proof.

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8.06a - Reduction formulae

Reducing an integral to a starting value

Establishing a one-step reduction

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