Reduction Formulae

Big Idea

To integrate general products of functions, we use integration by parts, a.k.a. what I call the IS-ID technique:

$\displaystyle \int u\, \mathrm dv = \underbrace{ \phantom- v \phantom- }_{\mathrm I}\, \underbrace{ \phantom- u \phantom- }_{\mathrm S} - \int \underbrace{ \phantom- v \phantom- }_{\mathrm I}\, \underbrace{ \phantom- \mathrm du \phantom- }_{\mathrm D}.$

The function $\mathrm I$ is taken from the easier-to-integrate function $\mathrm dv$ .

Questions

Question 1. For any non-negative integer $n$ , define

$\displaystyle I_n := \int_0^{\pi} \sin^n(x)\, \mathrm dx.$

Evaluate $I_0$ and $I_1$ . Hence, evaluate $I_n$ in terms of $n$ .

(Click for Solution)

Solution. To evaluate the first two integrals is straightforward:

$\begin{aligned} I_0 &= \int_0^{\pi} \sin^0(x)\, \mathrm dx = \int_0^{\pi}\, \mathrm dx = \pi, \\ I_1 &= \int_0^{\pi} \sin(x)\, \mathrm dx = \left[ - \cos(x) \right]_0^{\pi} \\ &= (-(-1) - (-1)) = 2. \end{aligned}$

For the general case, assume $n \geq 2$ , so that

$\begin{aligned} I_n &= \int_0^{\pi} \sin^n(x)\, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \cdot \sin^2(x) \, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \cdot (1-\cos^2(x)) \, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \, \mathrm dx - \int_0^{\pi} \sin^{n-2}(x) \cos^2(x) \, \mathrm dx \\ &= I_{n-2} - \int_0^{\pi} \sin^{n-2}(x) \cos^2(x) \, \mathrm dx. \end{aligned}$

To evaluate the integral,

$\begin{aligned} &\int_0^{\pi} \sin^{n-2}(x) \cos^2(x) \, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \cos(x) \cdot \cos(x) \, \mathrm dx \\ &= \left[ \underbrace{ \phantom- \frac{\sin^{n-1}(x)}{n-1} \phantom- }_{\mathrm I}\, \underbrace{ \phantom- \cos(x) \phantom- }_{\mathrm S} \right]_0^{\pi} - \int_0^{\pi} \underbrace{ \phantom- \frac{\sin^{n-1}(x)}{n-1} \phantom- }_{\mathrm I}\, \underbrace{ \phantom- (-\sin(x))\, \mathrm dx \phantom- }_{\mathrm D} \\ &= (0 - 0) + \frac 1{n-1} \int_0^{\pi} \sin^n(x)\, \mathrm dx = \frac 1{n-1} \cdot I_{n}. \end{aligned}$

Therefore,

$\displaystyle I_n = I_{n-2} - \frac 1{n-1} \cdot I_{n} \quad \Rightarrow \quad I_n = \frac {n-1}n \cdot I_{n-2}.$

For example,

$I_2 = \frac 12 \cdot I_0,\quad I_3 = \frac 23 \cdot I_1,\quad I_4 = \frac 34 \cdot \frac 12 \cdot I_0, \quad I_5 = \frac 45 \cdot \frac 23 \cdot I_1.$

More generally, if $n = 2k + l$ where $l = 0, 1$ ,

$\displaystyle I_n = I_{2k+l} = \frac{(n-1)(n-3) \cdot \cdots \cdot (n-(2k-1)) }{n(n-2) \cdot \cdots \cdot (n-2k)} \cdot I_l,$

where $I_l = \pi$ if $l = 0$ and $I_1 = 2$ if $l = 1$ .

Question 2. For any positive integer $n$ , define

$\displaystyle I_n := \int_{-1}^1 \frac{ x^n }{ \sqrt{2x+3} }\, \mathrm dx.$

Evaluate $I_0$ . Then evaluate $I_n$ in terms of $I_{n-1}$ . Finally, evaluate $I_1, I_2$ .

(Click for Solution)

Solution. Setting $n = 0$ ,

$\begin{aligned} I_0 &= \int_{-1}^{1} \frac 1{\sqrt{2x+3}}\, \mathrm dx = \int_{-1}^{1} (2x+3)^{-1/2}\, \mathrm dx \\ &= \left[ \frac 12 \cdot \frac{(2x+3)^{1/2}}{1/2} \right]_{-1}^1 = \left[ \sqrt{2x+3} \right]_{-1}^1 = -1 + \sqrt 5. \end{aligned}$

Writing $x^n = x \cdot x^{n-1}$ ,

$\begin{aligned} I_n &= \int_{-1}^1 \frac{ x^n }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \int_{-1}^1 x^n \cdot \frac{ 1 }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \frac 12 \int_{-1}^1 x^{n-1} \cdot \frac{ (2x+3-3) }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \frac 12 \int_{-1}^1 x^{n-1} \sqrt{2x+3}\, \mathrm dx - \frac 32 \int_{-1}^1 x^{n-1} \cdot \frac 1{\sqrt{2x+3}}\, \mathrm dx \\ &= \frac 12 \left( \left[\underbrace{ \phantom- \frac{x^n}{n} \phantom- }_{\mathrm I}\cdot \underbrace{ \phantom- \sqrt{2x+3} \phantom- }_{\mathrm S} \right]_{-1}^1 - \int_{-1}^1 \underbrace{ \phantom- \frac{x^n}{n} \phantom- }_{\mathrm I}\cdot \underbrace{ \phantom- \frac{1}{2\sqrt{2x+3}} \cdot 2 \, \mathrm dx\phantom- }_{\mathrm D}\right) - \frac 32 \cdot I_{n-1} \\ &= \frac 12 \left( \frac 1n \cdot (\sqrt 5 - (-1)^n) - \frac 1n \cdot I_n \right) - \frac 32 \cdot I_{n-1} \\ &=\frac 1{2n} \cdot (\sqrt 5 - (-1)^n) - \frac 1{2n} \cdot I_n - \frac 32 \cdot I_{n-1}. \end{aligned}$

Doing a little bit more algebruh,

$\begin{aligned} I_n &= \frac {1}{2n+1} \cdot (\sqrt 5 - (-1)^n) - \frac {3n}{2n+1} \cdot I_{n-1}.\end{aligned}$

Finally,

$\begin{aligned} I_1 &= \frac 13 \cdot (\sqrt 5 -(-1)) - \frac 33 \cdot I_0 = \frac 43-\frac 23 \sqrt 5 ,\\ I_2 &= \frac {1}{5} \cdot (\sqrt 5 - 1) - \frac {6}{5} \cdot I_{1} = - \frac {9}{5} + \sqrt 5 . \end{aligned}$

—Joel Kindiak, 4 Sept 25, 1727H

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