# Integration Reduction Formula

For Integration Recurrence Formula or reduction formula, it is important to set a relationship between two consecutive terms using mostly the integration by parts.

### Question 1

For any integer $m \ge 0$ let $\displaystyle I_m = \int_{0}^{1} x^m (x^2-1)^5 dx$.
Prove that for $\displaystyle m\ge 2, I_m = \frac{m-1}{m+11} I_{m-2}.$

\begin{aligned} \require{AMSsymbols} \require{color} \displaystyle \frac{d}{dx}(x^2-1)^6 &= 6(x^2-1)^5 \times 2x \\ \frac{d}{dx} \frac{(x^2-1)^6}{12x} &= (x^2-1)^5 \\ I_m &= \int_{0}^{1} x^m \times \frac{d}{dx}\frac{(x^2-1)^6}{12x} dx \\ &= \int_{0}^{1} x^{m-1} \times \frac{d}{dx}\frac{(x^2-1)^6}{12} dx \\ \color{red} u = x^{m-1},\ &\color{red}v’=\frac{d}{dx}\frac{(x^2-1)^6}{12} \\ \color{red} u’ = (m-1)x^{m-2},\ &\color{red}v=\frac{(x^2-1)^6}{12} \\ I_m &= \Big[x^{m-1} \times \frac{(x^2-1)^6}{12}\Big]_{0}^{1}-\int_{0}^{1} (m-1)x^{m-2} \times \frac{(x^2-1)^6}{12}dx &\color{red} \int uv’ = uv-\int u’v \\ &= 0-\frac{m-1}{12} \int_{0}^{1} x^{m-2}(x^2-1)^6 dx \\ &= -\frac{m-1}{12} \int_{0}^{1} x^{m-2}(x^2-1)^5(x^2-1) dx \\ &= -\frac{m-1}{12} \int_{0}^{1} \Big[x^{m}(x^2-1)^5-x^{m-2}(x^2-1)^5 \Big] dx \\ &= -\frac{m-1}{12} \int_{0}^{1} x^{m}(x^2-1)^5 dx + \frac{m-1}{12} \int_{0}^{1} x^{m-2}(x^2-1)^5 dx \\ I_m &= -\frac{m-1}{12} I_{m} + \frac{m-1}{12} I_{m-2} \\ I_m + \frac{m-1}{12} I_{m} &= \frac{m-1}{12} I_{m-2} \\ \frac{m+11}{12} I_{m} &= \frac{m-1}{12} I_{m-2} \\ \therefore I_m &= \frac{m-1}{m+11} I_{m-2} \end{aligned}

### Question 2

Show that $\displaystyle nA_n = \frac{2n-1}{2}A_{n-1}$ for $\displaystyle A_n = \int_{0}^{\frac{\pi}{2}} \cos^{2n}x dx, n \ge 1$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} A_n &= \int_{0}^{\frac{\pi}{2}} \cos^{2n-1}x \cos x dx \\ &\color{red} u = \cos^{2n-1},\ \color{red}v’=\cos x \\ &\color{red} u’ = (2n-1) \cos^{2n-2}(-\sin x),\ \color{red} v= \sin x \\ &= \Big[\cos^{2n-1} \sin x\Big]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} (2n-1) \cos^{2n-2} x (-\sin x) \sin x dx \\ &= \int_{0}^{\frac{\pi}{2}} (2n-1) \cos^{2n-2} x \sin^2 x dx \\ &= \int_{0}^{\frac{\pi}{2}} (2n-1) \cos^{2n-2} x (1-\cos^2 x) dx \\ &= (2n-1) \int_{0}^{\frac{\pi}{2}} \cos^{2(n-1)} x dx-(2n-1) \int_{0}^{\frac{\pi}{2}} \cos^{2n} x dx \\ A_n &= (2n-1)A_{n-1}-(2n-1)A_n \\ 2n A_n &= (2n-1)A_{n-1} \\ \therefore A_n &= \frac{2n-1}{2n} A_{n-1} \end{aligned}

Discover more enlightening videos by visiting our YouTube channel!

## Mastering Integration by Parts: The Ultimate Guide

Welcome to the ultimate guide on mastering integration by parts. If you’re a student of calculus, you’ve likely encountered integration problems that seem insurmountable. That’s…

## Trigonometry Made Easy: Integration by Parts Demystified

Integration by Parts is made of the product rule of differentiation. The derivative of $uv$ is $u’v + uv’$ and integrates both sides. \( \begin{aligned}…

## High School Math for Life: Making Sense of Earnings

Salary Salary refers to the fixed amount of money that an employer pays an employee at regular intervals, typically on a monthly or biweekly basis,…

## Master Trigonometric Integration by Substitution Now

Substitution of Angle Parts Example 1 Find $\displaystyle \int{(2x+3) \sin (x^2+3x)}dx$. \( \begin{align} \displaystyle\text{Let } u &= x^2+3x \\\dfrac{du}{dx} &= 2x + 3 \\du &=…

## The Best Practices for Using Two-Way Tables in Probability

Welcome to a comprehensive guide on mastering probability through the lens of two-way tables. If you’ve ever found probability challenging, fear not. We’ll break it…