For Integration Recurrence Formula or reduction formula, it is important to set a relationship between two consecutive terms using mostly the integration by parts.
Worked Example of Integration Reduction Formula
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 \( \displaystyle A_n = \int_{0}^{\frac{\pi}{2}} \cos^{2n}x dx, n \ge 1\).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
A_n &= \int_{0}^{\frac{\pi}{2}} \cos^{2n-1}x \cos x dx \\
&\color{red} u = \cos^{2n-1},\ \color{red}v’=\cos x \\
&\color{red} u’ = (2n-1) \cos^{2n-2}(-\sin x),\ \color{red} v= \sin x \\
&= \Big[\cos^{2n-1} \sin x\Big]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} (2n-1) \cos^{2n-2} x (-\sin x) \sin x dx \\
&= \int_{0}^{\frac{\pi}{2}} (2n-1) \cos^{2n-2} x \sin^2 x dx \\
&= \int_{0}^{\frac{\pi}{2}} (2n-1) \cos^{2n-2} x (1-\cos^2 x) dx \\
&= (2n-1) \int_{0}^{\frac{\pi}{2}} \cos^{2(n-1)} x dx-(2n-1) \int_{0}^{\frac{\pi}{2}} \cos^{2n} x dx \\
A_n &= (2n-1)A_{n-1}-(2n-1)A_n \\
2n A_n &= (2n-1)A_{n-1} \\
\therefore A_n &= \frac{2n-1}{2n} A_{n-1}
\end{aligned} \)
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