# Integration by Parts

Integration by Parts is made of product rule of differentiation. The derivative of $uv$ is $u’v + uv’$ and integrate both sides.
\begin{aligned} \require{color} (uv)’ &= u’v + uv’ \\ uv &= \int u’v + \int uv’ \\ \int u’v &= uv – \int uv’ \text{ or } \int uv’ = uv – \int u’v \\ \end{aligned} \\

### Worked Examples of Integration by Parts

$\displaystyle \text{Let }A_n = \int_{0}^{\frac{\pi}{2}} \cos^{2n} x dx \text{ and } B_n = \int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n} x dx, \text{where } n \text{ is an integer, } n \ge 0 \text{ and } A_n > 0, \ B_n > 0. \\ \displaystyle \text{The recurrence formula is } n A_n = \frac{2n-1}{2} A_{n-1}. \\$

$\displaystyle (a) \ \ \ \text{Show that } A_n = 2n \int_{0}^{\frac{\pi}{2}} x \sin x \cos^{2n-1} x dx \text{ for } n\ge 1.$

$\displaystyle (b) \ \ \ \text{Show that } \frac{A_n}{n^2} = \frac{2n-1}{n} B_{n-1} – 2 B_n \text{ for } n \ge 1. \\$

$\displaystyle (c) \ \ \ \text{Show that } \frac{1}{n^2} = \frac{2 B_{n-1}}{A_{n-1}} – \frac{2 B_n}{A_n} \text{ for } n \ge 1. \\$

$\displaystyle (d) \ \ \ \text{Show that } \sum_{k=1}^{n} \frac{1}{k^2} = \frac{\pi^2}{6} – \frac{2 B_n}{A_n} \\$

$\displaystyle (e) \ \ \ \text{Given } \sin x \ge \frac{2x}{\pi} \text{ for } 0 \le x \le \frac{\pi}{2} \text{ show that } B_n \le \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1 – \frac{4x^2}{\pi^2} \Bigg)^n dx. \\$

$\displaystyle (f) \ \ \ \text{Show that } \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1 – \frac{4x^2}{\pi^2}\Bigg)^n dx = \frac{\pi^2} {8(n+1)} \int_{0}^{\frac{\pi}{2}} \Bigg(1 – \frac{4x^2}{\pi^2}\Bigg)^{n+1} dx.$

$\displaystyle (g) \ \ \ \text{Show that } B_n \le \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt \le \frac{\pi^3}{16(n+1)} A_n \text{ using the substitution } x = \frac{\pi}{2} \sin t. \\$

$\displaystyle (h) \ \ \ \text{Show that } \frac{\pi^2}{6} – \frac{\pi^3}{8(n+1)} \le \sum_{k=1}^{n} \frac{1}{k^2} \lt \frac{\pi^2}{6}. \\$

$\displaystyle (i) \ \ \ \text{Find } \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k^2}.$