Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 491: 56

Answer

See the explanation.

Work Step by Step

We need the reduction formula. $\int_0^{\pi/2}\sin^{2n}x dx=\frac{2n-1}{2n}\int_0^{\pi/2}\sin^{2n-2} x dx$ (By the reduction formula) $=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\int_0^{\pi/2}\sin^{2n-4} x dx$ (By the reduction formula) $=\cdots$ (Continue the process) $=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \ldots \frac{3}{4}\cdot \frac{1}{2}\int_0^{\pi/2}\sin^0 x dx$ $=\frac{(2n-1)(2n-3)\ldots 3\cdot 1}{2n\cdot (2n-2)\ldots 4\cdot 2}\int_0^{\pi/2} 1dx$ $=\frac{(2n-1)(2n-3)\ldots 3\cdot 1}{2n\cdot (2n-2)\ldots 4\cdot 2}\cdot [x]_0^{\pi/2}$ $=\frac{(2n-1)(2n-3)\ldots 3\cdot 1}{2n\cdot (2n-2)\ldots 4\cdot 2}\cdot (\frac{\pi}{2}-0)$ $=\frac{1\cdot 3\ldots (2n-3)(2n-1)}{2\cdot 4\ldots (2n-2)(2n)}\cdot \frac{\pi}{2}$
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