Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 525: 37

Answer

$$\frac{22}{105} \sqrt{2}-\frac{8}{105}$$

Work Step by Step

Given $$ $$ Since \begin{align*} \int_{\pi / 4}^{\pi / 2} \cot ^{5} \phi \csc ^{3} \phi d \phi&=\int_{\pi / 4}^{\pi / 2} \cot ^{4} \phi \csc ^{2} \phi \csc \phi \cot \phi d \phi\\ &=\int_{\pi / 4}^{\pi / 2}\left(\csc ^{2} \phi-1\right)^{2} \csc ^{2} \phi \csc \phi \cot \phi d \phi \end{align*} Let $u=\csc \phi \ \ \to \ \ du=-\csc \phi \cot \phi d \phi$ and at $ \phi =\pi / 4\to u=\sqrt{2}$ at $ \phi =\pi / 2\to u=1$ , then \begin{align*} \int_{\pi / 4}^{\pi / 2} \cot ^{5} \phi \csc ^{3} \phi d \phi &=\int_{\pi / 4}^{\pi / 2}\left(\csc ^{2} \phi-1\right)^{2} \csc ^{2} \phi \csc \phi \cot \phi d \phi\\ &=\int_{\sqrt{2}}^{1}\left(u^{2}-1\right)^{2} u^{2}(-d u)\\ &=\int_{1}^{\sqrt{2}}\left(u^{6}-2 u^{4}+u^{2}\right) d u\\ &=\left[\frac{1}{7} u^{7}-\frac{2}{5} u^{5}+\frac{1}{3} u^{3}\right]_{1}^{\sqrt{2}}\\ &=\left(\frac{8}{7} \sqrt{2}-\frac{8}{5} \sqrt{2}+\frac{2}{3} \sqrt{2}\right)-\left(\frac{1}{7}-\frac{2}{5}+\frac{1}{3}\right)\\ &=\frac{22}{105} \sqrt{2}-\frac{8}{105} \end{align*}
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