Answered: V8 - 2r – x dx

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter6: Fractions

Section6.1: Simplifying Fractions

Problem 29WE

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Question

complete the square before using an appropriate
trigonometric substitution.

Expert Solution

Step 1

Given: $\int \sqrt{8 - 2 x - x^{2}} d x$

Step 2

Explanation:

Complete the square $8 - 2 x - x^{2} : - {(x + 1)}^{2} + 9$

$\int \sqrt{8 - 2 x - x^{2}} d x = \int \sqrt{- {(x + 1)}^{2} + 9} d x P u t x + 1 = u \Rightarrow d x = d u = \int \sqrt{- u^{2} + 9} d u P u t u = 3 \sin (v) \Rightarrow d u = 3 \cos (v) d v = \int \sqrt{- 9 \sin^{2} (v) + 9} 3 \cos (v) d v = \int \sqrt{9 (1 - \sin^{2} (v))} 3 \cos (v) d v = \int \sqrt{9 (\cos^{2} (v))} 3 \cos (v) d v = \int 3 \cos (v) \cdot 3 \cos (v) d v = \int 9 \cos^{2} (v) d v = 9 \cdot \int \cos^{2} (v) d v = 9 \cdot \int \frac{1 + \cos (2 v)}{2} d v = 9 \cdot \frac{1}{2} \cdot \int (1 + \cos (2 v)) d v = 9 \cdot \frac{1}{2} (\int 1 d v + \int \cos (2 v) d v) = \frac{9}{2} (v + \frac{1}{2} \sin (2 v)) + C = \frac{9}{2} (\sin^{- 1} (\frac{u}{3}) + \frac{1}{2} \sin (2 \sin^{- 1} (\frac{u}{3}))) + C = \frac{9}{2} (\sin^{- 1} (\frac{x + 1}{3}) + \frac{1}{2} \sin (2 \sin^{- 1} (\frac{x + 1}{3}))) + C$

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